A + B + C = 2 + 39 + 44 = 85. Ans.
2005
MATHCOUNTS CHAPTER
SPRINT ROUND
4. Shooting hoops for 30 minutes
1. We are given the following chart:
Bangkok
Bangkok
Cape
Town
Honolulu
London
Cape
Town
6300
Honolulu
6300
6609
5944
London
6609
11,535
11,535
5989
5944
5989
5. Olga purchases a rectangular mirror
that fits exactly inside a frame.
7240
7240
To find the distance between Honolulu
and Cape Town, go to the row labeled
Honolulu and look for the intersection
with the column labeled Cape Town.
The value there is 11,535. (Or go to the
column labeled Honolulu and look for
the row labeled Cape Town. The
answer is still the same.) 11,535. Ans.
2. We are asked to find the percent of
Canadian non-users for whom cost
is not the primary barrier in the pie
chart shown below (redrawn in
Excel).
27%
burns 150 calories. Do this for 7
days you will burn 7 * 150 = 1050
calories. 1050 Ans.
7%7%
13%
28%
18%
Other
No need
Lack of skills
Not enough time
Cost
Access to computer or internet
According to the chart, 28% said
cost was the barrier to using the
internet. Therefore 100 - 28 = 72%
did not say cost was a factor. 72
Ans.
3. It is 12:00:00 midnight. In 122
hours, 39 minutes and 44 seconds
the time will be A:B:C on a 12-hour
digital clock. 122 hours is 5 days +
2 hours (24 × 5 = 120). Thus the
clock will show 2:39:44. Therefore,
The shaded region represents the
mirror itself. The outer perimeter of
the frame measures 60 cm by 80
cm and the width of each side of the
frame is 10 cm. To determine the
width and height of the mirror,
subtract 20 (10 for each side of the
border) from the width of the frame
and 20 from the height of the frame.
80 – 20 = 60
60 – 20 = 40
Thus, the mirror itself is 60 cm wide
and 40 cm high. 60 × 40 = 2400
Ans.
6. The product of two positive whole
numbers is 2005. To determine what
those numbers are, find the factors
of 2005. Of course 1 is a factor, but
we’re told we can’t use it. Since the
number ends in 5, it is clear that 5 is
a factor.
2005 = 5 × 401
401 + 5 = 406 Ans.
7. Three friends each order a large
2
pizza. Shauntee eats of her
3
8
pizza. Carlos eats
of his and
9
26
Rocco eats
of his pizza. To find
27
out how much pizza is left, add up
all the remaining pieces. Shauntee
1
1
left. Carlos has
left and
3
9
1
Rocco has
left.
27
1 1 1
9
3
1 13
 




3 9 27 27 27 27 27
has
Ans.
8. It takes 24 minutes for Jana to walk
one mile. So at this rate how far will
she walk in 10 minutes?
If Jana walks 1 mile in 24 minutes
she must walk
1
of a mile in 1
24
minute. In ten minutes she must
walk
10 5

 0.416  0.4 Ans.
24 12
9. Roslyn has 10 boxes. 5 contain
pencils, 4 contain pens and 2
contain both pens and pencils.
If 2 boxes contain both pens and
pencils and 5 contain pencils then
5 – 2 = 3 boxes contain only pencils.
If 2 boxes contain both pens and
pencils and 4 contain pens, then
4 – 2 = 2 boxes contain only pens.
Thus, we have 2 containing only
pens, 3 containing only pencils and
2 containing both. 2 + 3 + 2 = 7.
Yet there are 10 boxes total, so 10 –
7 = 3 boxes do not have anything in
them.
3 Ans.
10. How many combinations of pennies,
nickels and/or dimes are there with
a total value of 25¢?
Start with dimes.
2D + 5 P #1
2D + 1 N #2
We’ve exhausted 2 dimes.
1D + 3N #3
1D + 2N + 5P #4
1D + 1N +10P #5
1D + 15P #6
We’ve exhausted 1 dime. So forget
about dimes. Now only nickels and
pennies.
