MATHCOUNTS

®
KJ School Competition #2 
Sprint Round Solutions
Problems 1 – 30
Name _______________________________________________
Grade :__________
Teacher ____________________________
DO NOT BEGIN UNTIL YOU ARE
INSTRUCTED TO DO SO
This round of the competition consists of 30 problems. You will
have 40 minutes to complete the problems. You are not allowed to
use calculators, books, or any other aids during this round. If you
are wearing a calculator wrist watch, please give it to your proctor
now. Calculations may be done on scratch paper. All answers
must be complete, legible, and simplified to lowest terms. Record
only final answers in the blanks in the right-hand column of the
competition booklet. If you complete the problems before time is
called, use the remaining time to check your answers.
In each written round of the competition, the required unit for the
answer is included in the answer blank. The plural form of the unit
is always used, even if the answer appears to require the singular
form of the unit. The unit provided in the answer blank is the only
form of the answer that will be accepted.
Total Correct
Scorer’s Initials
1.
Brage runs around the track at a steady rate, finishing 400
meters in exactly two minutes. At that rate, how many
minutes will it take him to run 4000 meters?
1.
2.
If
7.
2.
30
__
3.
In the below diagram two squares have been drawn. Given
that the larger square’s side length is equal to the smaller
square’s diagonal length and that the smaller square has a
side length of 6 cm, find the area of the shaded region.
7.
3.
54
cm_2
4.
1.75
5.
10
find
20
minutes
Lets draw the diagonal to help us and the diagonals of the
larger square as shown above. Now we have four congruent
right triangles each with area 18. We want three of them.
4.
Find the value of
as a decimal to the nearest hundredth.
5.
How many faces does an octagonal prism have?
There are 8 faces for the octagon and then 2 extra for the
bases. Total of 10 faces.
_
faces
6.
I will choose a random integer, Q, such that
.
What is the probability, expressed as a common fraction,
that Q is prime?
6.
3/10
_
. There are 10 numbers in total. How many are
prime? 13,17,19. That’s it. So there is a probability of
7.
40 times a number equals 30 times 1 more than the same
number. Find the number.
7.
3
__
8.
I roll a die 7 times and get a prime
number on every single roll. What is
the probability I will get another
prime number on my 8th roll. Express
your answer as a common fraction.
8.
1/2
_
9.
37
%
The probability of the 8th roll is independent of the first 7.
So we just need to find the probability of rolling a prime
number on one roll. There are 3 prime numbers 2,3,5 from
the 6 choices.
9.
What is the single discount that is equivalent to the two
consecutive discounts of 10% off followed by 30% off the
discounted price?
Let’s call the price . Now we are taking 90% of
70% of This is .9 and .7
and then
But we are looking for a discount so its
10. How many perfect squares can be found in the first 2011
positive integers?
and
. So we are looking at squares 1-44
which is 44 perfect squares in total.
10.
44
squares
11. Find the area of a trapezoid,
in square inches, with bases
of 3 inches and 5 inches and
a height 2 feet.
11.
96
in2
Tricky question. Height of 2
feet. We are looking for inches. 2 feet 24 inches.
So now we apply area of trapezoid formula to get
12. A palindrome is a number that reads the same way
forwards and backwards. How many 2-digit palindromes
are prime?
12.
1 palindromes
There are 9 two-digit palindromes (11,22…99). Only one of
these is prime.
13. I bought a house for $200,000 dollars in 2005. Given that
the house’s value depreciates at a
rate of 8%, in how many years
will the house first have a value
less than $150,000?
13.
4
years
This can easily be done by just continuously multiplying
200000 by .92. So after one year we have a price of 184000.
After two years it decreases to around 169000. In 3 years it
will become 155000. In the 4th year it will go below 150,000
so 4 years is correct.
14.
Find the area of a regular hexagon with
side length 2 cm. Express your answer in
simplest radical form.
So we need to know how to find the area of a hexagon. This
can be done in several ways. But in my opinion the best way
is dividing it up into 6 equilateral triangles, for a better idea.
Now we have 6 equilateral triangles we know that the area
of these are
. So each of these has an area of
are looking for 6 times this which is
and we
14.
cm2
15. What is the smallest positive integer, , such that the
product of 60 and is a perfect square?
