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2017 Mock Math Challengers
Full Solutions Manual
The following pages contain the questions and solutions for the 2017 Mock Math Challengers Event.
These questions and solutions are from the Blitz and Bull’s Eye rounds, all written in order.
Blitz Round
1. What is the area of a pizza with a diameter of 8 inches?
Solution: The diameter is 8 inches, therefore the radius is 4 inches. Using the formula for the
area of a circle, we get:
Area = πr2 in.2 = 42 π in.2 = 16π in.2
2. Let P be the probability that, when a fair coin is tossed 3 times, less than 3 tails will be flipped.
What is the value of P? Express your answer as a common fraction.
Solution: The chances of less than 3 coins showing tails is equivalent to the chances of not
having 3 tails at all. The coins have 12 chance of flipping tails each, so the chances of getting
7
three tails is ( 12 )3 , or 18 . Therefore, the chance of not flipping 3 tails is 1 − 18 . Therefore, P =
.
8
3. What is the smallest positive integer divisible by both 5 and 7?
Solution: lcm(5, 7) = 51 × 71 = 35 .
4. Let x ⊗ y = x + xy . What is the value of 6 ⊗ 63 ?
Solution: Some algebra:
6⊗
6
3
=6⊗2=6+
6
2
=6+3= 9
5. How many ways are there to select 2 representatives from a pool of 5 candidates?
Solution:
5
2
=
5!
2!3!
=
5×4
2
= 5 × 2 = 10 ways .
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6. What is the average of all of the two-digit positive integers? Express your answer as a decimal
rounded to the nearest tenth.
Solution: First, the number of two digit numbers to add together must be found. Let 10 be
the 1st term, 11 be the 2nd term, and etc. It can be noticed that the two digit number n is the
(n − 9)th number. Then 99, the highest two digit number, is the 90th number.
Second, the total sum of all the two digit numbers must be found. Adding 10+11+12...+98+99 is
tedious. However, when adding the first and last numbers of the equation, you get 10+99 = 109,
and adding the second and second last numbers also gives you 11 + 98 = 109. Doing this results
in 45 pairs (90 total numbers give 45 pairs of numbers) of 109. The total is then 45 × 109.
The average of all these two digit numbers is the total divided by the number of terms:
109×45
90
= 54.5 .
An alternate way of arriving at this answer is realizing that the average is the "halfway point"
between all the integers, which also happens to be the average of 10 and 99, giving 109
2 = 54.5 .
7. Let N = 102 − 92 + 82 − ··· +22 − 12 . What is the value of N?
Solution: When the entire equation is factored out by the difference of squares, the following
equation remains:
(10 + 9)(10 − 9) + (8 + 7)(8 − 7) ... +(2 + 1)(2 − 1)
= (19)(1) + (15)(1) ... +(3)(1)
= 19 + 15+ ... +3
= 55 .
8. Two fair dice are rolled. What is the probability that the sum of the numbers on the top of the dice
is 7? Express your answer as a common fraction.
Solution: After making a table of all the possible combinations of the two dice, a sum of 7
1
6
appears 6 times out of the 36 total possibilities. The probability is 36
, which simplifies to
6
A table of the sum of the dice results.
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9. What is the sum of the positive factors of 2017?
Solution: 2017 is a prime number, so the only factors of 2017 are 1 and itself. Therefore, the
sum of the two positive factors is 2018 .
10. In regular hexagon√
ABCDEF , with side length 2 cm and centre O, if M is the midpoint of side
CD, then M O = 3 cm. What is the area of the hexagon, in square centimetres? Express your
answer in simplest radical form.
Solution: The regular hexagon can be separated into 6 equal triangles. The base of each triangle
is the side length of the hexagon, which is 2 cm.√ The heights of each triangle is M O, which is
√
3 cm. The individual area of each triangle is 3 cm2 . 6 of these triangles with equal area of
√
√
3 cm2 makes the whole hexagon have an area of 6 3 cm2 .
11. If 3 students from Buium Secondary can write 3 Math Challengers problems in 3 minutes, then
how many minutes will it take 1 student to write 224 problems?
Solution: Assuming the three students are working simultaneously and at the same rate, this
means 1 student can work on 1 problem in 3 minutes. With each problem taking 3 minutes, it
will take a student 3 × 244 minutes, or 672 minutes, to solve 224 problems.
12. Andrew, Bill, and Cleo are each asked to pick a number between 1 and 10, inclusive. What is
the probability that the 3 of them picked distinct prime numbers? Express your answer as a
common fraction.
Solution: Between 1 and 10, there are only four prime numbers: 2,3,5, and 7. Each person
has 10 choices of numbers and can choose the same numbers. That results in 103 or 1000 total
possible number of choices among the three different people. For one of the three people, they
have 4 choices of prime numbers to choose from. The second person has only 3 prime numbers
left to choose from that is distinct to the first person’s choice. The third and last person has 2
choices left of distinct primes from the previous two people. This means there are 4 × 3 × 2
or 24 possible choices of distinct primes out of the 1000 total possible combination of choices.
3
24
Thus, the probability of choosing three distinct primes is 1000
or
.