5N #7
4N + 5P #8
3N + 10P #9
2N + 15P #10
1N + 20P #11
No more nickels. Now only pennies.
25P #12
12 Ans.
11. Let’s insert parentheses into the
expression in the following manner:
(1 – 3) + (5 – 7) + (9 – 11) + …
(41 – 43) + (45 – 47) + 49
Note: The value of the expression
does not change. We realize that
there are 25 odd numbers in the
integers 1 through 50 and every two
of them in this expression sum up to
-2. There are then 12 pairs that
each sum to –2 and the 49 is left
over.
(-2 × 12) + 49 = -24 + 49 = 25 Ans.
12. A rectangular tile measures 3 inches
by 4 inches. What is the fewest
number of these tiles that are
needed to completely cover a
rectangular region that is 2 feet by 5
feet? Well, you could draw it like
this:
But it’s easier to see that it takes
six 4-inch lengths to make 2 feet,
and it takes twenty 3-inch lengths to
make 5 feet. So we have 6 of these
tiles going down and 20 of these
tiles going across.
6 × 20 = 120 Ans.
13. Trapezoid ABCD has vertices
A(1, -2), B(1, 1), C(5, 7) and D(5, 1).
To find the area of the trapezoid,
draw it
and draw the line BD. Two right
triangles have been formed. The
area of triangle BDC is ½ × 6 × 4 =
12. The area of triangle BAD is
½ × 4 × 3 = 6.
12 + 6 = 18 Ans.
14. Trey receives a 5% commission. If
he sells a coat for $60 his
commission is $60 × .05 or $3. If,
however, the coat were discounted
by 20%, its cost would be $60 × 0.8
or $48. Taking the commission on
$48 we get $48 × .05 or $2.40.
$3 - $2.40 = $.60, which is 60 cents.
Ans.
That’s all the factors since if you go
any further than the square root of
144 (i.e., 12), you’ll only repeat the
factors we have, but switched (i.e.,
16 × 9, 18 × 8, etc.)
Looking at all the factors we can see
that only 7 are two-digit numbers.
12 + 16 + 18 + 24 + 36 + 48 + 72 =
226 Ans.
18. The points B(1, 1), I(2, 4) and G(5,
15. Five distinct points A, B, C, D and E
lie on a line. E is the midpoint of
segment AB or AE = EB.
A-------E-------B
D is the midpoint of segment AE.
AD = DE = ½ EB
A---D---E-------B
Both C and E are the same distance
from B.
A---D---E-------B-------C
EB = BC
The distance from D to B is 9 units.
DE + EB = DB and DE = ½ EB
1
EB + EB = DB = 9
2
3
EB = 9
2
3EB = 18
EB = 6
DC = DE + EB + BC =
3 + 6 + 6 = 15 ANS.
16. 4 farthings = 1 pence
12 pence = 1 shilling
20 shillings = 1 pound
1 pound + 5 pence =
20 shillings + 5 pence =
20 × 12 pence + 5 pence =
240 pence + 5 pence = 245 pence =
245 × 4 farthings = 980 farthings
Ans.
17. What is the sum of all the distinct
positive two-digit factors of 144?
First, start by figuring out just what
the factors of 144 are…
144 = 1 × 144
144 = 2 × 72
144 = 3 × 48
144 = 4 × 36
144 = 6 × 24
144 = 8 × 18
144 = 9 × 16
144 = 12 × 12
1) are plotted to form triangle BIG.
The triangle is translated five units
to the left and two units upward to
form triangle B’I’G’ as in the
diagram below:
Translating to the left means
subtracting 5 from the x-coordinate.
Translating two units upward means
adding 2 to the y-coordinate.
If B is (1, 1), then B’ is (1-5, 1+2) or
(-4, 3). If G is (5, 1) then G’ is (5-5,
1+2) = (0, 3). Thus the midpoint
between (-4, 3) and (0, 3) is
40 
,3  = (-2,3). Ans.

 2

19. The positive difference of the cube
of an integer and the square of the
same integer is 100. Let x be the
integer. Then:
x³ - x² = 100
x²(x – 1) = 100
Factor 100 looking for squares.
