Math Forum - Problem of the Week
Submissions for The Christmas Gifts
Student Short Answer Long Answer
Student 1
The old gentleman
would receive 39 hugs,
since there were 20
boys and 21 girls, 2 of
whom were newborns.
To solve this problem, I set up an equation where X was the number of
boys and (X+1) was the number of girls.
Since the 2 newborn girls made the girls have more than the boys, I
knew that there were now either one or two more girls than boys.
There would be one more now if the girls had previously had one fewer
and now the two newborns pushed them one ahead, or there could be two
more girls if there was an even number before the birth of the
newborns. The solution where there were two more girls didn't work
out to have a whole number of people, so I stayed with the idea that
there was now one more girl than the number of boys.
If each grandchild got the number of dollars that there were of that
gender, then X grandsons would total X times X, or X^2 dollars. If
there were (X+1) granddaughters, they would total (X+1)^2 dollars.
Together:
X^2 + (X+1)^2 = 841 dollars
X^2 + X^2 + 2X + 1 = 841
2X^2 + 2X + 1 =841
2X^2 + 2X - 840 = 0 (subtract 841 from each side)
X^2 + X - 420 = 0
Now to factor that equation: What numbers multiply to give -420 and
add to give 1? (-20 and 21):
(X-20) (X+21) = X^2+X-420 = 0
So, to make the numbers represented by (X-20) and (X+21) multiply to
0, either (X-20) or (X+21) must equal 0. This means that X=20 or X=21. There is no way to have -21 grandsons (which is what X
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 1 of 32
Student 2
Grandfather
Pennywise received 39
hugs from his 41
grandchildren.
represents), so there must be 20 grandsons, 21 granddaughters, two of
whom were newborn. This means 41-2 hugs for the grandfather.
Let's assume s= the number of grandsons
and
d= the number of granddaughters
Because Grandfather Pennywise give each grandson as much money in
dollars as there are grandsons, the total money recieved by grandsons
can be expressed as s^2 dollars.
For the same reason, the total money recieved by granddaughters can be
expressed as d^2 dollars.
The sum of the the girls' and boys' money could be expressed:
s^2 + d^2 = 841
With the birth of the twin girls, there is one more girls than there
are boys, so:
d = s+1
When I substitute d with (s+1), I can solve for s.
s^2 + (s+1)^2 = 841
s^2 + (s+1) (s+1) = 841
s^2 + s^2 + 2s+ 1 = 841
2s^2 + 2s = 840
s^2 + s = 420
-1 ± square root[1 + 4(420)] ÷ 2
-1 ± square root[ 1681 ] ÷ 2
[-1 ± 41] ÷ 2
s= 20 or s= -21
s = 20
s+1 = 21
s* a negative number, so:
There are 20 grandsons and 21 granddaughters, or 41 chidren and all.
Student 3
The grandfather
received 39 hugs.
© 1994-2016 Drexel University
http://mathforum.org/pows/
HOWEVER, the two newborns could not rush up to give Grandad a hug, so
he received 41-2, or 39 hugs.
First of all I deturmined that girls could be represented by b + 1
since this is the first year there were more girls than boys. Then we
know that the boys recieve the as many dollers as there are boys or
Page 2 of 32
Student 4
There were 20 boys
and 21 girls so, the
grandfather got a total
of 39 hugs.
b^2 and likewise with the girls g^2.
b^2 + g^2 = 841
Then I substituted g = b + 1 in for the girls and solved for b.
b^2 + (b + 1)^2 = 841
b^2 + b^2 + 2b +1 = 841
2b^2 + 2b - 840 = 0
b^2 + b - 420 = 0
(b - 20) (b + 21) = 0
b - 20 = 0 or b + 21 = 0
b = 20 or b = -21
There can't be a negitave # of people so there must be 20 boys, which means there are 21 (g = b + 1) girls with a total of
41 grandchildren. 41 minus the two who can't give hugs equals 39 hugs.
x=boys multiplyed by x times an dollar
y=girls multiplyed by y times a dollar
x squared + y squared=841
y=x+1
By using guess and check and making sure there was more girls than
boys. I placed numbers in and came up with 20 and 21. The 20 was the
amount of boys and squared it came to 400.Then the girls is 21 and
when u square it u get 441.When u add them together u get the total
amount he handed out in money and then u have a total of also 41
grandchilderen but, he did not get two hugs from the youngest
twins.So, the final answer is that he recieved 39 hugs.
Student 5
He received 39 hugs.
y=x+1
y=20+1
y+21girls
x=20 boys
41 grandchildern -2(twins)=39 hugs
The money he gave each child is how many of that sex that exists.
Therefore the amount of money given away equals the number of
grandson and grandaughter squared. Let g = girl and b = boy
g^2 + b^2 = 841
Girls number one more than boys.
g = b+1
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 3 of 32
I then used the second equation to substitute g in the first equation.
(b+1)^2 + b^2 = 841
Then I solved the equation.
b^2 + 2b + 1 + b^2 = 841
2b^2 + 2b - 840 = 0
2(b^2 + b -420) = 0
2(b+21)(b+20) = 0
b = -21 or b = 20
Since a negative number of boys is impossible, there are 20 grandsons.
Then I used the second equation to solve for the number or girls.
g = b+1
g = 20 + 1
g = 21
There are 21 girls.
The number of boys plus the number of girls is all the grandchildren.
20 + 21 =41
He has 41 grandchildren. Since the twins were not able to hug him,
the total number of hugs he received is two less than the number of
grandchildren.
41 - 2 = 39
He received thirty-nine hugs in all.
Student 6
This year granddad
received 41 hugs this
year.
© 1994-2016 Drexel University
http://mathforum.org/pows/
In the begaining of the problem, I made a chart for me to
understand the problem of the week better:
~~~~~~~~~~~~~~~~~~~~~~~~~~
There are X grandsons
Each grandson is given X ammount of dollars
Page 4 of 32
Student 7
The kind old
gentleman recieved 41
hugs. There are 21
girls, and 20 boys
© 1994-2016 Drexel University
http://mathforum.org/pows/
---------------------------------------------------------------------There are Y granddaughters
Each granddaughter is given Y ammount of dollars
~~~~~~~~~~~~~~~~~~~~~~~~~~
Since the total ammount of money that granddad spent this year
was $841 dollars, I setted the equation to equal $841.
X^2 + Y^2 = 841
X^2 stands for the ammount of money given to ALL the grandsons.
Y^2 stands for the ammount of money given to ALL the granddaughters.