125
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13. Four squares, each of side length 2 units, are cut from a rectangular piece of cardboard with
length 9 units and width 8 units, as shown below. The resulting object is folded to create a box
(with no lid). What is the volume of the box?
Solution: After removing the four 2x2 squares, the resulting net will form a box with the
dimensions of 2 units by 4 units by 5 units. The volume of the box is 40 units3
14. An arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant. How many different arithmetic sequences of positive integers exist such
that the first term is 1 and that the number 7 appears in the sequence at least once?
Solution: In order to have all the terms in the arithmetic sequence to be all integers, the starting
number and the common difference must be integers. The first term, 1 is already an integer, but
it is 6 away from 7. In order for the common difference to be an integer that is able to reach 7,
it must be able to add to itself a number of times so it is equal to 6. This means the common
difference must be divisible by 6. The number 6 has four factors: 1,2,3 and 6. This means
there are 4 different possible common differences that allow the sequence to have all integers
and reach 7, meaning there exists 4 different possible sequences.
15. In a movie theatre, a group of 5 students must seat themselves in a row of 5 seats. However, two
of the students, Max and Kevin, are best friends and must sit next to each other. Additionally,
another pair of students, Austin and Rhiannon, are sworn enemies and cannot sit next to each
other. How many seating arrangements satisfy these conditions?
Solution: When seating the 5 people, we can guarantee the two friends sit beside each other by
considering the pair as one person. From either left or right, there are 4! ways to arrange the
people. However, the actual number is 4! × 2, since for each of the 4! ways to seat the people
there are also two ways to seat the friends relative to each other; Max to the right of Kevin, or
vice versa. Now, the number of arrangements where the enemies sit apart can be found as the
difference between the 4! × 2 arrangements where the friends sit together and the total number
of ways the friends sit together and the enemies too. The difference will be the number of
arrangements where the friends sit together, but enemies are NOT sitting together. The latter
can be found if the enemies are also seen as one group, like the friends, so there are 3! × 2 × 2
ways to seat both the friends and enemies together. The multiplication of the two 2s takes into
account the cases where one friend sits to the left or right of the other or vice versa, along with
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the enemies who are sitting together like the friends. Then, the total number of arrangements
that fulfill the stated conditions is:
(4! × 2) − (3! × 2 × 2)
= 48 − 24
= 24 ways
16. An ice cream cone has a height of h centimetres and a radius of 4cm. A lump of ice cream is
placed in the cone such that it fills both the cone’s volume and a hemisphere sitting on top of
the cone, which has the same radius as the cone. The total volume of the ice cream is 96π cubic
centimetres. Find the value of h, in centimetres.
Solution: The formulas for the volume of the cone and the hemisphere can be used to solve for
the radius algebraically.
Volume of Cone: 13 πhr2
Volume of Hemisphere: 32 πr3
We know that the sum of the cone and hemisphere is the total volume of 96π cm3 .
1 2
3 r πh
+ 23 πr3 = 96π
Placing the known radius of 4 cm in the place of r, we can solve for the height h using algebra.
2×43
42
3 πh + 3 π = 96π
128
16
3 πh + 3 π = 96π
16
3 π(h + 8) = 96π
8 + h = 18
h = 10
The algebra shows that the height of the cone h is 10 cm .
17. What is the units digit of 22017 + 02017 + 12017 + 72017 ?
Solution: The units digit of the sum of the numbers above is equal to the units digit of the sum
of the four numbers’ unit digits.
Keeping track of the units digits of the exponents of 2, starting from 21 , the following 4 unit
digits are always looped in a pattern: 2,4,8, and 6. Tracking the units digit of the exponents of
7, starting at 71 , four more unit digits are found in this pattern: 7,9,3, and 1. The units digit of
12017 and 02017 are both 1 and 0, respectively.
From the pattern of the units digit of the exponents of 2, it shows that every fourth exponent of
2, where the exponent is positive, the units digit of the number is 6. This means that 22016 has a
unit digit of 6 (2016 is a multiple of 4), but it also means that the unit digit of 22017 is 2. Using
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the same logic for the exponents of 7, every fourth exponent of 7, such as 72016 , has the unit
digit of 1. Now, it means that 72017 has the units digit of 7. Adding up the unit digits of the four
numbers, we can find the sum of the units digit, and therefore the units digit itself:
2 + 0 + 1 + 7 = 10
The units digit of the sum is 0.
It can be concluded that the unit digit of 22017 + 02017 + 12017 + 72017 is 0 .
18. Given that positive integers x, y satisfy x2 + y 2 = 100, and xy = 22, what is the value of x + y?
Solution: Knowing that the square of the sum of two numbers can be represented algebraically
as followed:
(x + y)2 = x2 + 2xy + y 2
(x + y)2 − 2xy = x2 + y 2
Placing the given solutions to some sets of the variables,a new equation is made:
(x + y)2 − 2(22) = 100
(x + √
y)2 = 144
x + y = 144 = +12
x + y = 12
Note: There was a mistake in the wording of the problem as x and y weren’t integers, but more
experienced contest writers would have ignored the fact as the method to solving this problem
is rather a common one.
19. As part of a school fundraiser, a group of students sold x medium-sized poinsettias at $12 each,
and y extra-large poinsettias at $32 each. If 500 poinsettias were sold, and a total of $9000 was
raised, how many medium-sized poinsettias were sold?