1 × 100
100 = 10 × 10 and 10 – 1 = 9
100 × 9 = 900; doesn’t work.
Continuing: 2 × 50, 4 × 25
25 = 5 × 5 and 5 – 1 = 4. 5 Ans.
20. If you take the sheet of paper and
fold it twice, then unfold and mark
the quarters 1 through 4, it looks like
this:
1 + 2 + 2 = 5 Ans.
23. If a number is 25 or less, double the
Turn the paper over
and number the side behind 2 as 5,
the side behind 1 and 6, the side
behind 4 as 7 and the side behind 3
as 8. Now refold and cut. The final
image you get is:
Thus side 8
is directly behind quarter #3. The
bottom right hand corner is actually
all 4 corners of the big page.
Cutting the wedge in it cuts wedges
in all 4 corners. This means we are
down to choices A, D, and E. B has
the corner cut from the bottom, not
from the side. C and F have the
edge triangle cut, not a wedge. The
top right corner encompasses the
center edge between 1 and 3, and 2
and 4. Cutting wedges in there
provides two cuts symmetrical
around the horizontal fold. This
removes choice D (no cuts at all in
the middle) and finally choice A
because the type of cut is not a
wedge. This leaves us with only E
as the choice. E Ans.
21. Henry took five tests and got an
average of 57. This means a total of
57 × 5 = 285 points. He scored at
least 50 points (out of a possible
100) on each test. To find the
highest score, we must minimize the
scores on the other four tests.
Suppose Henry scored 50 on each
of the 4 tests. That sums to 200
points. 285 – 200 = 85 Ans.
22. How many ordered pairs (x, y)
satisfy BOTH conditions?
Condition I: x = 1 or y = 0 or y = 2
Condition II: x = 0 or x = 2 or y = 1
From condition I, choose x = 1.
Then from condition II we are forced
to choose y = 1 or (1,1). Now from
condition I, choose y = 0. From
condition II, x must be either 0 or 2
or (0,0) or (2,0) for two more
choices. Finally choose y = 2 from
condition I. Again, x must be either
0 or 2 for (0, 2) and (2,2) for two
more choices.
number. If a number is more than
25, subtract 12 from it. If the third
number in Zan’s sequence is 36, we
need to find the four distinct
numbers that could have been the
first number in her sequence.
Suppose the second number in the
sequence is less than 25. Then the
second number is doubled to give
36. Therefore, the second number
is 18. What would the first number
be? If it is less than 25, it would
have been doubled, so the first
number could be 9. But if the first
number is greater than 25, 12 is
subtracted from it to get 18. 18 + 12
= 30, so the first number could also
be 30. Now go back to the third
number and assume the second
number is more than 25 so 12 is
subtracted from it to get 36. 36 + 12
= 48 So the second number is 48.
Now assume, the first number is
less then 25 so it is doubled to get
48. Thus, the first number could
also be 24. Now assume the first
number is more than 25 so 48 is 12
less. Thus, the final possibility is 48
+ 12 = 60. The four possibilities are
9, 30, 24 and 60. Their sum is:
9 + 30 + 24 + 60 = 123 Ans.
24. My age has two digits (x and y). If I
reverse them, divide by three and
add 20, I get my age back again. I
can represent my age as 10x + y.
Reverse that and it becomes 10y +
x. Therefore,
10 y  x
 20  10 x  y
3
10y + x + 60 = 30x + 3y
7y + 60 = 29x
Since x is a single digit, it must be
large enough that 29x is greater
than 60. Thus x ≥ 3. If x = 3, then
29x = 87 and 7y = 27. That won’t
work. If x = 4, then 29x = 116 and
7y = 56 so y = 8. So I must be 48
years old. 48 Ans.
25. The sequence of integers in the row
of squares and in each of the two
columns of squares form three
distinct arithmetic sequences.
and then a 10-foot plank from D to
C, where D is the midpoint of AB .
Since D is the midpoint AD = 8. DC
= 10. CD is perpendicular to AB and
AC is the radius of the circle. Due to
the Pythagorean Theorem,
AC² = CD² + AD² = 8² + 10² = 164
Area of the circle = π r² = π AC² =
164π Ans.