15.
15
_
16.
24
codes
17.
6
factors
18.
13
19.
30560
We can do this using guess and check or we could use the
prime factorization of 60. For a number to be a perfect
square, the exponents of the prime factorization of the
number must be even.
. We need the 3 and
the 5 to have even exponents. The closest is 2. So we need
one 3 and one 5 which is 15.
16. Miles has a 3-digit locker code. If he knows there are no 0’s
in the code and exactly 2 5’s in it, how many codes could
Miles possibly have?
Lets place the 5’s first. We can place then in the first and
second positions, the first and third positions, or the second
and third positions. 3 ways. Now for the remaining letter it
can be any of the 10 digits except 5 (only 2 5’s allowed) and
0. So 8 choices.
codes.
17. How many positive factors does 333 have?
Once again we can proceed with the prime factorization of
333 or we can list out the factors. Well in this case, 333 can
Be factored into
. To find the number of factors we
add one to the exponents and multiply.
factors.

18. In the MATHCOUNTS Summer Camp, exactly of the
participants are boys. Exactly
boys
of all participants have
glasses and exactly of all the female participants have
glasses. Find the least number of boys that have glasses.
In this problem we can see that the total number of campers
must be a multiple of 15. And that the total number of
campers must be a multiple of 18
. The smallest
possible number of campers is therefore 90. Of these 90, 18
wear glasses. And of the 30 girls, 5 wear glasses. This means
boys wear glasses
19. Find
_
20. What is the measure of the smaller angle,
rounded to the nearest whole number,
formed by the hour and minute hand when a
standard clock reads 8:31?
20.
70
degrees
Use the formula
to find that the angle is 69.5
degrees which rounds up to 70.
21. Find the sum of the first 50 terms of the sequence:
21.
9850
_
22.
4
_
1,9,17,25,33,41…
Its very easy to just use pairing in the problem. First we need
to find the 50th term. This is 49 8’s after the first term. So its
(the first term is 1). We can now see
that there will be 25 pairs of numbers that will add up to the
same amount (1st and 50th, 2nd and 49th…). This amount is
393+1=394. So our answer is
22. Find the smallest positive integer, , such that
by exactly 3 of the first 5 positive integers.
is divisible
Another tricky question. You would think that it would be
the first number divisible by 1,2, and 3 (which is 6).
However we want it to be as small as possible so we want
numbers with common factors. 2 and 4. And we throw in the
1 because 1 divides all positive integers. The LCM of 1,2, and
4 is simply 4.
23. Find the area of a regular octagon with a side length of 2
feet. Express your answer in simplest radical form.
2
For these octagon area problems I always draw 45-45-90
degree triangles, rectangles, and a square as shown above.
Now we can see 4 right isosceles triangles with side length
, a square with area 4, and 4 rectangles each with area
. In total this is an area of
23.
ft2
24. If
find
24.
23
_
25.
61
_
.
25. Yujian has a two digit number in the form AB. He knows
that
is a four digit number and can be written in the
form CDEF where
,
,
, and
. Find Yujian’s number.
Fun problem. Since
we are looking for 2-digit
numbers whose units digit are the same as their square’s
units digit. So B can equal 1, 5, or 6. So suppose
(we’ll
start small). This means C=3. So we are looking for squares
in the 3000’s. So
. The only square that
will work is 61. Let’s try it.
. Does it satisfy the
requirements? Yes it does.
26. In Mr. Yodice’s Social Studies class there are a total of 45
students. Every single student in the class likes at least one
of three sports: football, baseball, or cricket. Exactly of the
class likes football, 28 people like
cricket, and 21 people like baseball.
If exactly 3 students like all three
sports, how many students like
exactly 2 sports?
We could draw a Venn Diagram and proceed but I like to do
these problems with simple algebra. Well first we know that
exactly 15 kids do football, 28 like cricket, and 21 like
baseball. Then there are 3 that like all the sports. This is a
total of 67 kids. So there are
over counts.
