Problem Solving(part 2)
I Hope That You Are A Digital Computer?
26. What is the ones digit of 1! 2! 3!  999!?
{Hint: See what happens to the one’s digits:
Factorial One’s-digit One’s digit of the sum
1!
1
1
2!
2
3
3!
6
9
4!
4
3
}
27. Find the exact value of
The Collapse Of Rationalism
1
1
1
1


 
.
1 2
2 3
3 4
999,999  1,000,000
{Hint: Rationalize the denominators. For example:
1
1 2
1 2


 2  1 .}
1
1 2 1 2
The Beast With Many Fingers And Toes
28. How many digits does the number 86,666  520,000 have?
{Hint: Zeroes come from factors of 10. Factors of 10 come from 5’s and 2’s.}
Don’t Get Stumped; Use The Fundamental Theorem Of Arithmetic.
29. Forrest stump heard that there are only two numbers between 2 and 300,000,000,000,000
which are perfect squares, perfect cubes, and perfect fifth powers. He decided to look for
them, and so far he has checked out every number up to about 100,000 and is beginning to
get discouraged. What are the numbers he is trying to find?
{Hint: Every positive whole number greater than 1 can be written as a product of prime
factors.
If N is a positive whole number greater than 2, then
n1
n2
N  2  3  5n3   pk nk . In order for N to be a perfect square, all the positive
exponents would have to be multiples of 2; in order for N to be a perfect cube, all
the positive exponents would have to be multiples of 3; and in order for N to be a
perfect fifth power, all the positive exponents would have to be multiples of 5. So all
the positive exponents would have to be common multiples of 2, 3, and 5.}
Sorry, I Can’t Give You Change For A Dollar.
30. What is the largest amount of money in U.S. coins(pennies, nickels, dimes, quarters, but no
half-dollars or dollars) you can have and still not have change for a dollar?
{Hint: It’s more than 99 cents. For instance: 3 quarters and 3 dimes is $1.05, but you can’t
make change for a dollar.}
Destination Cancellation.
31. Express as a fraction, in lowest terms, the value of the following product of 1,999,999
1


 1  1  1
factors 1    1    1     1 
.
2
3
4
2,000,000

 
 