The first information we have is 14
and 18. This says we have an
arithmetic sequence where values in
the vertical column in the middle
differ by 4. This leads us to:
27. f(n) = f(n – 1) + f(n – 2)
Now we have 21, ?, ?, 6, where
each value in the sequence differs
by the same number. If x is the
difference between terms in the
sequence, we get:
21, 21 – x, 21 – 2x, 21 – 3x and
21 – 3x = 6
15 = 3x and x = 5
Therefore, the difference between
terms in this sequence is 5 and we
can fill in some more of the squares.
f(1) = 3
f(3) = 10
f(3) = f(2) + f(1) = f(2) + 3 = 10
f(2) = 7
f(4) = f(3) + f(2) = 10 + 7 = 17
f(5) = f(4) + f(3) = 17 + 10 = 27
f(6) = f(5) + f(4) = 27 + 17 = 44 Ans.
28.
a 3 b 8 c 2
 ,  , 
b 4 c 9 d 3
4a = 3b
9b = 8c
3c = 2d
c=
2
d
3
9b = 8c = 8 ×
2
16
d=
d
3
3
27
b
16
3
a= b
4
3
27
81 2
b b
b
ad 4
16
64

 2 
b2
b2
b
81
Ans.
64
d=
We have the final sequence where
we let y be the difference between
terms in the sequence. So we have:
N, N + y, N + 2y, N + 3y, N + 4y,
N + 5y
From these terms we know
N + 5y = -17 and
N
+ y = -9
4y = -8 and so y = -2
N - 2 = -9
N = -7 Ans.
26. To be able to walk to the center C of
a circular fountain, a repair crew
places a 16-foot plank from A to B
29. A play has two male roles, two
female roles and two roles that can
be either gender. Men must be
assigned to male roles and women
to female roles. There are 5 men
and 6 women. The different
situations are:
A – 2 women play the either/or roles
B – 2 men play the either/or roles
C – 1 man and 1 women play the
roles
Situation A: 4 women get roles and
2 men get roles.
6×5×4×3×5×4=
30 × 20 × 12 = 7200 ways.
Situation B: 2 women get roles and
4 men get roles
6×5×5×4×3×2=
30 × 20 × 6 = 3600
Situation C: 3 women get roles and
3 men get roles
6×5×4×5×4×3=
30 × 20 × 12 = 7200 ways.
This is for when the male takes the
first role and the female takes the
second. We also have to do this
one more time when the female
takes the first role and the male
takes the second for another 7200
ways.
7200 + 3600 + 7200 + 7200 = 25200
Ans.
30. What is the arithmetic mean of all of
the positive two-digit integers with
the property that the integer is equal
to the sum of its first digit plus its
second digit plus the product of its
two digits? Let x be the tens digit
and y be the ones digit. Then:
10x + y = x + y + xy
9x = xy
y=9
Does this mean that x can be
anything?
19 = 1 + 9 + 9
29 = 2 + 9 + 18
Apparently so.
19, 29, 39, …, 89, 99 is an
arithmetic sequence. Since there
are an odd number of values the
mean is just the median or 59. Ans.
TARGET ROUND
1. Francisco starts with the number 5,
doubles it, adds 1, doubles the
result, adds 1, doubles the result
and continues this pattern of two
alternating calculations. Francisco
does 8 total calculations.
Calc 1: 5 × 2 = 10
Calc 2: 10 + 1 = 11
Calc 3: 11 × 2 = 22
Calc 4: 22 + 1 = 23
Calc 5: 23 × 2 = 46
Calc 6: 46 + 1 = 47
Calc 7: 47 × 2 = 94
Calc 8: 94 + 1 = 95
Phong starts with 5, adds 1, doubles
the result, adds 1, doubles the
result, adds 1, doubles the result,
and continues this pattern of two
alternating calculations. Phong also
does 8 calculations.