Some of these are the 2 sports kids and some are the three
sports kids. Remember that for the 15,28, and 21 we have
already counted the three sports kids. So we need to
subtract 3 3 times from 22 to get the number of kids that do
exactly 2 sports.
students.
26.
13 students
27. What is the value of
27.
7
_
28.
39
cents
29.
162
This problem can be made much easier if we assign a
variable to this expression.
Squaring both sides gives
Solving this gives
.
.
28. In the land of BLOINK there are two coins in the currency.
There is the ploink which is the 5¢ coin and there is the
kzoink which is the 11¢ coin. Find the largest number of
cents that cannot be made exact change for in the land of
BLOINK.
This problem requires what is known as the Chicken
McNugget Theorem. It says that the largest number that
cannot be made using relatively prime positive integers
and is
. In this case its
29. What is the largest possible product of a set of positive
integers, not necessarily distinct, that have a sum of 14?
Playing around with the digits a little we can see that the
maximum product is obtained by using the maximum
number of 3’s. Here we can use 4 3’s and 1 two to give
_
30. Daniel loves bunnies. He has a huge rectangular backyard
which he wants to fill with bunnies. However the bunnies
he wants to buy will kill each other if put in the same
enclosed region. To make the regions he buys 11 long
straight fences which can extend as far as he wants (but
remember they are straight). If he wants each bunny to
have a separate region, and he uses all of the fences, what is
the greatest number of bunnies Daniel can buy?
Let’s break this down into smaller cases. Suppose we have 1
fence. That’s simple we can divide the area into 2 regions.
What about 2? The fences can only intersect in one point
giving 4 regions. Maybe they increase by 2 each time.
However, 3 fences can give 7. Hmm 2,4,7. Well if it started at
1 and then went to 2 and then to 4 and then to 7 we could
easily see the pattern. Wait! What if we have 0 fences? This
would provide just one big region. Now we can see a pattern
(+1,+2,+3…) but just to be sure let’s draw a picture to find
the greatest number of regions for 4.
Sure enough the greatest number of regions is 11. And now
the pattern makes sense. For each step there is exactly one
more line which we can divide into 2 regions.
Now it’s easy we can make a table and keep going till we get
our answer
30.
67 bunnies
Fences
Regions
0
1
2
3
4
5
6
7
8
9
10
11
1
2
4
7
11
16
22
29
37
46
56
67
MATHCOUNTS

®
KJ School Competition #2 
Target Round Solutions
Problems 1 and 2
Name _______________________________________________
Grade :__________
Teacher ____________________________
DO NOT BEGIN UNTIL YOU ARE
INSTRUCTED TO DO SO
This round of the competition consists of eight problems,
which will be presented in pairs. Work on one pair of
problems will be completed and answers will be collected
before the next pair is distributed. The time limit for each
pair of problems is six minutes. The first pair of problems
is on the other side of this sheet. When told to do so, turn
the page over and begin working. Record your final answers
in the designated space on the problem sheet. All answers
must be complete and legible. This round assumes the use
of calculators, and calculations may be done on scratch paper,
but no other aids are allowed.
In each written round of the competition, the required unit for the
answer is included in the answer blank. The plural form of the unit
is always used, even if the answer appears to require the singular
form of the unit. The unit provided in the answer blank is the only
form of the answer that will be accepted.
Total Correct
Scorer’s Initials
1.
Haolin’s pet turtle has a square plot of land with an area of
10 square feet. If Haolin doubles the
length of each side of the square, how
many square feet will her turtle
have?
So
2.
and we are asked to find
1.
40
ft2
. Expanding:
In the game of cricket, runs can be scored in 5
ways: singles, doubles, triples, 4’s, or 6’s. In
the Scholars Academy Cricket Match: Nabu
scored 7 6’s, 4 4’s, and 21 singles. If he scored
a total of 180 runs, find the greatest number
of triples Nabu could’ve hit.
So in total (without doubles and triples) Nabu has hit
runs. So only using doubles and triples:
runs. We want as many three’s as possible.
So what is the closest multiple of 3. Well the closest multiple
of 3 is 102 but that’s over, so we want 99. And this works
because
is a multiple of 2.
triples
2.