{Hint: Look for a pattern:
 1
1  
 2
 1  1 1 2
1    1    
 2  3 2 3
 1  1  1 1 2 3
1    1    1     
 2  3  4 2 3 4
1
2
1
3
1
4
}
Officer, I Got The License Plate Number, But I Was Lying On My Back.
32. The number on a license plate consists of five digits. When the license plate is turned
upside-down, you can still read a number, but the upside-down number is 78,633 greater
than the original license number. What is the original license number?
{Hint: The digits that make sense when viewed upside-down are 0, 1, 6, 8, and 9.
1st digit
2nd digit
3rd digit
4th digit
5th digit
7
8
6
3
3
Upside-down plate
Original plate
Difference of the plates
}
One Smokin’ Good Problem
33. Mrs. Puffem, a heavy smoker for many years finally decided to stop smoking altogether.
“I’ll finish the 27 cigarettes I have left,” she said to herself, “and never smoke another one.”
It was Mrs. Puffem’s practice to smoke exactly two-thirds of each complete cigarette(the
cigarettes are filterless). It did not take her long to discover that with the aid of some tape,
she could stick three butts together to make a new complete cigarette. With 27 cigarettes on
hand, how many complete cigarettes can she smoke before she gives up smoking forever,
and what portion of a cigarette will remain?
{Hint: With 27 complete cigarettes, she can smoke 27 complete ones and assemble 9 new
complete ones…, keep going.}
Just gimme an A.
34. A class of fewer than 45 students took a test. The results were mixed. One-third of the
class received a B, one-fourth received a C, one-sixth received a D, one-eight of the class
received an F, and the rest of the class received an A. How many students in the class got
an A?
{Hint: The number of students in the class must be a multiple of 3, 4, 6, and 8, and must be
smaller than 45.}
Working Backwards In Notsuoh
35. A castle in the far away land of Notsuoh was surrounded by four moats. One day the castle
was attacked and captured by a fierce tribe from the north. Guards were stationed at each
bride over the moats. Johann, from the castle, was allowed to take a number of bags of gold
as he went into exile. However, the guard at the first bridge took half of the bags of gold
plus one more bag. The guards at the second third and fourth bridges made identical
demands, all of which Johann met. When Johann finally crossed all the bridges, he had just
one bag of gold left. With how many bags of gold did Johann start?
{Hint: Sometimes working backwards is a good idea. If Johann has 1 bag of gold left, then
how many did he have when he approached the fourth guard?
x
 12 x  1  1 }
approached the 4th guard
taken by the 4th guard
bags left
Grazin’ In The Grass Is A Gas, Baby, Can You Dig It?
36. A horse is tethered by a rope to a corner on the outside of a square corral that is 10 feet on
each side. The horse can graze at a distance of 18 feet from the corner of the corral where
the rope is tied. What is the total grazing area for the horse?
{Hint:
18
18
8
8
}
Life Is Like A Box Of Chocolate Covered Cherries.
37. Assume that chocolate covered cherries come in boxes of 5, 7, and 10. What is the largest
number of chocolate covered cherries that cannot be ordered exactly?
{Hint: If you can get five consecutive amounts of cherries, then you can get all amounts
larger. Here’s why: Suppose you can get the amounts 23,24,25,26,27 , then by the
addition of the box of size 5, you can also get 28,29,30,31,32 , and another addition
of the box of size 5 produces 33,34,35,36,37 and so on. This would also be true of
seven consecutive amounts and ten consecutive amounts, but five consecutive
amounts would occur first. So look for amounts smaller than the first five
consecutive amounts.}
Easy Come, But Not So Easy Go.
38. A man whose end was approaching summoned his sons and said, “Divide my money as I
shall prescribe.” To his eldest son he said, “You are to have $1,000 and a seventh of what
is left.” To his second son he said, “Take $2,000 and a seventh of what remains.” To the
third son he said, “You are to take $3,000 and a seventh of what is left.” Thus he gave each
son $1,000 more than the previous son and a seventh of what remained, and to the last son
all that was left. After following their father’s instructions with care, the sons found that
they had shared their inheritance equally. How many sons were there, and how large was
the estate?
{Hint: Let M be the total value of the estate.
First Son
Second Son
M  1,000
M  1,000
M  3,000 
1,000 
7
2,000 
7
7
Since each son’s share is the same, solve an equation to determine the value of M,
and use it to find the number of sons.}
Round And Round With Sarah And Hillary
39. Sarah and Hillary are racing cars around a track. Sarah can make a complete circuit in 72
seconds, and Hillary completes a circuit in 68 seconds.
a) If they start together at the starting line, how many seconds will it take for Hillary to pass
Sarah at the starting line for the first time.
{Hint: Every time Sarah reaches the starting line must be a multiple of 72 seconds, and
every time Hillary reaches the starting line must be a multiple of 68 seconds. So
they will both be at the starting line at common multiples of 72 and 68.}
b) If they start together, how many laps will Sarah have completed when Hillary has
completed one more lap than Sarah?