In the third paragraph, it mentioned someabout how the
granddaughters were especially happy this year, due to the birth of
twin girls. It had also mentioned something about the granddaughter
population finally beating the grandson population for the first time
in quite a spell. This information then told me that the
granddaughters had one more than the grandson population. So, some
how they are evenly tied.
In order to solve this, took the half of 841, and square rooted
it. (20.5) That information gave me the ammount of granddaughters or
grandsons. But since the answer was 20.5, there couldn't be half of a
child. Next, I increased the ammount of y to 21. 21 squared equaled
441. $841 dollars minus 441 gave me the value of X^2. After sloving
for X, it gave me the value of 20.
Since the granddaughters beat the grandson population by 1, the
ammount of granddaughters is 21 and the ammount of grandsons is 20.
Therefore, granddad received 41 hugs this Christmas.
Define our variables:
x=girls
y=boys
We know that there are more girls than boys, and that this only
happened recently due to the fact that there were 2 twin girls born a
month before. We also know that the boys would always get more money
than the girls. This means that before there much have been one less
girl than boy, so now that there are 2 more girls, it must be that
there is one more girl than there are boys. So, the new variables much
be:
x=girls
x-1=boys
We also know that the amount of girls, times the amount of girls, is
equal to the amount of money the girls got; and that the amount of
boys, times the amount of boys is equal to the amount of money the
Page 5 of 32
Student 8
Grandpa received 39
hugs.
boys got. We know the total amount of money all the children got, so
our equation is:
(x*x)+(x-1)^2=841
To further understand this equation, I broke it down:
x^2+(x-1)(x-1)=841
x^2+x^2-x-x+1=841
2x^2-2x+1=841
2x-2x=841-1
2x^2-2x/2=840/2
x^2-x=420
As I studied this more and more, I realized I needed to use the
quadratic formula. In order to do this, I transformed the equation
into the below equation:
x^2-x-420=0
The quadratic formula is:
x1=(-b+(b^2-4ac)^1/2)/(2a)
I then plugged in my equation:
x1=(-(-1)+(1^2-4(1)(-420))^1/2)/2(1)
x1=(1+(1+1680)^1/2)/2
x1=(1+(1681)^1/2/2
x1=(1+41)/2
x1=21
The amount of girls is equal to 21. Therefore,the amount of boys is
equal to:
x-1=y
21-1=y
20=y
There are 20 boys. Therefore:
21+20=41
There are 41 g
This year the number of girls is more than boys, and there are two
possible ways, one is girls number is one more than boys, the other
is two girls more than boys. Therefore, I assumed "x" is the number
of boys, and "x+1" or "x+2" are the numbers of girls. There are two
equations as follows:
x^2+(x+1)^2=841
x^2+x^2+1=841
2(x^2)=840
x^2=420
© 1994-2016 Drexel University
http://mathforum.org/pows/
x^2+(x+2)^2=841
x^2+x^2+4=841
2(x^2)=837
x^2=418.5
Page 6 of 32
x=20.49
x=20.46
Since the number of boys would not be a decimal number, so I drop the
decimal, and round up the number to 20. If the number of boys is 20,
than the number of girls would be 20+1, or 20+2.
So we replace the number to the equations above, the check the
correct answer.
20^2+(20+1)^2=841
400+441=841
841=841
Student 9
Grandfather
Moneybags recieve 27
hugs from his
grandchildren.
20^2+(20+2)^2=841
400+484>841
884>841
From the above calculations, I find out the first equation is
correct, and the total number of boys is 20, and the total number of
girls is 21. The total number of grandchildren would be 20+21=41,
since two of them are newborns, and they did not hug their grandpa,
therefore, grandpa received only 39 hugs (41-2) from his
grandchildren.
Let x=the number of female grandchildren, and let y=the number of male
grandchildren.
We know that x girls recieved x dollars because each one recieved as
many dollars as there were girls. Therefore, x(x) dollars were given
out to girls alone, or x^2.
In the same way, there were y boys and each boy recieved y dollars, so
y(y) dollars were given to boys alone, or y^2.
Since $841 is the total amount of money given out, x^2+y^2 must equal
$841. Therefore:
x^2+y^2=841
x+y=29
We find the square root of each side.
We know that x+y grandchildren were present, but 2 could not give
their grandfather a hug. So x+y-2=hugs given.
Since we know that x+y=29, we can substitute 29 in for x and y and
simplify:
x+y-2=hugs
29-2=hugs
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 7 of 32
27=hugs
Student 10
The kind old
gentleman received 39
hugs.
27 hugs were give to Grandfather Moneybags.
The critical information needed to find the number of hugs the kind
old gentleman received is to find out how many granddaughters and
grandsons he has.
Let x=number of granddaughters
y=number of grandsons
First, we can write the equation x^2+y^2=841 from the given
information in the problem. It states the that kind old gentleman
gives to each of his grandsons as many dollars as there were grandsons
and to each of his granddaughters as many dollars as there were
granddaughters. This means that x number of granddaughters will
receive x number of dollars, therefore we have x^2 as the total amount
of money he gave to his granddaughters. The same for the boys results
in y^2. And since we know that he gave $841 this year, the equation
x^2+y^2=841 is valid.
Second, we can write the equation x-y=1. This equation comes from the
information given in the problem and from reasoning. Since the
problem states that this year the girls will receive more per person
than the boys for the first time in quite a spell due to the birth of
twin (2) girls, we know that x has to be greater than y by one or two.
From trial and error reasoning, we know that the difference between x
and y must be one to satisfy the equation x^2+y^2=841. If the
difference between x and y is 2, there is no way for x^2+y^2=841 to be
satisfied.
Therefore, the two equations needed to solve the critical information
for this problem are x^2+y^2=841 and x-y=1.
Solving for x and y
1). From x-y=1 we can rearrange to get x=y+1
2). From x^2+y^2=841 we can rearrange to get y^2=841-x^2
3). Substitute x from 1) into x from 2) we get y^2=841-(y+1)^2
Then expand y^2=841-(y+1)^2 to get y^2=841-(y^2+2y+1)
Moving everything to the left side of the equation, we get
y^2+y^2+2y+1-841=0
Simplifying 2y^2+2y-840=0
Now solve for y using the quadratic formula
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 8 of 32
(-b+-(b^2-4ac)^(1/2))/2a
From 2y^2+2y-840=0 a=2, b=2, c=-840
Substituting the values into the quadratic formula we get the
following y=20, y=-21
From the y values, we know that 20 is the only correct answer
because the number of grandsons cannot be negative.