Calc 1: 5 + 1 = 6
Calc 2: 6 × 2 = 12
Calc 3: 12 + 1 = 13
Calc 4: 13 × 2 = 26
Calc 5: 26 + 1 = 27
Calc 6: 27 × 2 = 54
Calc 7: 54 + 1 = 55
Calc 8: 55 × 2 = 110
The positive difference of their
results is:
110 – 95 = 15 Ans.
2. All vertices of the cube are to be
colored such that no two vertices on
the same edge of the cube are the
same color. Clearly, one color won’t
work. But two does.
2 Ans.
3. A store purchases a television from
a factory for $87.89. The store
normally charges 225% of this.
$87.89 × 2.25 = 197.7525
The coupon gives 25% off.
197.7525 × 0.75 = 148.314375 ≈
$148.31 Ans.
4. Four different symbols represent
unknowns (‘ya know, this is just
making it take a couple of seconds
longer!!!).
Let a represent the triangle.
Let b represent the square.
Let c represent the hexagonallyshaped symbol.
Let d represent the upside down
trapezoid.
Then the 3 equations are:
a+b=d
5.
a+a=c+c+c+c+c
a+a=d+c
Let’s rewrite that:
a+b=d
2a = 5c
2a = d + c
Since 2a = 5c and a and c must both
be integers less than 10, we can see
that this equation is only true if a = 5
and c = 2.
Let’s go back to 2a = d + c.
We now have 2(5) = d + 2, so
d = 8. Using the equation a + b = d,
we now know 5 + b = 8, so
b = 3. Ans.
What is the greatest whole number
that MUST be a factor of the sum of
any four consecutive positive odd
numbers? Let x be the first odd
number. Then:
x+x+2+x+4+x+6=
4x + 12 = 4(x + 3)
x is an odd number. Adding 3 to it
will make it even. Thus, x + 3 is
divisible by 2.
4 × 2 = 8 Ans.
6. In the figure below, the smaller circle
has a radius of two feet and the
larger circle has a radius of four feet.
We are asked to find the total area
of the shaded regions.
First, one can create rectangles by
connecting the tangent lines with the
diameters of each circle. Each
rectangle has a length equal to the
radius and a width equal to the
diameter. The area of the smaller
rectangle is 4 × 2 = 8. The area of
the larger rectangle is 8 × 4 = 32.
32 + 8 = 40.
Now this area includes the shaded
area so subtract the areas of the
semicircles. The area of the smaller
semicircle is:
½πr² = 2π
The area of the larger semicircle is
8π. So the total area to be
subtracted is 2π + 8π = 10π
40 - 10π ≈ 8.5841 ≈ 8.58 Ans.
7. Jamie has a jar of coins. The jar
contains the same number of
nickels, dimes and quarters. The
total value of the coins in the jar is
$13.20.
Let x be the number of coins of each
type. Then:
5x + 10x + 25x = 1320
(i.e., move to cents; it’s easier)
40x = 1320
x = 33 Ans.
8. John, Mike and Chantel divide a pile
of pennies amongst themselves
using the following process:
- If the number of pennies in the pile
is even, Mike will get half of the pile.
- If the number of pennies in the pile
is odd, one penny will be given to
Chantel, and John will get half the
pennies in the pile.
This process repeats until the pile is
empty. They start with 2005
pennies.
-1- The number of pennies is odd so
Chantel gets one and John gets half
or 1002.
J: 1002, M: 0, C: 1
There are 2005 – 1003 = 1002
pennies left.
-2- The number of pennies is even
so Mike gets half or 501.
J: 1002, M: 501, C:1
There are 501 pennies left.
-3- The number of pennies is odd so
Chantel gets one and John gets half
of 500 or 250.
J: 1252, M: 501, C: 2
There are 250 pennies left.
-4- The number of pennies is even
so Mike gets half or 125.
J: 1252, M: 626, C: 2
There are 125 left.
-5- Chantel gets 1 and John gets
half of 124 or 62.
J: 1314, M: 626, C: 3
There are 62 left.
-6-: Mike gets half of 62 or 31.
J: 1314, M: 657, C: 4
There are 31 left.
-7-: Chantel gets 1, John gets 15
and 15 are left.