Solution: Two equations can be written from the given information. Let x be the number of
medium poinsettias sold and y be the number of extra-large poinsettias sold.
(
x+y=500
12x+32y=9000
Solving this system of equations gives the results x = 350 and y = 150. 350 medium sized
poinsettias were sold.
20. Given that 2x + 2x = (2y )y = 22025 , and that y is positive, find the value of x + y.
2
Solution: 2x + 2x = (2y )y = 22025 can be simplified to 2x+1 = 2y = 22025 . Since the bases of
the powers are equal, then the value of the exponents are equal.
x + 1 = 2025
x = 2024
2
y = 2025
y = +45
x + y = 2024 + 45 = 2069
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21. Each side of an isosceles triangle is coloured either red, yellow, green, or blue. Two colourings
are considered distinguishable only if the triangle cannot be reoriented, including flipping it
over, in such a way that the two colourings look the same. Find the number of distinguishable
colourings of the triangle.
Solution: Let Side 1 and 2 be the equal sides in the isosceles triangle and Side 3 be the remaining
side. There are two cases on how Sides 1 and 2 are coloured. They can either be coloured the
same, or differently.
If they are coloured the same, there are four colours to choose from, allowing 4 possible
colourings with Side 1 and 2 being the same colour. Sides 1 and 2 could also be different
colours, but for the colouring choices to be considered different, they cannot be reorientated or
flipped to match another colour choice. This means which colour Sides
1 and 2 each are does
4
not matter, but which two colours are used. This means there are 2 or 6 ways to colour the
two sides with different colours. The sum of these two cases show that there are 10 total ways
to colour Sides 1 and 2.
Side 3, however, can be any colour, since it is the side not equal to sides 1 and 2. There are 4
choices for how to colour side 3.
With 10 ways to colour the pairs Side 1 and 2, and 4 ways to colour side 3, there are a total of
4 × 10, or 40 unique ways of colouring the isosceles triangle.
22. Rectangle ABCD has an area of 2016 square units. Point Z is inside the rectangle and point
H on AB such that ZH is perpendicular to AB and ZH:CB= 4 :7. Find the area of pentagon ADCBZ.
Solution: The area of the entire rectangle ABCD is 2016, where the side lengths are AB and
CB. Since ZH is 47 of CB, the area of the smaller rectangle formed by the side AB with the
length ZH as its height is 47 of 2016, or 1152 square units. 4ABZ has half of the area of the
rectangle formed by AB and the length ZH, so the triangle’s area is 576. With the area of
rectangle ABCD and the triangle ABZ known, this means the area of the pentagon ADCBZ is
the difference between the two shapes. Where ZH is placed on AB does not matter.
2016 − 576 = 1440
The area of the pentagon ADCBZ is 1440 square units .
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23. The numbers 3, 6, 9, ..., 300 are given in a list. At each step of a process, you may take two
distinct numbers from the list, and replace the larger integer by their positive difference. This
process ends when no more positive differences can be made. What is the smallest possible sum
of the numbers in the list at any point in this process?
Solution: These numbers are part of an arithmetic sequence where the common difference is
3. There are 100 numbers in this sequence. Each step of the process replaces the larger number
with the positive difference, which will always be smaller than the larger number itself. If
repeated until this process cannot continue, then all the larger numbers have gotten as small as
they can. Then the sum of all the numbers is at its minimal. Any of two numbers selected and
placed through the steps of the process. will result in the replacement of the larger number with
a multiple of three. This process can continue until both numbers that are chosen will become
3s, which then has a difference of zero, so the process involving the two numbers must stop
and move to other numbers. Eventually, this process will reach a point where it has turned all
the numbers in the sequence the the lowest number possible, which is 3. Then, the process can
continue no more and all the numbers have been reduced as low as possible. Since there are
100 terms, and at the end of the process they are all 3s, then the sum of all the numbers in the
sequence is 3 × 100 or 300 .
24. How many integers n, between 1 and 10, inclusive, exist such that the fraction
n!
n2
is an integer?
Solution: In order for the fraction to be an integer, n!, when written in prime factors, must
contain all the prime factors found in n2 . The only ways to solve this problem is to do it case
by case for the ten integers. For each case, the types and amounts of each prime factor for the
numerator and denominator has to be counted and compared. When n is 2 and 3, however, the
denominator n2 is larger than n!. 4,5, and 7 result in improper fractions, due to the prime factors
of n! not being able to completely cancel out the prime factors of n2 . 1,6,8,9, and 10 are the
only 5 integers that make the fraction a integer.
1
25. In Miniland, 15
of the population has a disease called Science Fair Stress Disorder. A test exists
for the disease; if a person is sick, it will detect it 90 percent of the time, whereas the other 10
percent of the time, it will falsely say the person is healthy. On the other hand, if a person is
healthy, the test will correctly say the person is disease-free 90 percent of the time, but there’s
also a 10 percent chance it will say the person has the disease. The test has determined that
Ally, a resident of Miniland, has Science Fair Stress Disorder. What is the probability that Ally
actually has the disease? Express your answer as a common fraction.