Therefore, y=number of grandsons=20
4). Because we know x=y+1 and we know y=20, we can find x very easily.
x=y+1, substitute in y, we get x=20+1=21
Therefore, x=number of granddaughters=21
Therefore, the critical information has been solved, the kind old
gentleman has 21 granddaughters(x) and 20 grandsons(y). Thus he has a
total of 41 grandchildren.
To double check our work, we can substitute x and y into x^2+y^2=841.
(21)^2+(20)^2=841
441+400=841
841=841
Student 11
He received about 62
hugs.
To find the final answer (the number of hugs), we substract 2(number
of newborns) from the total number of grandchildren(41) because the
problem states that all the grandchildren gave the old kind gentleman
a hug except the newborns. Therefore, he got 41-2=39 hugs.
Let's let x=amount of granddaughters.
Let's let y=amount of grandsons.
Therefore, x+y-2=amount of hugs.
If x is greater than y and x-2 is less than y, x-1=y.
xx+yy=841
xx+(x-1)(x-1)=841
xx+xx-2x+1=841
2xx-2x=840
xx-x-420=0
x=-b-SQRT(4ac)/2a
x=1+SQRT(1680)/2
x=1+4(SQRT105)/2
y=x+1
y=1+4(SQRT105)/2+1
y=1.5+4(SQRT105)
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 9 of 32
Student 12
Grandfather received
39 hugs. He has 41
grandchildren, 21 girls
and 20 boys. (The
twins didn't give him a
hug)
Student 13
The hugs the kind
gentlemen received
are b+(g-2)
Student 14
He recieved 39 hugs.
© 1994-2016 Drexel University
http://mathforum.org/pows/
HUGS=x+y-2
HUGS=[1+4(SQRT105)/2]+1.5+4(SQRT105)-2
HUGS=12(SQRT105)/2
HUGS=ABOUT 62
First, I knew that the equation involved squared numbers. Grandfather
gives each child as much money as there are grandsons/daughters. For
example, if there were 10 grandaughters, he would give them each $10
or 10x10=100 or 10squared.
Second, I knew that last year, the girls were each received $1 less
than the boys. If there were 10 girls last year and there were 12
boys, then this year they would be even. Whereas, if there were 11
girls last year and 12 boys, this year there would be 13 girls and 12
boys, it is the only way that the girls could get more money than the
boys for the first time.
Then I set up the equation, letting x=the number of boys this year.
so x^2+(x+1)^2=841(the total amount of money). If you FOIL the (x+1)
^2 part you get x^2+ x^2+2x+1=841. Simplify=2x^2+2x+1=841. Set the
equation =to 0 so 2x^2+2x-840=0. You can factor a 2 out so 2(x^2+x420)=0. x=20 and x+1=21.
b = the amount of boys
g = the amount of girls
The twins had to be excluded so that was why I put g-2. 2 means the
twins.
In this problem entitled "The Christmas Gift" our objective was to find out how many granddaughters and grandsons
Grandfather pennywise had in order to find out how many hugs he recieved. I used the given information to help me
come up with some equations.......
I know that the total number of grandsons (x=grandsons) was less than the total number of granddaughters
(y=granddaughters) this year due to the birth of twin girls. I figured that each grandson recieves x dollars; therefore,
the total amount given to all the grandsons was x^2. The amount given to each granddaughter was y dollars; therefore,
the total amount given to all the granddaughters was y^2. Last year there were more grandsons than granddaughters
and now that 2 more granddaughters were born (we are given that this year the granddaughters outnumbered the
grandsons) the number of granddaughters must have exceded the number of grandsons by one.
I was able to come up with 2 equations by using the above given information........
x+1=y
x^2+y^2=841
Next I substituted x+1 in for y. I was able to come up with this equation...
2x^2+2x-840=0
Which simplifies to..
x^2+x-420=0
Page 10 of 32
x=20
Then i substituted 20 in for x in this equation...
x+1=y
20+1=y
21=y
To find out how many hugs he got I added 20+21=41
Since the two new twins cannot yet give their loving grandfather hugs quite yet he only recieved 39 hugs.
Student 15
The old fellow will
receive 39 hugs from
his grandchildren this
christmas.
To solve for this problem I first came up with the following
simultaneous equation:
x^2 + y^2 = 841
y=x+1
Then by elimination by substitution I came up with:
x^2 + x - 420 = 0
Student 16
Grandfather
Pennywise received 39
hugs, even though
there were 41 children
(because of the
newborns). Bonus: No
Bonus
© 1994-2016 Drexel University
http://mathforum.org/pows/
Whose root is 20. Then solving for "y" I came up with 21. Since the
twins were still infants I subtracted 2 from the sum of x and y to
come up with the number of hugs which is equal to 39.
First, I realized that "because of the birth of twin girls, the girls
would have more than the boys" meant the girls had 1 or 2 more than
the boys. 2 didn't work in solving, but 1 more than the boys did.
Now, I set the number of boys equal to 'x'. Since they each received
as many dollars as there were boys, the total given to the boys is (x)
(x) or x^2. Now, there were (x+1) girls, because the girls had one
more, because of the newborn twins. They each received as many
dollars as there were girls, or (x+1)^2. Now I have the equation,
since $841 were given out:
(x^2) + ((x+1)^2) = 841
Simplified:
x^2 + x^2 + 2x + 1 = 841
Reduce:
2(x^2) + 2x = 840
Divide both sides by 2:
x^2 + 2x = 420
Subtract 420:
x^2 + 2x - 420 = 0
Page 11 of 32
Student 17
Grandfather
Pennywise got a total
of 39 hugs and has a
total of 41
grandchildren.
Factor:
(x-20)(x+21) = 0
Therefore 'x' is 20 or -21. You can't have a negative number of
boys, so 'x' (the number of boys) must equal 20. Thusly, (x+1) (the
number of girls) must equal 21. Therefore, there are a total of 41
children, but the problem says the two newborns did not hug Grandad,
so he received only 39 hugs.
First I stated the variables:
H = hugs
G = girls
B = boys
THen I made my equations:
B2 + G2 = 841
B+G-2=H
G=B+1
(this is true because if the girls had 2 more girls than
boys then Grandfather Pennywise would have given out an even number
of money)
Then, knowing that there was 1 more girl than boys, I started to
solve my equations:
B2 + G2 = 841
B2 + (B+ 1)2 = 841
B2 + B2 + B + B + 1 = 841
B2 + B2 + 2B + 1 = 841
2B2 + 2B + 1 = 841
2B2 + 2B = 840
B2 + B = 420
Now I took the square root of 420 and got close to 20 1/2 so then I
tried 20 boys and 21 girls and got right! Then I worked out my last
equation:
B+G-2=H
20 + 21 - 2 = H
41 - 2 = H
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 12 of 32
39 = H My answer!