33
triples
MATHCOUNTS

KJ School Competition #2 
Target Round Solutions
Problems 3 and 4
Name _______________________________________________
Grade :__________
Teacher ____________________________
DO NOT BEGIN UNTIL YOU ARE
INSTRUCTED TO DO SO
Total Correct
Scorer’s Initials
®
3.
A circle, a square, and a line are drawn in a plane. What is
the greatest number of distinct intersections the three
shapes could have?
3.
12 intersections
The diagram below shows 4 intersections (which isn’t the
most).
This problem can really be done by experimentation and
knowing that a circle can intersect a square in 8 places. Then
the line can pass through each the square and the circle
twice to give a total of 12 intersections as shown below
4.
I draw a circle such that its area, in square meters, is
exactly of its circumference. Find the area of the circle, in
square meters. Express your answer as a decimal to the
nearest thousandth.
Plain and simple algebra. Nothing else.
therefore the radius of the circle is and the
area is
4.
1.396
m2
MATHCOUNTS

KJ School Competition #2 
Target Round Solutions
Problems 5 and 6
Name _______________________________________________
Grade :__________
Teacher ____________________________
DO NOT BEGIN UNTIL YOU ARE
INSTRUCTED TO DO SO
Total Correct
Scorer’s Initials
®
5.
When 3 standard 6-sided die are rolled, what is the
probability that the sum of the upward faces is 6? Express
your answer as a common fraction. One instance is shown
below:
5.
5/108
_
There are
possible rolls. How many of them have a
sum of 6? Well how can we get a sum of 6? We can have
.
can be done in 6 ways
can be done in 1 way
can be done in 3 ways.
This is a total of 10 ways.
6.
A cube is inscribed in a sphere of radius
inches. The
surface area of the cube, in square inches, can be expressed
in the form
where
and are positive
integers. Find
The diameter of the
sphere is
. This is
the space diagonal
length of the cube. The
space diagonal of a
cube with side length
is s . So in this case
and the surface area of the cube is
.
6.
11
_
MATHCOUNTS

KJ School Competition #2 
Target Round Solutions
Problems 7 and 8
Name _______________________________________________
Grade :__________
Teacher ____________________________
DO NOT BEGIN UNTIL YOU ARE
INSTRUCTED TO DO SO
Total Correct
Scorer’s Initials
®
7.
The numbers 1-9 will be put in the circles shown below
such that the sums of the triangle’s sides (4 circles) are
equal. What is the largest possible value of this sum?
7.
23
_
8.
125
_
To obtain the largest possible sum we need to use the 9
twice, the 8 twice, and 7 twice. After placing these we have
sums of 15,16, and 17. So the remaining 21(
must be divided into 3 parts, ,
,
. Solving for
we find that
. Adding this to the largest number (17)
gives a final answer of 23.
8.
Find the sum of all two digit numbers which satisfy the
following requirement: the product of the number’s digits
minus the number itself is equal to
.
For example: the number 25 would become
Fun problem if you like algebra. Let’s call the digits of the
number and where
and
. So now
we can get the following equation:
How can we make the left side factorable? By adding 10 to
each side:
We know that one of these terms must be negative, and we
also know which one because
. So now we scout for
pairs that work.
and
are the only ones that
give positive solutions for and . These pairs give solutions
51 and 74 respectively.
MATHCOUNTS

®
KJ School Competition #2 
Team Round Solutions
Problems 1 – 10
School ________________________________________
Chapter________________________________________
Team Members______________________________________ , Captain
_______________________________________________
_______________________________________________
_______________________________________________
DO NOT BEGIN UNTIL YOU ARE INSTRUCTED
TO DO SO.
This section of the competition consists of 10 problems which the team
has 20 minutes to complete. Team members may work together in any
way to solve the problems. Team members may talk during this section
of the competition. This round assumes the use calculators, and
calculations may also be done on scratch paper, but no other aids are
allowed. All answers must be complete, legible, and simplified to lowest
terms. The team captain must record the team’s official answers on his/her
own problem sheet, which is the only sheet that will be scored.
Total Correct
Scorer’s Initials
1.
What is the area of the triangle with vertices
and
?
1.