Solution: There are two cases when the test reveals a person as having the disease: one where the
1
person is correctly identified to be sick and when a healthy person is misdiagnosed as sick. 15
of
the population is sick, and the test has a 90% success rate, meaning the chances of a sick person
1
9
9
diagnosed correctly is 15
× 10
or 150
, meaning that out of 150 people, 9 are correctly diagnosed
as sick. The chance of a person being healthy is 14
15 and the chance of being misdiagnosed is
1
14
10 , so there is a chance of 150 chance of a person being falsely diagnosed. The fraction means
that out of 150 people, 14 are actually healthy, but misdiagnosed. Combining the two fractions,
9
14
150 and 150 , means that out of 150 people, 23 are diagnosed to have the disease, but out of the
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23, 9 actually have it, while the other 14 are false alarms. So the chances of Ally having the
9
disorder is
.
23
26. How many positive integers, less than or equal to 1000, are not divisible by 2, 3, nor 7?
Solution: To find the number of integers under 1000 that are NOT divisible by 2,3 or 7, we need
to find the number of integers under 1000 that ARE divisible by 2,3, or 7 first. To determine how
many integers under or equal to n is divisible by m, the number can be found by by dividing n
by m. When you divide n by m,you get p, either a whole number or not. When it is not, then p
will be rounded down. This means that p × m is the smallest multiple of m under or equal to n,
and is the pth multiple of m.
Drawing a 3-circled Venn Diagram, we can first find out how many numbers are divisible by
all of 2,3, and 7. It is asking the same task as how many numbers under 1000 are divisible by
42(2 × 3 × 7). There are 23 integers under 1000 divisible by all of 2,3, and 7, so in the Central
common area, is the number 23. For the common areas between two of the circles, we will find
the number of integers under 1000 divisible by 2 and 3, 2 and 7, and 3 and 7. There are 71
integers divisible by 2 and 7, but 23 of them are also divisible by 2,7, and 3. This means there
are 71 − 23 or 48 integers divisible by 2 and 7, but not 3. Using the same process for the other
common areas between two circles, there are 143 integers divisible by only 2 and 3, and 24
divisible by only 3 and 7. Now, there are 500 integers under 100 divisible by 2, but 23 or them
are divisible by all 2,3, and 7, 48 others are divisible by only 2 and 7, and 143 more divisible
by only 2 and 3. This leaves 500 − 23 − 48 − 143 or 286 numbers under 100 divisible by 2
only. Doing the same way of calculating, there are 143 numbers only divisible by 3 and 47 only
divisible by 7.
Adding all these numbers together, we get the total number of integers that are divisible by 2,3,
or 7, which is 714. This means there are 1000 − 714 or 286 numbers under or equal to 1000
that are NOT divisible by 2,3, or 7.
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Bull’s-eye Round
1. A drawer contains 6 blue, 4 black, and 5 white T-shirts. Jacob is blindfolded and takes T-shirts
from the drawer at random. How many T-shirts does he need to remove to guarantee that he has
taken at least 2 T-shirts of each colour?
Solution: To find the number of minimum number of shirts Jacob need to take, the worst
possible scenario has to be considered. Suppose Jacob takes out 6 shirts, but in the worst case
scenario, all of those 6 shirts are blue, so he is not even close to 2 shirts of colour. He picks
out 5 shirts again, but these five shirts are all white. He still has not gotten any black shirts yet.
Picking two more shirts would guarantee the last two are black. In this worst case scenario,
Jacob has first picked 6 blue shirts, then 5 white shirts, and finally 2 black shirts. He has picked
13 shirts overall, which means 13 shirts is the minimum Jacob can take to guarantee he has at
least 2 shirts of every colour.
2. A ferry ticket costs $23.10, which includes a 5% tax. What is the cost of the ticket without tax?
Express your answer in standard monetary format (with two digits after the decimal point).
Solution: Let x be the cost of the ticket before tax.
1.05x = 23.10
x = 23.10
1.05 = 22
The cost of the ticket before tax is $22.00 .
3. After buying n candies at a price of 11 cents per candy, Lucy notes that the total price, in cents,
of all n candies was of the form 19A0, where A represents a digit. Find the value of A.
Solution: 19A0 cents is the price of n candies at 11 cents each, so 19A0 is also divisible by
11.For an integer to be divisible by 11, the sum of the second digit from the left and second digit
from there in the number, when subtracted from the sum of the first digit and every second digit
from there in the number, is a number that is divisible by 11. This means if 19A0 is divisible by
11, then (1 + A) − (9 + 0), simplified to A − 8, is divisible by 11. Since A can only be a single
digit number, A can only be 8, for no other single digits when subtracted by 8 gives a number
divisible by 11. When A is 8, A − 8 becomes 0 which is divisible by 11. A must be 8 .
4. Usain Bolt starts a 100-metre race stationary at the start line. Once the race begins, he increases
his velocity, at a constant rate, until 5 seconds have elapsed. At this point, he is travelling
at x metres per second. He then maintains this speed until he crosses the finish line, 10 seconds after the start of the race. What is the value of x? Express your answer as a common fraction.