Student 18
There were 29 children
in all.
I am not completely familiar with the methods of algebra, but I think
that I have got the right.
To begin with, I denoted the number of dollars each child received as
x and the number of children who received money as x also, because it
was stated in the problem that they were equal.
Then, I denoted the total amount of money given out, as the number of
people who received money multiplied by the amount each received,
since each individual received the same amount as each other
individual: xx or x^2.
It was also mentioned that Grandpa Moneybags gave out a total of $841.
Thus we have:
x^2 = 841
x is the square root ot x^2, and because we can do the same thing to
each side of the equation (since it is an equation), we have:
x = square root(841)
When the aritmetic is worked out, we end up with the solved equation:
Student 19
The grandsons gave
their grandfather 20
hugs and his
grandaughters gave
him 19 hugs, giving a
total number of 39
hugs.
x = 29
Square of x is represented by: x(2)
Let x=the number of grandsons.
Each grandson will get x dollars.
The total the grandsons will get =x(2)dollars
x+1=the number of grandaughters.
Each grandaughter will get (x+1) dollars
The total the grandaughters will get:(x+1)(x+1)dollars.
Therefore, the equation giving the total sum distributed becomes:
x(2)+(x+1)(x+1)=841
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 13 of 32
x(2)+x(2)+1+2x=841
2x(2)+2x=840
x(2)+x=420
x(2)+x-420=0
Let us apply the quadratic formula to solve for x:
-b+-square root of(b(2)-4ac)/2a
-1+-square root of (1+1680)/2
Solving for x=(-1+41)/2=20
Student 20
Grandfather Penywise
receved 39 hugs.
Student 21
Grandfather recieved
39 hugs from his
grandchildren (not
including the
newborns).
© 1994-2016 Drexel University
http://mathforum.org/pows/
The number of grandsons is x=20 is also equal to 20 hugs.
The number of the grandaughters is x+1=21. The number of hugs by the
grandaughters is equal to 21-2=19. The total number of hugs is equal
20+19=39.
I first made the equation d*d+S*S=841
where d= amt of grandaugters and s= amt of grandsons
because we know that this was the first time there was more
grandaugters then grandsons, and that was because 2 twin grandaugters
were born, we know that there is one more grandaughter then grandson.
So that d=s+1
I then substituted d for s+1:
(s+1)*(s+1)+s*s=841
I then figured out that s+1 * s+1 = s*s+2s+1
so I substituted s*s+2s+1 for (s+1)*(s+1):
s*s+2s+1+s*s=841
2s*s+2s+1=841
2s*s+2s=840
s*s+s=420
Becasue we know there is one more grandaughter then grandson, then
s(s+1)=420
So that therefore, there must be 21 grandaughters, and 20
grandsons.so that all together Grandfather Penywise has 41
grandchildren, but because it says that the two twins did not give
him a hug, then he receved 41-2=39 hugs.
First, I assigned variables.
B=# of boys, and $s each boy gets THIS YEAR
G=# of girls, and $s each girl gets THIS YEAR
N=# of newborns (2)
Page 14 of 32
Then, I wrote down the important info and turned it into
equations/inequalities, and combined/simplified them.
Every boy gets $B
There are B boys
$B^2 go to boys
Every girl gets $G
There are G girls
$G^2 go to girls
$B^2+$G^2=841
Student 22
The kind, old
gentleman recieved 39
hugs.
Student 23
He received 39 hugs
======================================================================
Last year there were more boys than girls
This year there are more girls than boys
2 girls were born
Last year
This year
B>G-N
BG-2
B
First, I tried guess and check but I decided that was going to take me
forever. Then I relized that each group, the boys and the girls
almost had the same amount, so I split $841 in half and got 420.5. I
took the square root of 420 and it is approximitly 20.4939 and then I
did 20 times 20 because each grandson got as many dollars as there
were grandsons and I got 400. Since there were twins, there would be
21 girls and 21 times 21 (because each granddaughter got as many
dollars as there were granddaughters) and I got 441. I added 400 and
441 together and got 841 and knew I had the right amount of people.
Then I added 21 and 20 toghter because that is the number of
grandchildren the olod man had and I got 41. But i subtracted 2 from
41 because the newbornsdidn't give their grandfather a hug.
Let x = of grandsons gifts
Let x+1 = granddaughters gifts since the two twins let the girls be
more numerous than the boys.
We know that the amount of money the boys receive is the number of
boys (x). Each boy receves the amount of money that is same to the
amount of boys of boys. So the amount of money the grandfather gave
to the boys is the number of boys (x) times the amount of money each
boy recieved (the amount each boy received is the amount of boys so
it is x). Therefore the amount the grandfather gave to the boys is
(x)(x) or x^2 and by reapeating this proces for the girls we know the
amount of money that the grandfather gave to the girls is (x+1)^2
We also know the grandfather gave out a total of $841
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 15 of 32
Now we can set up an equation. It is the total amount of money given
to the girls plus the total amount of money given to the boys = the
total amount given out.
It is x^2 + (x+1)^2 = 841
Now we solve
x^2 + x^2 + 2x + 1 = 841
2x^2 + 2x - 840 = 0
Know I used the quadratic formula to sol've for x
x = {-2 +/- sqrt of [4 - 4(2)(-840)]}/4
x = {-2 +/- sqrt of [4 + 6720]}/4
x = [-2 +/- sqrt of (6724)]/4
x = (-2 +/- 82)/4
x = -84/4 or 80/4
x = -21 or 20
-21 does not work because he can't have negative grandchildren.
Student 24
Student 25
The kind old
gentleman would
recieve 39 hugs. He
gets 19 hugs from girls
and 20 hugs from
boys. He doesn't
recieve hugs from the
newborn twin girls but
he still gives them
money.
The kind old
gentleman received 39
hugs.
© 1994-2016 Drexel University
http://mathforum.org/pows/
20 works, now we know there are 20 grandsons and 21 granddaughters.
20 + 21 = 41
he has 41 grandchildren
however he receieved 39 hugs since the two twins couldn't hug him.
The amount of granddaughts must be more than the amount of
grandsons. The amount of granddaughts squared plus the amount of
grandsons squared must add up to 841. That's how I got 21
granddaughters and 20 grandsons.