J: 1329, M: 657, C: 5
-8-: Chantel gets 1, John gets 7 and
7 are left.
J: 1336, M: 657, C: 6
-9-: Chantel gets 1, John gets 3 and
3 are left.
J: 1339, M: 657, C: 7
-10-: Chantel gets 1, John gets 1
and 1 is left.
J: 1340, M: 657, C: 8
-11: Chantel gets 1.
J: 1340: M: 657, C: 8
657 Ans.
TEAM ROUND
assembled Tworks in the room (and
no partially assembled or
disassembled Tworks). We have to
find how long it took. The least
common multiplier of 10 and 8 is 40,
so in 40 minutes 5 Tworks have
been assembled and 4 Tworks
disassembled. Every 40 minutes
one more Twork gets assembled.
We need to have 35 so 5 × 40 = 200
minutes to get 5 more Tworks. 200
Ans.
4. How many triangles are in the
figure? Start with the obvious ones.
1. A competition problem requires one
hour to develop. 30,000 students
work on the problem for an average
of 24 seconds each. The total time
spent by the students to solve the
problem is 30,000 × 24 = 720000
seconds. One hour is 60
minutes/hour × 60 minutes/sec or
3600 seconds. The ratio of the
development time to the total time
spent by the students is:
There are 16 triangles that contain
no smaller triangles within them.
Now look for triangles that are made
out of 2 smaller triangles.
3600
36
1


Ans.
720000 7200 200
2. Select a three-digit multiple of 3.
Calculate the sum of the cubes of
the digits of that number. Calculate
the sum of the cubes of the new
number and continue doing so until
you arrive at a number that is equal
to the sum of the cubes of its digits.
I.e.,
x³ + y³ + z³ = xyz
Start with something easy.
111 is a multiple of 3.
1³ + 1³ + 1³ = 3
3³ = 27
2³ + 7³ = 8 + 343 = 351
3³ + 5³ + 1³ = 27 + 125 + 1 = 153
1³ + 5³ + 3³ =1 + 125 + 27 = 153.
Ans.
3. A room contains 30 assembled
Tworks. There are enough pieces
to assemble 100 more Tworks. A
Twork takes 8 minutes to assemble
and 10 minutes to disassemble.
Emma starts assembling Tworks as
Ed begins disassembling the ones
already made. They both continue
to work until there are exactly 35
There are 10 of those.
There cannot be any triangles made
out of 3 smaller ones but there are
triangles made out of 4 smaller
ones.
and
for a total of 8 triangles there.
The only other triangular shape is
made out of 8 smaller triangles.
and
for a total of 2 more.
16 + 10 + 8 + 2 = 36 Ans.
5.
6.

x  2 25 x
2



1
For a power of (x – 2) to equal 1, x –
2 must be 1 or x – 2 must be -1 and
25 - x² must be even.
Start with x – 2 = 1. Then x = 3 and
25 - x² = 25 - 3² = 16. So x can be
3. Now suppose x – 2 = -1. x = 1
and that will work as well since
25 –12 = 24 and is then the even
power we would need for a base
of -1. But there’s one other thing:
Any value to the 0th power is 1 so 25
- x² = 0 will give us 2 other values, x
= 5 and -5 does that. This gives us
4 values. 4 Ans.
The minute hand of a clock (that
should be a 12-hour clock!)
measures 10 cm from its tip to the
center of the clock face. The hour
hand measures 5 cm from its tip to
the center of the clock face. We are
asked to find the sum of the
distances traveled by the tips of both
hands in one 24-hour period. The
minute hand has a length of 10 cm
so as it goes round it creates a circle
with radius 10 cm. Its
circumference is:
2πr = 20π. The minute hand goes
round the circumference 1 time per
hour or 24 times per day.
24 × 20π = 480π
The hour hand measures 5 cm so
as it goes round it creates a circle
with radius 5 cm. Its circumference
is 2πr = 10π
The hour hand also goes round the
circumference 2 times per day.
2 × 10π = 20π
The total distance traveled is:
480π + 20π = 500π cm but we
were asked for meters which means
that the total distance traveled is 5π
meters.