{Hint: Let n be the number of laps completed by Sarah, then 72n  68  n  1 .}
Some divisibility rules for positive integers:
1) A positive integer is divisible by 2 if its one’s digit is even.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as
100a  10b  c , but 100a  10b  c  2  50a  5b   c , so if the one’s digit, c is even(divisible
by 2) then the integer abc will also be divisible by 2.
2) A positive integer is divisible by 3 if the sum of its digits is divisible by 3. This process may
be repeated.
Examples: 243 is divisible by 3, but 271 is not.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as
100a  10b  c , but 100a  10b  c  99a  9b  a  b  c  3  33a  3b    a  b  c  , so if the
the sum of the digits
sum of the digits, a  b  c , is divisible by 3 then the integer abc will also be divisible by 3.
3) A positive integer is divisible by 4 if the ten’s and one’s digits form a two-digit integer
divisible by 4.
Examples: 724 is divisible by 4, but 726 is not.
Here’s why:
Suppose we have the four-digit integer abcd, then its value can be expressed as
1000a  100b  10c  d , but
1000a  100b  10c  d  1000a  100b   10c  d   4  250a  25b  
10c  d 
, so if
the two-digit integer cd
the ten’s and one’s two-digit integer, cd, is divisible by 4 then the integer abcd will also be
divisible by 4.
4) A positive integer is divisible by 5 if its one’s digit is either a 5 or a 0.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as
100a  10b  c , but 100a  10b  c  5 20a  2b  c , so if the one’s digit, c is divisible by 5,
then the integer abc will also be divisible by 5, but the only digits divisible by 5 are 0 and 5.
5) A positive integer is divisible by 6 if it’s both divisible by 2 and divisible by 3.
6) A positive integer is divisible by 7 if when you remove the one’s digit from the integer and
then subtract twice the one’s digit from the new integer, you get an integer divisible by 7.
This process may be repeated.
Examples: 714 is divisible by 7 since 71 – 8 = 63, but 423 is not since 42 – 6 = 36.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as
100a  10b  c , but
100a  10b  c  90a  9b  3c  10a  b  2c 
new integer minus twice one's digit
 9 10a  b  2c   21c  10a  b  2c 
 10 
10a  b  2c 
 7   3c 
new integer minus twice one's digit
, so if the new integer minus twice the one’s digit, 10a  b  2c , is divisible by 7 then so is the
original integer abc.
Or
A positive integer is divisible by 7 if when you remove the one’s digit from the integer and
then subtract nine times the one’s digit from the new integer, you get an integer divisible by
7. This process may be repeated.
Examples: 714 is divisible by 7 since 71 – 36 = 35, but 423 is not since 42 – 27 = 15.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as
100a  10b  c , but
100a  10b  c  90a  9b  10c 
10a  b  9c 
new integer minus 9 times one's digit
 9 10a  b  9c   91c  10a  b  9c 
 10 
10a  b  9c 
 7  13c 
new integer minus 9 times one's digit
, so if the new integer minus nine times the one’s digit, 10a  b  9c , is divisible by 7 then so
is the original integer abc.
7) A positive integer is divisible by 8 if the hundred’s, ten’s, and one’s digits form a three-digit
integer divisible by 8.
Examples: 1240 is divisible by 8, since 240 is, 3238 is not, since 238 is not even divisible by
4.
Here’s why:
Suppose we have the five-digit integer abcde, then its value can be expressed as
10000a  1000b  100c  10d  e , but
10000a  1000b  100c  10d  e  10000a  1000b   100c  10d  e 
 8 1250a  125b   100c  10d  e  , so if the hundred’s,
the three-digit integer cde
ten’s, and one’s three-digit integer, cde, is divisible by 8 then the integer abcde will also be
divisible by 8.
8) A positive integer is divisible by 9 if the sum of its digits is divisible by 9. This process may
be repeated.
Examples: 243 is divisible by 9, but 9996 is not.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as
100a  10b  c , but 100a  10b  c  99a  9b  a  b  c  9 11a  b    a  b  c  , so if the
the sum of the digits
sum of the digits, a  b  c , is divisible by 9 then the integer abc will also be divisible by 9.
9) A positive integer is divisible by 11 if when you remove the one’s digit from the integer and
then subtract the one’s digit from the new integer, you get an integer divisible by 11. This
process may be repeated.
Examples: 1001 is divisible by 11 since 100 – 1 = 99, but 423 is not since 42 – 3 = 38.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as
100a  10b  c , but
100a  10b  c  90a  9b  2c 
10a  b  c 
new integer minus one's digit
 9 10a  b  c   11c  10a  b  c 
 10 
10a  b  c 
 11c
new integer minus one's digit
, so if the new integer minus the one’s digit, 10a  b  c , is divisible by 11 then so is the
original integer abc.
Or
A positive integer is divisible by 11 if when you subtract the sum of the ten’s digit and every
other digit to the left from the sum of the one’s digit and every other digit to the left you get a
number divisible by 11. This process may be repeated.
Examples: 9031 is divisible by 11 since 1  0    3  9   11, but 423 is not since
 4  3   2   5 .
Here’s why:
Suppose we have the three-digit integer abcde, then its value can be expressed as
10,000a  1,000b  100c  10d  e , but
10,000a  1,000b  100c  10d  e   9,999  1 a  1,001  1 b   99  1 c  11  1 d  e
 9,999a  1,001b  99c  11d   a  c  e    b  d 