Because the total amount of money given out is $841, it is logical to
write an equation such that x is boys and y is girls and x squared
plus y squared equals 841. We square each x and y because depending
on the number of boys or girls, that determines the amount given to
each child. By solving graphically or through guess and check
methods, we come to the conclusion that there are 20 boys and 21 girls
which is concurrent with the observation that there is more money for
Page 16 of 32
Student 26
The grandpa receives
39 hugs.
girls than boys. Thus 20 squared plus 21 squared is 841. However,
the problem asks how many hugs the man gets and since two of the girls
can't walk yet, he doesn't get 41 hugs, but rather 39.
G = number of girls
B = number of boys
So I made one equation.
G=B+1
The equation works because the girls finally receive more money per
person this year because of the twins. Last year the boys had to have
one more than the girls, so that this year the girls could have more
people than the boys by only adding two.
G^2 = amount of money all the girls together receive
B^2 = amount of money all the boys together receive
G^2 works because each girl gets the same amount of money. The amount
of money they each get is the number of girls there are. So if there
are 2 girls each girl gets $2. Since there are 2 girls the total
amount of money the girls get is $4. If you put 2 in for G in the
expression G^2 you still get $4. You can try 5 girls also. Each girl
gets $5. The total amount of money the girls get is $25. If you put 5
in for G in the expression G^2, you still get $25. So the expression
works. The same thing for the boys and B^2.
So I made another equation.
G^2 + B^2 = 841
The total amount of money the boys and girls get together is $841.
Then I substituted B + 1 from the first equation for G in the second
equation.
G^2 + B^2 = 841
(B+1)^2 + B^2 = 841
B^2 + 2B + 1 + B^2 = 841
2B^2 + 2B + 1 = 841
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 17 of 32
2B^2 + 2B + 1 - 1 = 841 - 1
2B^2 + 2B = 840
(2B^2 + 2B)/2 = 840/2
B^2 + B = 420
I couldn't go any farther from here, so I used a little common sense
and guess and check. I figured out that you need a number plus that
number squared to equal 420. I decided to try 20 because I knew that
20 squared is 400 and that 420 was close to 400.
B^2 + B = 420
20^2 + 20 ? 420
400 + 20 ? 420
420 = 420
So B is equal to 20. Then you go back to the first equation, put 20 in
for B, and solve for G.
G=B+1
G = 20 + 1
G = 21
So the number of girls is 21. Then you add 20 and 21 to get the number
of grandchildren.
20 + 21
41
So there are 41 grandchildren. But since two of them are newborns they
can't give hugs yet, so you subtract two from 41 and you are left with
39. So the Grandpa receives 39 hugs.
Student 27
Student 28
The final solution is
that Grandfather
Moneybags received
39 hugs.
The old gentleman
received 39 hugs.
© 1994-2016 Drexel University
http://mathforum.org/pows/
I recognized that there was more granddaughters than grandsons. The equation I used was F^2 + M^2 = 841. I then
looked at the square root of 420 or roughly half of 841. Based on that, I realized F=21 and M=20. Therefore, there were
41 grandchildren. Grandfather Moneybags got hugs from all of his grandchildren except for the twin newborn girls.
Therefore, Grandfather Moneybags received 39 hugs.
I first determined from the problem that the number of girls or
granddaughters one plus the number of boys or grandsons (g=1+b). And
Page 18 of 32
because the amount of money given to the girls and boys equaled as
many dollars as there were girls and boys, I formed this equation:
(1+B)^2 + (B)^2= 841
I then multiplied this equation out, and found the new equation to
be:
2B^2 + 2B + 1= 841
I then subtracted the 841 from both sides and got this equation:
2B^2 + 2B - 840= 0
I then factored the above equation and got this new equation:
(B+21)*(B-20)=0
I then set (B+21) and (B-20) from the above and found the factors
to be -21 and 20. I know the answer to this problem can't be
negative, so I got rid of the -21. So the number of boys or
grandsons equals 20.
I then put the number of boys,20, back into the equation G=(B+1),
and found the number of granddaughters to be 21.
20 grandsons plus 21 granddaughters equals 41 hugs, minus the 2
babies hugs, equals 39 hugs.
Student 29
The gentleman
received 39 hugs.
© 1994-2016 Drexel University
http://mathforum.org/pows/
Since we were trying to find out how many hugs the grandfather
received, we first needed to find out how many boys and girls there
are.
Let x = # of boys, and
y = # of girls.
Since each boy received as many dollars as there were grandsons and
each girl received as many dollars as there were granddaughters,
x^2+y^2=841(the total amount of money given out)
This year there are more girls than boys due to the birth of twin
girls. So lets assume: y=x+1, therefore...
x^2+(x+1)^2=841
Page 19 of 32
Student 30
The frandfather was
given 39 hugs from his
grandchildren.
Student 31
The granfather
received 27 hugs.
© 1994-2016 Drexel University
http://mathforum.org/pows/
x^2+x^2+2x+1=841
2x^2+2x+1=841
To solve for x the quadratic equation needs to be factored.
1/2(4x^2+4x+1) +1/2=841
-1/2 -1/2
1/2(2x+1)^2=840.5
(2x+1)^2=1681
Then you take the square roots of each side.
2x+1=41
-1 -1
2x=40
x=20, number of boys
Plug this into the equation for the girls.
y=20+1, y=21, number of girls
Therefore, 21 girls and 20 boys, and 41 hugs
-2 hugs for the newborn twins
ANSWER: 39 HUGS
I decided to call the girls, x; and the boys, y. x^2 + y^2 = 841
because the boys and girls are given as much money as there are
children. (i.e. five boys= ive dollars each= 25 dollars). It was
stated in the problem that twin girls were born so the girls finally
outnumbered the boys. The girls were behind and to have twins bring
them to the lead they must have only been one behind. Therefore
x=y+1. Now there are two equatioons and we can put our amazing
algebra skills to work with the process of substitution. I
substituted (y+1) into the x in the first equation and came up with:
(y+1)^2 + y^2 =841. Then I expanded (y+1)^2 and combined to come up
with 2y^2 + 2y + 1=841 I subtracted the 1 and then divided everything
by two to get: y^2 + y=420 and I subtracted the 420 to the left side
and F.O.I.Led to get (y+21)(y-20) or y=-21 or 20. I picked the
positive amount of people which means that there are 20 boys. So I
then took the 20 and plugged it back into the original equation and
got the girls to equal 21. Then to find how many hugs
Grandpa "Moneybags" got I added the boys and girl and subtracted the
two newborns because it said in the problem that the newborns did not
give him a hug.
The number of how many girls there are squared is the number of money
he gave away and smae for the boys. For example if there were five
girls he would have given away 25 dollars.