Solution: We can graph the velocity of Usain Bolt’s velocity during the race on a Velocity vs
Time graph. Think of Velocity as similar to speed. The area under the velocity line of the graph
represents the distance travelled of the object in the time frame. Mr. Bolt starts at a velocity
of 0 and ends at a velocity of x over the duration of 5 second. The remaining 5 seconds of the
race, Mr. Bolt maintains a constant speed and won the 100m race. Looking at the graph, the
velocity line forms a trapezoid that is made up of a triangle (When Mr.Bolt is accelerating) and
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a rectangle (When Mr.Bolt is finishing the race at a constant speed.)
The height of the triangle is x, the final velocity of Mr.Bolt after the first 5 seconds, for which
the base of the triangle also represents 5 seconds, the time passed since the start of the race on
the graph. The area of the triangle is therefore 5x
2 . The rectangle has a height of x, the constant
velocity, and a base of 5, representing the remaining 5 seconds it took for him to finish the race,
with an area of 5x. The sum of these areas is equal to the distance travelled. Algebra is all that
is needed:
5x
2
+ 5x = 100m
15x
2 = 100
2
= 40
x = 100 × 15
3
From the algebra, it is shown that Mr. Bolt has accelerated to
40
metres per second .
3
5. Alice and Bob have 3 children. If each child is equally likely to be a boy or a girl, what is the
probability that the couple have 2 boys and 1 girl? Express your answer as a common fraction.
Solution: Given that the couple has three children, and each child can either one of two genders,
the total number of possible combination of children is 23 , or 8. How many of these ways contain
2 boys and one girl is the number of ways with the order of the children being born. The girl can
be born first, followed by the two boys, or she can be born second, or third and last, giving three
possible birth orders, and therefore three ways. Three ways out of the possible eight allows two
3
boys and one girl means the chances the couple has those children is .
8
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6. An ant is located at the bottom-left corner of the grid below. It can only move up or to the right,
along the grid lines. How many distinct paths can it take to get to the top-right corner?
Solution: The total number of paths the ant can take is the number of combinations of up and
right paths toward the top right corner. To get there, the ant three right paths and two up paths.
The problem can be simplified if we let R be a right path and U be an up path. Now, the problem
is an arrangement problem of how many ways are there to arrange 3 Rs and 2 Us. There are 5
choices for the first path, 4 for the second, etc, giving 5! ways of arrangements. However, the 3
Rs and the 2 Us are identical. There are 3! ways the Rs are arranged in the same combination,
and 2! ways the Us are arranged too. Dividing 5! by 3! and 2! gives the number of unique
5!
or 10 paths .
arrangements, which will be 3!2!
7. How many 3-digit numbers have exactly 2 digits the same (while the third is different)?
Solution: There are three cases to count:


#1 − T hree digit numbers without 0 present.
#2 − T hree digit numbers with one 0.

#3 − T hree digit numbers with two 0s.
The first case is relatively straightforward. Out of 9 single-digit, non-zero integers, 2 will be
chosen, and there are two possibilities which of the two will be a pair and which will be alone.
Then, there are 3 ways to arrange the two integers: thelonely digit in the hundreds, tens, or units
digit. So the number of numbers in the first case is 92 × 2 × 3, or 216 ways.
The second case is when there is a zero in the number. Since one of the digits is guaranteed
to be 0, then there are 9 choices left to choose from the non-zero digits. The composition of
the numbers will then be a non-zero integer in the hundreds place, while there are two ways to
arrange the remaining 0 and non-zero in the tens and units place. There are 9 × 2, or 18 numbers
that fit the case.
The third case has two zeroes, meaning there can only be one non-zero digit in the hundreds
place, and there are 9 choices for that digit. This means there are 9 numbers that fall in this case.
The total number of three digit numbers that fit the conditions specified in the question will be
the sum of the three cases, which is 216 + 18 + 9, equivalent to 243 total numbers.
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8. In a swimming pool, there are 2 pumps and a drain. Pump A can fill the pool in 30 minutes,
Pump B can fill the pool in 20 minutes, and the drain can empty the pool in 15 minutes. After
filling the pool for 5 minutes only with Pump A (and with the drain closed), Pump B joined in.
5 minutes later, the drain was accidentally opened. After that, the pumps continue to fill the
pool until it was full. How long (in minutes) was the entire process, starting from when Pump
A started pumping and ending when the pool was full?
Solution: We would need to find the speed of the water pumps and the speed of draining in
order to calculate the time.
1
of the pool
Pump A has a speed of pumping 1 pool in 30 mins, so it has a speed of filling 30
1
per minute. Pump B can fill the pool in 20 mins, so it fills 20 of the pool per minute. The drain
1
takes 15 mins to empty the pool, resulting in 15
of the pool draining per minute. In the first 5
1
5
minutes, Pump A would have filled 30 or 6 of the pool. In the next 5 minutes, Pump B and A
1
of the
worked together, so the rate the pool was filling up is the sum of their speeds, which is 12
5
1
pool filled per minute. During this time frame, 12 of the pool is filled in with the 6 of the pool
7
of the pool will be filled in total at the end of the time period when Pumps
already filled, so 12
A and B worked together, before the drain is opened. When the drain is opened, the rate which
1
1
1
the pool fills slows. The new rate of filling is 12
− 15
, or 60
of the pool being filled per minute.