 11   909a  91b  9c  d    a  c  e    b  d  


 alternating from one's alternating from ten's 
, so if the sum of the alternating digits from the one’s digit minus the sum of the alternating
digits from the ten’s digit,  a  c  e    b  d  , is divisible by 11 then so is the original
integer abcde.
Divisibility And Conquer
40. Of the following three numbers determine which are divisible by 2, which are divisible by
3, which are divisible by 4, which are divisible by 6, which are divisible by 8, and which are
divisible by 9: 3  10999  736 , 3  10999  534 , 3  10999  952 .
{Hint: Divisibility rules.}
Keep On Dividing And Conquering.
41. Of the following three numbers determine which are divisible by 3, which are divisible by
6, which are divisible by 7, which are divisible by 9, and which are divisible by 11:
1358024680358024679, 864197523864197523, 964197523864197522.
{Hint: Divisibility rules.}
Seven Come Thirteen, Not Eleven.
42. Show that the second divisibility rule for 7 can also be used as a divisibility rule for 13.
Modify the explanation to show why it works for 13.
Cogswell Cogs Or Spacely Sprockets?
43. In a machine, a small gear with 45 teeth is engaged with a large gear with 96 teeth. How
many more revolutions will the smaller gear have made than the larger gear the first time
the two gears are in their starting position?
{Hint: A revolution of the smaller gear is a multiple of 45 teeth, and a revolution of the
larger gear is a multiple of 96 teeth. So the gears are again in the starting positions
at common multiples of 45 and 96.}
My Tile Cutter Is Broken, So What Now?
44. The figure shows that twenty-four 8"12" rectangular tiles can be used to tile a 48" 48"
square without cutting tiles.
a) Is there a smaller sized square that can be tiled without cutting using 8"12" tiles? If so,
find it.
{Hint: The dimension of the square tile must be a multiple of both dimensions of the
rectangular tiles.}
b) What is the smallest square that can be tiled without cutting using 9"12" tiles?
{Hint: See the previous hint.}
Solving Without Completely Solving
45. If a  b  2 and a  b  3 , then find a 4  b 4 .
2
2
{Hint: Square the two equations.}
Odds, Evens, What’s The Difference?
46. What do you get if the sum of the first 8,000,000,000 positive odd integers is subtracted
from the sum of the first 8,000,000,000 positive even integers?
{Hint:  2  4  6  8 
 16,000,000,000   1  3  5  7 
 15,999,999,999  }
The Great Luggage Caper
47. Great Aunt Christine is going for her annual holiday to Barbados. She sends her butler John
down to the airport with her collection of suitcases, each of which weighs either 18 or 84
pounds, and is informed that the total weight checked-in is 652 pounds. Show that this is
impossible without listing and checking all the possible combinations of 18 and 84 pound
suitcases.
{Hint: The total weight of the suitcases is of the form 18x  84 y , where x and y are
nonnegative integers. So the expression must be divisible by the greatest common
factor of 18 and 84.}
Gerry Benzel’s Favorite Problem
48. A bottle and a cork together cost $1.10. If the bottle costs $1.00 more than the cork, what
does the cork cost?
{Hint: Let x be the cost of the cork and y the cost of the bottle.}
Getting Solutions Without Actually Solving
49. Notice that
 x  a  x  b   x 2   a  b  x  ab
 x  a  x  b  x  c   x3   a  b  c  x 2   ab  ac  bc  x  abc
 x  a  x  b  x  c  x  d   x 4   a  b  c  d  x3   ab  ac  ad  bc  bd  cd  x 2
  abc  abd  acd  bcd  x  abcd .
a) Use inductive reasoning to determine the value of the coefficient of x n1 and the constant
term in the expansion of the following product:  x  a1  x  a2   x  an  .
b) Use the previous result to determine the sum of the seventeenth powers of the 17
solutions of the equation x17  3x  1  0 .
{Hint: The Fundamental Theorem of Algebra guarantees that the equation
x17  3x  1  0 has seventeen solutions(counting duplicates). The seventeen
solutions of  x  a1  x  a2   x  a17   x17  3x  1  0 are a1 , a2 , , a17 . So
adding the seventeen equations together yields:
a117  3a1  1  0
a17
2  3a2  1  0
. Now use the previous result.}
17
 a17
 3a17  1  0
a
17
1
 a17
2 
17
 a17
  3 a1  a2 
 a17   17  0
Dots And Dashes, But It’s Not Morse Code.
50. The diagram shows a sequence of shapes T1 , T2 , T3 , T4 , . Each shape consists of a number
of squares. A dot is placed at each point where there is a corner of one or more squares.
T1
Shape
T1
T2
T3
T4
T2
number of rows, n
1
2
3
4
T3
number of squares, S
1
4
9
16
T4
number of dots, D
4
10
18
28
a) Use inductive reasoning to find a formula for S in terms of n.
b) How many squares are in shape T25 ?
c) Use inductive reasoning to find a formula for D in terms of n.
{Hint: Notice that D  n2 in the last column is always a multiple of 3.}
d) How many dots are in shape T25 ?
D  n2
3
6
9
12