So, g squared + b squared =841. The number of girls excedes the
Page 20 of 32
Student 32
Kind old "Moneybags"
got 39 hugs from his
loving grandkids.
number of boys by 1 because of the birth of the twin girls. It
couldn't excede the boys by 2 because then in the past they would
have had the same number but it said the boys always had more.
So, g= b + 1. (b+1)squared + bsquared = 841
Take the square root of each number and get (b+1) + b = 29
The grandfather has 29 grandchildren, 14 boys and 15 girls. He has 29
grandchildren but only received 27 hugs because the twins were too
young.
x=grandsons
y=granddaughters
841=(x*x+(y*y)
There was less girls than boys last year and now there are more girls
this year with the addition of the twins, so the difference was only
one. There are only 1 more girls than boys this year.
(y*y)= (x*x)+2x+1
841=(x*x)+(x*x)+2x+1
841=2(x*x)+2x+1
840=2((x*x)+x)
420=(x*x)+x
(x*x)+x-420=0
(split into 2 problems now)
(x-20)=0
x=20
Student 33
The old gentleman
received 39 hugs total.
(x+21)=0
x=-21
x can't equal a negative so there are 20 grandsons ans 21
granddaughters. That is 41 hugs but the twins are too young and that
equals 39.
In this problem I was already given that the total amount of money
give out by the grandfather was $841 dollars. I was also given twin
baby girls were born, so that made the total number of girls greater
than the total number of boys. I let the number of boys be n and the
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 21 of 32
number of girls be (n+1). Also, I made n be the amount of money per
boy and (n+1) be the amount of money for each girl. I then used
these variables and set up my equation:
n(n)+((n+1)(n+1))=841
Next I multiplied out the variables and got:
(n^2)+(n^2+2n+1)=841
I combined like terms:
2n^2+2n+1-1=841-1
2n^2+2n=840
Next I divided all the terms by 2:
n^2+n=420
I then moved the 420 to the other side of the equation in order to
make a quadratic equation. I then solved for n:
n^2+n-420=0
(n+21)(n-20)=0
n=20 and -21
In this problem though, you cannot have a negative number so n=20.
I plugged 20 in for the number of boys, which was n, and since the
n=20, there were 20 boys.
Next I plugged 20 in for the equation for the number of girls, which
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 22 of 32
was (n+1), I got 21 girls. Since though, the newborns didn't hug
their grandpa, I simply subtracted 2 from 21 and got 19. 20 boys and
19 girls added together equals 39, 39 hugs for Rich Grandfather
Student 34
Grandfather
Pennywise recieved a
total of 39 hugs from
his grandchildren.
Moneybags.
I began by determining the number of grandchildren (including the
twins). Now the problem states that the girls had more than the boys
due to the birth of twins. The means that there must be either one
more girl, or two more girls. In order to determine which was true,
I thought about how I was to total up the number $841. Since each
grandson got the same amount in dollars for each grandson and the
same applied to the granddaughters, I deduced that by squaring the
number of grandsons/grandaughters, I would come to each genders
total. Then I could add those two figures up to come to my total of
$841. So I found that if you square an even number you get an even
number (i.e. 2*2=4), and likewise found that by squaring an odd
number you get an odd number (3*3=9). Since 841 is odd, it must
therefore be the sum of an even number and an odd number. This means
that the difference between genders could not be two because
otherwise you would either have all evens or all odds.
So now that I knew that I must have just one more granddaughter, I
could create my equation. I assigned the variable 'b' for the number
of grandsons (or boys). I then assigned the variable 'b+1' for the
girls since they had one more than the boys. Therefore adding
b+(b+1) would give you the total number of grandchildren. So in
order to solve for 'b', I squared 'b' and 'b+1' so that I could set
them equal to $841, thus forming the equation b^2+(b+1)^2=841. This
equation shows that amount of money given to the boys plus the amount
given to the girls will equal the given amount of $841.
Begin the solving process by squaring 'b+1' to get b^2+2b+1. Now
combined the like terms on the left side of your equation and also
subtract both sides by one to get the equation: 2b^2+2b=840 . Next,
divide both sides by two and get b^2+b=420. Since this is a
quadratic, you must get all values on one side so subtract 420 from
both sides and recieve: b^2+b-420=0. So now we can factor the
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 23 of 32
trinomial found on the left side of the given equation. After I
factored I had (b-20)(b+21). I now set each of these binomials equal
to 0 and solved. My answers were 20 and -21. Obviously, -21 is not
feasible since that would indicate a negative amount of boys.
Therefore the answer must be 20 boys. This means that there were 21
girls since 20+1=21. In order to check these numbers square them and
add. The square of 20 is 400 and the square of 21 is 441. Add them
together and you will recieve the proper amount of $841.
Student 35
The old "rich"
Grandfather
"Moneybags" recieved
39 hugs from his
grandchildren.
Now we can finally determine how many hugs Grandfather Pennywise
recieved. He recieved 20 hugs from the grandsons, but since the
twins were much too young, we must subtract two from the number of
female hugs to get 19 (21-2=19). The sum of 20 (boy hugs) and 19
(girl hugs) is 39. So in conclusion the kind old grandfather
recieved a total of 39 loving hugs.
In the problem I was given that he gave out 841 dollars. I was also given that there were twin girls which made the total
number of girls more than the total number of boys.
I let x stand for the total number of boys and (x+1) stand for the total number of girls.
I then let x stand for the total amount of money for the boys and (x=1) stand for the total amount of money for the
girls.
I then made an equation for the total amount of money:
x(x) + ((x+1)(x+1)) = 841
Then I multiplied the variables.
x^2 + x^2 + 2x + 1 = 841
I combined like terms.
2x^2 + 2x + 1 = 841
-1
-1
2x^2 + 2x = 840
You then have to divide by 2.
2x^2 + 2x / 2 = 840 / 2
x^2 + x = 420
I then made it into a quadratic equation.
x^2 + x + 420 = 0
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 24 of 32
(x+21)(x-20) = 0
x = -21 x = 20
You can't have a negative number, so I have to plug 20 in for x.
x = 20
so there are 20 boys that gave hugs.
(20) + 1 = 21
so there are 21 girls, but the 2 newborns didn't give hugs.
21 - 2 = 19
so there are 19 girls that gave hugs.
20 + 19 = 39 total hugs.
Student 36
There are 21 girls and
20 boys, so the
grandfather will
receive 39 hugs
because the two
Newborns will not hug
him.