5
The pool still has 12 of it left to fill.
Let x be the minutes it takes for the rest of the pool to fill.
1
5
60 x = 12
5
x = 12
× 60
1 = 25 minutes
It will take 25 minutes to finish filling the pool.
The first 5 minutes was Pump A by itself, until joined by Pump B for the next 5 minutes, until
the drain is opened. It takes another 25 minutes to fill the pool completely. In this situation,
35 minutes has passed from the start to end.
9. A diagonal of a polygon is a line that connects two non-adjacent vertices of the polygon. How
many diagonals of a regular hexagon do not pass through the centre of the hexagon?
Solution: First, we can choose
pairs of the hexagon’s vertices that will represent a diagonal.
6
The number of pairs is 2 , which results in 15 pairs. However this counting also counts the
side lengths of the hexagon too. There are 6 sides of a hexagon, which are not diagonals, so
subtracting 6 from 15 gives you 9 diagonals. However, 3 of the diagonals will pass through the
center. This means that a hexagon has 9 − 3 or 6 diagonals that do not pass through the center.
These solutions are written by the members of CSSMA.
10. Bobby notes that a 500 metre tall skyscraper in front of him is currently casting a 20 metre
shadow. If Bobby is 1.5 metres tall, how long, in centimetres, is his shadow?
Solution: The ratio of the shadow length to actual height for the building should also be the
same for the shadow length and height of a person. Let x be the length of Bobby’s shadow. Then
20m
xm
500m and 1.5m should be equal. Solving the algebra, x would be 0.06 metres. Bobby’s shadow
length would then be 6 centimetres .
11. A parallelogram has 3 consecutive vertices at (0,0), (1,5), and (4,5). What is the length of the
shorter diagonal of this parallelogram? Express your answer as a radical in simplest form.
Solution: If a line is drawn from (1,5) down to the line between (0,0) and (3,0) such that it is
perpendicular, a right triangle is formed. Using the Pythagorean Theorem, the side lengths of
√
the new triangle will be 5 and 2, so the length of the hypotenuse will be 29 .
These solutions are written by the members of CSSMA.
12. Given a circle with center O, and chords AB and CD of the circle, let E be the foot of the
perpendicular from O to AB, and let F be the foot of the perpendicular from O to CD. If AB=
8, CD= 6, and EF = 1, find the radius of the circle.
Solution: It is a fact that if a chord in a circle is perpendicular with a line that passes through
the circle’s center, then that line bisects the chord. This means AE and EB are equal, and both
has a length of 4. CF and FD are also equal, with each having a length of 3. OB and OC are
the radii of the circle, and are the hypotenuses of the right triangles OEB
p and OFC, respectively.
Let x be the length of OE. Using the Pythagorean Theorem, OC is (x + 1)2 + 32 and OB is
√
x2 + 42 . OB and OC are also equal, so algebra can be used to solve for x.
√
(x + 1)2 + 32 = x2 + 42
(x2 + 2x + 1) + 9 = x2 + 16
2x + 10 = 16
2x = 6
x=3
p
Since OE is 3, using the Pythagorean Theorem again, shows that OB is 5. The radius of the
circle O is 5 .
These solutions are written by the members of CSSMA.
Co-op Round
1. What is the positive difference between the largest two-digit prime and the least three-digit prime?
Solution: The largest two digit prime is 97. The least three digit prime is 101. The positive
difference between the two primes is 4 .
2. If 5 students have 15 problems to do in 36 minutes, then, assuming the students divide the work
evenly, how many hours does each student have, on average, per problem? Express your answer
as a decimal rounded to the nearest tenth.
Solution: If the 15 problems are evenly divided among 5 students, then each student gets 3
problems each. Each of those students would then have 36 minutes to do 3 problems, which is
also the same as one problem per 12 minutes. 12 minutes is 15 of 60 minutes, which is also one
hour. This means each student has 51 or 0.2 hours to work on one problem.
3. A random perfect square, between 1 and 10000 (inclusive) is chosen. What is the probability that
the last digit is a 6? Express your answer as a common fraction.
Solution: 1 is 12 and 10000 is 1002 . This tells us that there are 100 squares between 1 and
10000, which are 12 , 22 , 32 , etc. For a number to have a units digit of 6 when squared, then
the square root must have an units digit of either 4 or 6. Only those digits work. We can start
counting how many squares have square roots with a units digit of 6 or 4.
For the squares with a root with units digit 6, 62 is the smallest, and 962 is the largest, there are
10 squares total. For the squares with roots with units digits of 4, 42 is the smallest and 942 is
the largest, resulting in 10 squares once again. Combining the number of the squares, we have
20 squares with a units digit of 6. There are 100 perfect squares to choose from. Then, the
1
20
chances of randomly selecting a square with a units digit of 6 is 100
or
5
4. One day, Albert suddenly realized that he will turn x years old in the year x2 . Given that he was
born in the 20th century, in what year was Albert born?
Solution: If Albert is x years old on x2 , this means that his birth year is x2 − x. For his birth
year to be in the 20th century, this means that the earliest year he can be born in the 20th century
is 1900 and the lastest year he can be born in is 1999. This means that his birth year x2 − x must
fall between these two years.