18
_
We can either plot these points out draw a rectangle and
subtract the area of the 3 right triangles remaining, or use
Shoelace method. Either way results in an answer of 18.
2.
Walter has the below configuration of rectangles and
squares. In total how many rectangles are there in Walter’s
configuration?
Combination problem. We have to choose 2 of the 11 vertical
segments and 2 of the 3 horizontal segments to make a
rectangle. This is
3.
Below is a
rectangular grid of unit squares with a
segment containing the midpoints of two squares. Find the
area of the shaded region. Express your answer as a
common fraction.
So its 4.5 across and 1.5 up. Lets use similar triangles! So we
can set up the following proportion
. This solves to
. Now we can use similar
triangles to find the other base of the trapezoid and we find
that it is . So using these bases and a height of 1 we find
that the area of the trapezoid is .
2.
3.
165 rectangles
2/3
_
4.
Kavi has all the letters of BANANA in a bag (1 B, 3 A’s, and 2
N’s). He spills them on a table, decides to pick up exactly 2
letters, and make a 2-letter word with them. How many
distinct 2-letter words can Kavi make?
4.
8 2-letter words
For this problem it is actually convenient to list them out and
count. So we can have a B and an A. 2 ways to arrange those.
We can have a B and an N. 2 ways there. A and an N –
another 2 ways. Then we can have NN and AA, each
arranged in 1 way. Adding
5.
What is the coefficient of the
is multiplied out?
term when
5.
270
_
6.
4/5
_
Here we use the binomial theorem or actually multiply it
through. Either way would result in an answer of 270.
Binomial theorem way:
6.
The Cavaliers and the Celtics play a best of 3 series. Because
the Cavaliers are ranked higher than the Celtics, they are
given 1 more home game than the Celtics. The Cavs play at
home in the 1st game and the 3rd
game (if there is a 3rd game). The
Cavaliers win an away game with
the probability of and a home
game with the probability of .
Find the probability that the
Cavaliers will win the series.
Express your answer as a
common fraction.
Casework here. 3 cases. I denote W – A Cavaliers win and L –
a Celtics win. First case: WW. This is one home game win and
one away game win so a probability of
. Second
case: WLW. Both home game wins.
. Last case
LWW. A home game loss, an away game win, and a home
game win.
Adding these we get a final
probability of .
7.
How many
cube to make a
cubes are needed to adjoin a
cube?
7.
19
cubes
This is a fairly easy computation, and is quite easy for a #7
(so is #8 – they are there just to throw you off).
.
This is based off of a problem 2 years ago that won the 2009
National Countdown.
8.
Find the product of the greatest common divisor and least
common multiple of 1337 and 343.
8.
458591
_
The product of the greatest common divisor and least
common multiple of 2 numbers is always their product.
.
9.
In Circle O,
and the
radius of the circle
is 6. Find the area
of the shaded
region. Express
your answer in
simplest radical
form in terms of .
For this problem we need to draw a few segments. If we
connect the point on the circle, ill call this point A (not on the
diameter), with the diameter we have a
degree triangle with hypotenuse of 12. That means that
the altitude of this triangle is
. So now we can find the
area of
which is
. The remaining portion is simply
of the circle’s area. So this is . So our final answer
is
9.
_
10. ABCDEFGHIJKL is a regular dodecagon (12-sided polygon)
with side length
cm. If the vertices of the dodecagon
are labeled clockwise, find the area of triangle FGH, to the
nearest whole
number.
It may seem like this problem requires trigonometry but it
really doesn’t. First we need to find out what the interior
angle of this polygon is. Using the formula
we find
that each angle is exactly
. So now we can concentrate
on our triangle. The triangles’ vertices are adjacent so we
have a
degree triangle. Ouch! For these
types of problems it’s much easier to just play around with
them a bit. We want to derive a right triangle out of this
mess but how…we could draw an altitude but that still gives
us a
angle. But what if we tried taking
from the
and adding that to the
angle. Then we could get a
triangle. It would look something like the
triangle shown on the following page. Now we have a
with a
beside it to make
one giant
triangle. Now we can find the area
of the largest triangle and subtract the area of the
to get the area of the
. So
subtracting
10.
37
cm2