Student 37
Grandpa "Moneybags"
received 39 hugs
I wrote down that x^2 +(x +1)^2 = 841. I said x+1 because it said the
boys
usually got more money until the 2 girls were born, so there were
less girls, but not equal. 2 made the girls a larger number, meaning
there was originally only one less girl.
I knew that the numbers would have to be perfect squares because I
used an example and saw that if he had 7 girls and gave them each 7
dollars, then he would have given $49, it would always be a perfect
square. I then divided 841 by 2 and got 420, I took the square root
of 420 and got 20., I then said 20 would be
x. 20^2 is 400 and 21^2 is 441, when added together this equals 841.
x^2= grandsons
(x+1)^2= granddaughters
the x quantities are squared because "moneybags" gave to each of his
grandsons/granddaughters as many dollars as there are
grandsons/granddaughters. Mathmatically speaking:
x(number of grandsons/grandaughters)multiplied x(amount each received)
x^2 + (x+1)^2=841 (amount of $ in total given to grandchildren)
first simplify what's in parenthesis
x^2 + x^2 + 2x + 1=841
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 25 of 32
second add like terms
2x^2 + 2x + 1=841
third subract 841 from each side of the equation so you can solve the
equation using the quadratic fromula
fourth using the quadratic formula perform the proper operations
2x^2 + 2x - 840=0
x= -b +/- the square root of b^2-4ac divided by 2a
x= -2 +/- the square root of 4-4(2)(-840)divided by 2(2)
x= -2 +/- the square root of 4+6720 divided by 4
x= -2 +/- the square root of 6724 divided by 4
x= -2 +/- 82 divided by four
x= -2 + 82 divided by 4 or -2-82 divided by 4
x= 20 or x= -21
"moneybags" cannot have a negative amount of granchildren so x=20
there are 20 grandsons
there are 21 graddaugthers
there are 41 grandchildren
but two of the grandchildren can't walk so grandpa got only 39 hugs
Student 38
The grandfather got 39
hugs from his
grandchildren.
A=# of grandsons or dollars they recieved
A+1 =# of grandaughters or the dollars they recieved
A2 + (A+1)2 = 841
A2 + A2 + 2A + 1 = 841
2A2 + 2A + 1 = 841
2A2 + 2A = 840
A2 + 2A = 420
A2 + A - 420 = 0
(A+20)(A-21) = 0
20+21=41
41-2=39
Student 39
The grandfather will
receive 39 hugs.
We know that the grandchildren received a total of $841.
We know that there are 2 more girls than boys.
We will use b for boy and g for girl.
b + g must equal $841.
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 26 of 32
We know that if we have 20 boys they will get $400.
Now we have: 400 + g= $841
If you subtract these two figures you will get $441.
Since you know that there are 2 more girls there total will be
higher. We also know that there were normally more boys than girls
until this year. You would most likely choose 21 because they now
have more girls and normally there were more boys that girls. If 21
is used you would get $441 for the girls.
Add: $400+$441=$841
Student 40
The grandfather
recieved 39 hugs: 20
from his grandsons, 19
from his
granddaughters, and 0
from the pair of twin
girls born just one
month before.
Now you have 20 boys and 19 girls to give him hugs. This is a total
of 39 hugs.
Let Step
Let: b=the number of grandsons
Since each grandson received the amount of money equal to the
number grandsons
bxb or b²= amount of money given to grandsons
now for the granddaughters
This is a tricky part of the problem. Now if we look closely at the
problem we see that in many previous years there were more grandsons
then granddaughters. But with the birth of the twin girls, there
were more granddaughters then grandsons. This means that in previous
years there was one more grandson then granddaughter; with the birth
of the twin girls, there was one more granddaughter then grandson.
Thus,
Let: b+1= number of granddaughters, therefore
(b+1)(b+1) or (b+1)² = amount of money given to the
granddaughters,
again because each granddaughter recieved the amount of money equal
to the number of granddaughters.
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 27 of 32
So we can come up with the equation:
b²+ (b+1)²=$841 ($841 is the total amount of money that the
grandfather gave out, as indicated in the problem.)
b²+ (b+1)²=$841
the next step is to simplify the amount of money given to the
granddaughters. We do this by using the FOIL method
b²+(b+1)(b+1)=$841 ~~~Foil Method- (b+1)(b+1)=b²+2b+1 so..
b²+b²+2b+1=$841
now we simplify
2b²+2b+1=841
from here we continue to solve the equation
2b²+2b=840
2b²+2b+(-840)= 0
---------------divide the equation by 2
2
and we come up with:
b²+b-420=0
with:
we then use factoring in reverse and come up
(b-20)(b+21)=0
so he have 2 answers
b=20
answer
b=-21
imaginary answer
So the 20 grandsons each got $20
and the 21 granddaughters got $21
© 1994-2016 Drexel University
http://mathforum.org/pows/
20x$20=$400
21x$21=441
Page 28 of 32
Student 41
The grandfather
recieved 39 hugs
Student 42
he gets 39 hugs.
Student 43
My solution is that
there were 39 hugs
given.
Student 44
The old man received
39 hugs.
© 1994-2016 Drexel University
http://mathforum.org/pows/
$400+$441=$841, and going back to the beginning of this long holiday
problem we see that $841 is the amount of money that the grandfather
dished out for Christmas. So the answers check out. There were 20
grandsons and 21 granddaughters. But why then did he recieve only 39
hugs? Well as you remember, the new born twins were too young to run
up to good old Granpa Moneybags and give him a hug. So there ends
this complicated holiday problem, and yes I remembered to use my
algebra.
there are b boys and g number of girls.there is one more girl than
boy. g^2 + b^2=841 and g=b+1 ,so (b+1)^2+b^2=841. solving on my
calculator, b=20 and g=21. since the two newborns dont grandpa a hug,
he gets 20+19=39 hugs
If Grandpa "moneybags" gives out 1$ for each grandchild to each
grandchild, than that means he gives out x squared $s. let x be the #
of grandchildren he has. If the twin girls made it more girls than
boys, there is either 1 or 2 more girls than boys. Since he gives out
an odd # of $, there is 1 more girl than boys.