1900 ≤ x2 − x ≤ 1999
x would also have to be an integer since it is the age of Albert. Using Trial and Error, we find
the only integer x that can satisfy the inequality is 45. This means Paul will 45 years old in this
problem. If Paul will be 45 years old on the year 452 , which is 2025, it means his birth year is
2025 − 45, the year 1980 .
These solutions are written by the members of CSSMA.
5. What is the units digit of 32017 ?
Solution:If you were to write out the units digit of the powers of 3, you would see the following
pattern:
The units digit of the powers of three repeats in a cycle of 3,9,7, and 1, where every power of
three that has an exponent divisible by 4, has a units digit of 1. 2016 is a number divisible by 4,
so the units digit of the 32016 is 1. According to the cycle of the units digit from the chart above,
this means the units digit of 32017 is 3 .
6. Let f (x) be the exponent of the largest power of two that is a factor of x. For example, f (24)
would equal 3, since 23 = 8 is a factor of 24, but 24 = 16 is not. If n is a randomly chosen
positive integer, then what is the probability that f (n) is odd?
Solution: It is clear that f (k) = 0 for all odd values k. We see that f (k · 2m ) = m for all odd k
and positive integer m. From there it follows that f (k · 2m ) is odd when m is odd. Therefore,
all numbers n for which f (n) is odd can be written in the form k · 21 , k · 23 , k · 25 , and so on.
Since exactly half the numbers on the natural number line is odd, 21 · 211 or 14 of them can be
1
1
written in the form k · 21 , 21 · 213 or 16
of them can be written in the form k · 23 , 12 · 215 or 64
of them
1
5
in the form k · 2 , and so on. This creates an infinite geometric series starting with 4 and and
a
with a common ratio of 14 . Using the formula S = 1−r
, the forumula for the sum of an infinite
1
1
geometric series, we get S = 1−4 1 , which simplifies to .
4
3
7. A right isoceles triangle is inscribed in a circle with radius r. What is the area of this triangle, in
terms of r?
Solution: When a right triangle is inscribed inside a circle, the hypotenuse is always the diameter
of the circle. This means the hypotenuse of the right triangle is 2r. Since it is isosceles, the two
legs of the triangle must be equal, which we’ll call x. Using the Pythagorean Theorem, we get
that x2 + x2 = (2r)2 . Solving this equation gives:
2x2 = 4r2
x2 = 2r2
√
x=r 2
These solutions are written by the members of CSSMA.
However, the question asks for the area of the triangle. We know that for a right triangle, the
base and height is just the length of the two legs of the triangle, so:
√
(r 2)2
Area =
2
2r2
=
2
= r2
8. The moon is 400, 000km away from Earth, and has a radius of 1750km. If the Sun is 160, 000,
000km away from the Earth, but appears the same size as the Moon, viewed from Earth, then
what is the radius of the sun, in kilometres?
radius of moon
Solution: Because apparent size is inversely proportional to distance, we get that distance
of moon =
radius of sun
radius of sun
1750
distance of sun . This gives the equation 400,000 = 160,000,000 . Solving the equation for the radius
= 700, 000km .
of the sun gives radius of sun = 1750·160,000,000
400,000
9. What is the greatest integer, less than 1000, that is a palindrome when expressed in binary?
Express your answer in decimal. (Note: a palindrome is a number that is the same when read
forwards or backwards.)
Solution: We want the binary number to have the maximum number of place values, so we look
for the greatest power of 2 less than 1000. This happens to be 512, or 29 , which has 10 digits
in its binary representation. 1000000001 is a palindrome, but definitely not the greatest one we
can find. We can create a larger one by replacing the second and second to last 0 with 1, giving
us 1100000011, which is 771. Keep replacing the leftmost and rightmost 0 with a 1 until you
reach a number just under 1000.
1110000111 = 903
1111001111 = 975
1111111111= 1023
The last one exceeds 1000, so the largest binary palindrome less than 1000 is 1111001111, or
975 . It should be noted that we started replacing the 0’s closest to the outside in order to have
the leftmost digits be 1, as they are far more significant.
These solutions are written by the members of CSSMA.
10. Point A is 7 units away from point X, and point B is 5 units away from point Y . Lines AX and
BY are both perpendicular to line XY , which is a canal with length 16. Max needs to travel
from his school at point A, down to any point on the canal to fill up his bucket with water, and
then back up to his house on point B. What is the total length of the shortest such trip possible,
in units?
Solution: Assume Max’s school is on the opposite side of the canal as his house. We then
get that the shortest distance from his house to the school is a straight line path. The horizonal
distance from his house to the school is is length of the canal, which is 16, and the vertical
distance is 7 + 5 = √
12. Using the Pythagorean Theorem, we can find the distance from his house
to the school to be 162 + 122 which simplifies to 20 . Note that if we had assumed his house
was on the same side as his school, the answer would remain unchanged since the bottom part
of the path would just be reflected across XY .
11. AoPS-City has a new row of 5 blocks marked for new development. A house, apartment, and
hotel each take up 1 block of land, while a mansion and a school each take up 2 blocks. How
many ways are there to allocate these 5 blocks if these 5 buildings are the only options?