5π ≈ 15.70795 ≈ 15.708 Ans.
7. There are two shoelace patterns for
two identical shoes with 14 holes
each. Assume that the holes form a
rectangular grid and that each hole
is 1 cm from its nearest horizontal
and vertical neighbor-holes.
Determine the ratio of the total
length of the shoelace shown in the
first pattern to the total length of the
shoelace shown in the second
pattern.
The first pattern looks like this.
The straight lines are of length 1.
The diagonal line is the hypotenuse
of a 1 cm square so its length is
2 . There are 7 straight lines for a
length of 7 and there are 6 diagonal
lines for a length of 6 2 . So the
length of the shoelace to lace up
pattern 1 is 7 + 6 2 .
The second pattern looks like this:
It has one straight line and 12
diagonal lines for length of
1 + 12 2 .
7  6 2 15.485281374


1  12 2 17.970562748
0.861702640 ≈ 0.86 Ans.
8. A four-digit perfect square number is
created by placing two positive twodigit perfect square numbers next to
each other. We must find the
number. So what are the two-digit
perfect square numbers?
16, 25, 36, 49, 64, 81
What are the four-digit square
numbers that start around 1600?
1600 and 1681. 1681 looks
promising. How about around
2500? 2500, then 2601. That does
it. Each of the other squares can’t
start this number since the next
square will be at least 100 greater.
1681 Ans.
9. Ella rolls a standard six-sided die
until she rolls the same number on
consecutive rolls. What is the
probability that her 10th roll is her last
roll?
What’s the probability that the first
two rolls don’t get her the same
number twice?
Ella can roll any of 6 values the first
time and she can roll any of 5
values the second time. So the
probability is:
6  5 30 5


The probability
6  6 36 6
that Ella doesn’t get it on the third try
is the probability that she doesn’t get
it on the second try and she rolls
one of the 5 permissible values on
the third try. So similarly to go to the
10th try, the probability is:
5 5 5 5 5 5 5 5 1
       
6 6 6 6 6 6 6 6 6
The last time she must roll the same
number as her 9th roll and, therefore,
there is only one possible roll that
will work.)
58
390625


9
10077696
6
0.038761339 ≈ 0.039 Ans.
10. A deck of playing cards has 26 red
and 26 black cards. The deck is
split into two piles, each having at
least one card. There are 6 times
as many black cards as red cards in
the first pile. The number of red
cards is a multiple of the number of
black cards in the second pile.
Let x be the number of cards in the
first pile and y the number of cards
in the second pile.
x + y = 52
Let a = the number of red cards in
the first pile and b = the number of
black cards in the first pile. Then:
a + b = x and 6a = b
Let c = the number of red cards in
the second pile and d = the number
of black cards in the second pile.
Then:
c + d = y = 52 – x and
c = Nd where we do not know N.
a + c = 26 and b + d = 26.
Going back to the first pile:
7a = x
So the number of cards in the first
pile is a multiple of 7. The multiples
of 7 less than 52 are:
7, 14, 21, 28, 35, 42, 49
Remember that the number of cards
in the second pile must also be a
multiple of something so the value
better not be prime.
52 - each multiple of 7 is:
45, 38, 31, 24, 17, 10, 3
31, 17 and 3 are primes, ruling out
21, 35 and 49, leaving 7, 14, 28 and
42 as possibilities.
Consider 7 for the number of cards
in the first pile.
Then a = 1 and b = 6.
c = 26 – 1 = 25
d = 26 – 6 = 20
25 and 20 are not multiples of each
other.
Consider 14:
a =2 and b = 12
c = 26 – 2 = 24
d = 26 – 12 = 14
24 and 14 are not multiples of each
other.
Consider 28:
a = 4 and b = 24
c = 26 – 4 = 22
d = 26 – 24 = 2
c is definitely a multiple of d.
How about 42?
Then a = 6 and b = 36.
c = 26 – 6 = 20
d = 26 – 36 = -10 – I don’t think so!
The number of red cards in the
second pile is c or 22. Ans.