X^2+(X+1)^2=841
if you factor this out you get X^2+X^2+2X+1than you solve from here
2X^2+2X+1-1=841-1
(2X^2+2X)/2=840/2
X^2+X=420
(X+21)(X-20)=420
than you use zero product property, and since X cant be negative, X
must be 20. That means there are 21 girls and 20 bo
B2+G2=841
B+1=G
B2+B2+B+B+1=841
2B2+2B+1=841
2B2+2B+1/2=841/2
B2+B=420
B2=420-B
B=20
20 BOYS
21 GIRLS
39 HUGS
First you know that there are more girls than boys and because there
are twins, and the problem did not mention that girls and boys
received the same amount before the twins, x is boys and x+1 is
Page 29 of 32
Student 45
The grandfather
recieved 39 hugs.
girls. You also know that he gives out as much money to each
grandson as there are grandsons, so grandsons squared. Thus you set
the problem up: x(squared) + (x+1)squared = 841. Simplify to get
2xsquared + 2x +1 = 841. Subtract the one and: 2xsquared +
2x = 840. Divide by two: xsquared + x = 420. Next
factor: x(x+1)=420. From here I just thought of something times
that number plus one to equal 420. After some thought, I thought of
20 and 21 whose products equals 420. Thus x=20 and x+1=21 and there
are 20 boys and 21 girls. This means that the grandfather has 41
grandchildren, minus the two newborns who were unable to hug, so he
received 39 hugs.
The granddaugthers were happy because they would be receiving more
than the grandsons. This was the first time and it was due to the
birth of twins. This meant the boys have be recieving $1 more than
the girls and now the girls will recieve $1 more than the boys.
X=amount each boy recieved and number of boys
X+1=amount each girl recieved and number of girls
each X got X so amount all boys recieved is (X^2)
each (X+1) got (X+1) so amount all girls recieved is (X^2+2X+1)
X^2+X^2+2X+1=$841
2X^2+2X+1=$841
2X^2+2X=$840
X^2+X=$420
Using the guess and check method, I found X to be 20.
Student 46
The old gentleman
received twenty-nine
hugs.
© 1994-2016 Drexel University
http://mathforum.org/pows/
This means there were 20 boys and 21 girls. The grandfather recieved
39 hugs subtracting the newly born girls.
In order to figure out how many hugs, or grandchildren, the old man
had I used the equation 841=X(X)+Y(Y). In this equation X represents
the number of granddaughters and Y represents the number of
grandsons. I came up with this equation because each girl receives an
amount of money equal to the number of granddaughters there are. For
example;if there are two granddaughters each receives two dollars, a
total of four dollars. This means that if X represents the number of
granddaughters the grandfather gives out X squared dollars total to
his granddaughters. The same system is used for the grandsons, so X
Page 30 of 32
squared plus Y squared equals the total number of dollars the
grandfather gives to his grandchildren.
In solving this problem though I had to rearrange my equation
because I could not solve for a number with the equation 841=X(X)+Y
(Y). I did this by changing the equation to X+Y=the square root of
841. I changed it to this because finding the square root of a number
is the opposite of squaring a number. The square root of 841 is
29;therefore, the grandfather had twenty-nine grandchildren.
X=granddaughter Y=grandsons X+Y=square root of 841 --> X+Y=29
Student 47
27 children hugged
their grandfather.
Student 48
Grandfather
"Moneybags" recived
39 hugs
© 1994-2016 Drexel University
http://mathforum.org/pows/
Since you know there are more girls than boys you can say that there
were n amount of boys and n+1 amount of girls. There can't be too
many girls because they had never had as much as the boys before.
You can set up the equation:
(n+n+1)^2=841
you put the total amount of girls and boys together and multiply it
by themselves because the man was giving them each girl or boy the
amount of money of how many girls and boys there are.
So if there are 2 boys each would get 2 dollars and you see 2 boys
with 2 dollars each = 4 dollars all together. So if you total up how
many girls and boys there are and multiply it by how many girls and
boys there are you get how much money was given all together and that
is what 481 is.
From here on down you just do algebra:
(n+n+1)^2=481
Square root of ((n+n+1)^2)= square root of 481
n+n+1=29
n+n+1-1=29-1
2n=28
2n/2=28/2
n=14
and n+1=15
so there are 14 boys and 15 girls.
We then can say there were 29 in total but the two twins did not go
up and hug him so 29-2=27 children who went up to hug him.
The question was how many hugs did Grandfather "Moneybags" get,
I Started out with the things i already knew. He gave each grandson
the amount of grandsons in dollars, represented as S^2,and the same
with his Grandaughters, D^2. In the past there had been more boys
than girls(S>D), but this year there were more girls because of the
Page 31 of 32
Student 49
Grandfather
Pennywise recieved
thirty-nine hugs.
Student 50
Grandfather
"moneybags" recieved
20 hugs from the boys
and 21 hugs from the
girls, for a total of 41
hugs.
birth of 2 twin girls(S
First I decided to find out the information I needed to know. I needed to know how many grandsons and
granddaughters Grandpa Pennywise has. I knew that he gave out $841 total dollars to all the grandchildren, so I set up
the equation '(X*X)+(Y*Y)=$841.' X being the grandsons, and the amount of money each of them got, and Y being the
granddaughters and the amount of money each of them got. After trying this once I realized that I couldn't have two
variables and only one equation. So using some logic I figured out that there is one more girl than boy. (I found this out
by taking the square root of 841. The answer is 29. Since it is not an even number, there couldn't have been an even
number of guys and girls before the infants which would leave the girls with two more people after the births. The only
way the girls could have more is by there being just one more, because there was one less before the infants.) Since
there is one more girl than boy, I can make the equation, 'X+1=Y .' Next I used substitution and put 'X=Y+1' into the
origional equation and got, '(X*X)+((X+1)*(X+1))=$841.' I simplified and found the answer 'X=20.' Next I substituted 20
into 'X+1=Y,' and found that 'Y=21.' When I add the grandsons and granddaughters together I found out that 41 total
grandchildren recieved money. Finally I figured out that Grandfather Pennywise recieved only 39 hugs out of the 41
children, because the infants did not give him hugs.
I was not sure how to approch this problem so i did a guess and check
method. I figured that there had to be quite a large number of grand
kids so I started to play with numbers in the teens and twenties. I
figured that what ever number i choose I had to multiply it with
itself because he gave out the same amount of money as kids. I also
knew that there had to be either one or two more girls because of the
newborn twins. So I first tried numbers like 18x18=324 and
20x20=400, 324+400=724. 724 was to low a number so I tried bigger
numbers. 20x20=400, 22x22=484, 400+484=884. 884 was very close to
the answer(841) so i figured that maybe there was only one more girl
than boys. 20x20=400, 21x21=441, 400+441=841. 841 was the number I
was looking for so my answer means that there was 20 boys and 21
girls, therefore he recieved a total of 41 hugs.
© 1994-2009 Drexel University
http://mathforum.org/pows/
© 1994-2016 Drexel University
http://mathforum.org/pows/
Page 32 of 32