Solution: We’ll call the house, apartment, and hotel "small buildings" or S and the mansion
and school "large buildings" or L. To fill up the 5 blocks of land, we can either have 2L and 1S,
1L and 3S, or 5S. We’ll look at each case separately.
Case 1: 2L and 1S
We have 3 buildings in total in this case, of which L appears twice, and so there are 3!
2! = 3 ways
in which the buildings can be ordered based on size. There are 3 options for the S and 2 options
for each of the two L, giving us 3 · 22 = 12 options for choosing the 3 buildings. This gives
3 · 12 = 36 ways of allocating the 5 blocks.
These solutions are written by the members of CSSMA.
Case 2: 1L and 3S
We have 4 buildings in this case, of which S appears three times. There are 4!
3! = 4 ways of
arranging the buildings based on size. There are 2 options for the large L and 3 options for each
of the 3 S, giving us 2 · 33 = 54 ways of choosing the 4 buildings. This gives 4 · 54 = 216 ways
of allocating the 5 blocks.
Case 3: 5S
We have 5 buildings in this case, of which S appears five times. There is only one way
of arranging the buildings based on size. There are 3 options for each of the 5 S, giving us
35 = 243 ways of choosing the 5 buildings. In total, there are 1·125 = 243 ways of allocating the
5 buildings. Summing up the total number of ways from each case, we get 36+216+243 = 495
ways of allocating the 5 blocks.
12. For what base k is the equation 202k = 10010 true?
Solution: 202k can be written in its expanded form as 2k 2 + 0k 1 + 2k 0 , or simply 2k 2 + 2. This
gives the equation 2k 2 + 2 = 100. Solving the equation for the positive root gives:
2k 2 = 98
k 2 = 49
k=7
13. How many positive factors does the number 2016 have?
Solution:2016’s prime factorization is 25 × 32 × 71 . We can use these factors to construct every
factor of 2016, and thus we can count the number of constructions.
To make a factor, we can include 20 ,21 ... all the way up to 25 . The gives us 6 possible powers of 2
to include in the factor construction. Then, we can use 30 , 31 , or 32 as a next step in constructing
the prime. This has 3 powers of three to choose from. For the last part of factor construction,
there are only 70 and 71 to choose from. Choosing one power from each group of powers of the
prime factors will construct every factor of 2016, from as low as 20 × 30 × 70 , which is 1, to as
high as 25 × 32 × 71 , which is 2016. Since there are 6 powers of 2, 3 powers of 3, and 2 powers
of 7 to choose from to make a factor, then you can make 6 × 3 × 2 or 36 factors. This means
that 2016 has 36 factors in total.
14. How many positive integers, less than 2016, have exactly 3 positive factors?
Solution: Let x2 be any integer under 2016 that has three factors. The three factors of x2
must always be 1, x, and x2 . However, if x is composite, then we would have more than three
factors, the additional factors being the factors of x. Therefore, x must be a prime. We see that
452 > 2016, and so x < 45. There are exactly 14 primes less than 45, namely 2, 3, 5, 7, 11, 13,
17, 19, 23, 29, 31, 37, 41, and 43. Each prime can be squared to get an integer less than 2016
that have exactly 3 factors: 1, the prime, and the square of the prime.
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15. A chicken wants to cross a road (don’t ask why). The chicken, which is on one end of the road,
is on a grid at the point (0, 0). The other end of the road is (5, 0). Every minute, the chicken
randomly chooses to move up, down, right, or left 1 unit, and all four options are equally likely.
What is the probability that after exactly 9 minutes, the chicken will be at (5, 0)? Express your
answer as a common fraction.
Solution: In 9 minutes, the chicken can make 9 directional movements. Let U be the direction
of up, D be the direction of down, L be the direction of left, and R be the direction of right.
There are 4 possible directions the chicken can travel per minute, so there are 49 or 262144 total
possible paths the chicken can take in 9 minutes. For it to reach (5,0) in 9 minutes, at least 5 of
the 9 directions it chooses must be going to the right. The other 4 directions must be in pairs of
opposite directions to that cancel each other out, so that only 5 of the movements to the right
will matter. There are 3 possible pairs for the remaining 4 moves:
Case 1: 2U and 2D
In this case, our total set of moves is 5R, 2U , and 2D. There are 9 moves in total, of which R
is to be repeated 5 times, U twice, and D twice. As the chicken can make these moves in any
9!
= 756 possible arrangement of moves that will take the chicken to (5, 0).
order, we get 5!2!2!
Case 2: 1U , 1D, 1R, and 1L
In this case, our total set of moves is 6R, 1L, 1U , and 1D. Again, the chicken must move 9
times, this time repeating R 6 times, we get 9!
6! = 504 possible arrangement of moves.
Case 1: 2R and 2L
In the last case, our total set of moves is 7R and 2L. As before, evaluating the number of ways to
9!
arrange this set of moves is 7!2!
= 36 ways. Adding up all three cases, there 756+504+36 = 1296
ways the chicken could get from (0, 0) to (5, 0) in 9 moves, our of the 262144 total paths the
81
1296
.
chicken could have taken. Therefore, the probability is 262144
which simplifies to
16384
These solutions are written by the members of CSSMA.