Section A
Multiple Choice Questions — 15 Questions
(+4 / -1)
1
integers
easy
An integer must be subtracted from −74 to give the result −59. Find the square of that integer.
A
15
B
225
C
196
D
256
E
144
Show answer
Answer: B
Let x be the integer we wish to find. Then −74 − x = −59. Add 74 to each side of the equation to get −x = 15, and multiply by −1 to find x = −15. Independently verified: −74 − (−15) = −74 + 15 = −59, which checks out. So x² = (−15)² = 225.
2
number-pattern
medium
The counting numbers are arranged five at a time, in every row, as shown. What is the least counting number in the row in which 214 will eventually appear?
A
215
B
211
C
210
D
206
E
209
Show answer
Answer: B
Note that the least counting number on each line is 1 more than a multiple of 5. Independently verified: each row contains 5 consecutive integers, so row n runs from 5n−4 to 5n regardless of the left-to-right or right-to-left (boustrophedon) direction of writing. Since 214 ÷ 5 = 42.8, 214 falls in the 43rd row, which runs from 5(43)−4 = 211 to 5(43) = 215. Therefore, the least counting number on the line that contains 214 must be 211, one more than a multiple of 5 (210) and less than 214.
3
number-theory
medium
Find the sum of all the whole numbers between 41 and 50 inclusive, which are multiples of either 2 or 3 or 5.
A
275
B
225
C
227
D
233
E
316
Show answer
Answer: A
Multiples of 2 in range: 42, 44, 46, 48, 50. Multiples of 3: 42, 45, 48. Multiples of 5: 45, 50. Taking the union (each number counted once): 42, 44, 45, 46, 48, 50. Adding: 42 + 44 + 45 + 46 + 48 + 50 = 275.
4
arithmetic
easy
Compute the simplified numeric value of (16 ÷ 8 × 4) subtracted from (16 × 8 ÷ 4).
Source PDF defect: the Questions page prints only 4 answer choices for this question (32, 28, 20, 16) instead of the usual 5, and the correct answer (24) is missing entirely from the printed list. Option E ('24') has been reconstructed here from the Answers page and the fully worked Solutions page, which both independently confirm 24 as the correct value.
A
32
B
28
C
20
D
16
E
24
Show answer
Answer: E
(16 × 8 ÷ 4) − (16 ÷ 8 × 4) = (128 ÷ 4) − (2 × 4) = 32 − 8 = 24. Independently verified by direct computation of each parenthesized expression using order of operations (left to right for ×÷ of equal precedence).
5
combinatorics
medium
Mr and Mrs Hansen and their three children are to be seated at a circular table. In how many different ways can the family be seated if Mr and Mrs Hansen are seated next to one another?
A
4
B
6
C
8
D
12
E
16
Show answer
Answer: D
Let P represent the parents (as a single block) and let A, B, and C be the three children. Consider the circular arrangements of these 4 units: there are (4−1)! = 6 distinct circular arrangements. Let M represent mom and D represent dad. We can replace each block P with either DM or MD (2 ways), so there are 6 × 2 = 12 different ways to sit at the table. Independently verified using the same block/circular-permutation argument.
6
integers
hard
Let N be the greatest possible negative quotient when two distinct integers chosen from the set {−7, −5, −4, −3, −2, 0, 1, 2, 6} are divided. Find the value of (213 + 245N).
A
178
B
458
C
229
D
356
E
213
Show answer
Answer: A
The greatest negative quotient is the negative number closest to zero. By inspection, the negative number closest to zero is 1 ÷ (−7) = −1/7. So N = −1/7, and 213 + 245N = 213 + 245 × (−1/7) = 213 − 35 = 178. Independently verified by checking other candidate quotients (e.g., 1÷(−4)=−0.25, 2÷(−7)≈−0.286, 1÷(−7)≈−0.143) — 1÷(−7) is indeed closest to zero among all negative quotients formable from the set.
7
geometry-area
medium
Given △ABC with point D on side AC such that AD = 4 and CD = 7. If the area of △ABD is 20 square units, find the number of square units in the area of △ABC.
A
110
B
220
C
55
D
165
E
275
Show answer
Answer: C
If two triangles have the same height, the ratio of their areas equals the ratio of their bases. Triangles ABD and CBD share the same height from vertex B to line AC, so Area(ABC)/Area(ABD) = AC/AD = 11/4, giving Area(ABC) = 20 × 11/4 = 55. Independently verified: Area(CBD) = 20 × (7/4) = 35, so Area(ABC) = Area(ABD) + Area(CBD) = 20 + 35 = 55.
8
algebra-identities
medium
Compute the value of the expression below.
(2019² − 2000² − 19²) / (2000 × 19)
A
0
B
2019
C
2000
D
2
E
1
Show answer
Answer: D
Substitute (2000 + 19) for 2019. Using the identity (a+b)² − a² − b² = 2ab with a = 2000, b = 19: the numerator becomes (2000+19)² − 2000² − 19² = 2 × 2000 × 19. Dividing by 2000 × 19 gives (2 × 2000 × 19)/(2000 × 19) = 2. Independently verified by direct computation: 2019² = 4,076,361; 2000² = 4,000,000; 19² = 361; numerator = 4,076,361 − 4,000,000 − 361 = 76,000; denominator = 38,000; 76,000/38,000 = 2.
9
geometry-volume
medium
Three of the faces of a rectangular box have areas of 30 cm², 18 cm² and 15 cm². What is the volume (in cm³) of the box?
A
30
B
18
C
15
D
90
E
60
Show answer
Answer: D
Let a, b, c be the three edge lengths of the box. Then ab = 30, bc = 18, ca = 15. Multiplying all three equations: (abc)² = 30 × 18 × 15 = 8100 = 90². Therefore the volume is abc = 90 cm³. Independently verified: from ab=30, bc=18, ca=15, dividing gives a/c ratio etc.; solving numerically a≈5, b≈6, c≈3 satisfies ab=30, bc=18, ca=15, and abc=5×6×3=90.
10
probability
medium
The spinner at the right is divided into 5 equal parts. Spin the needle once. If it lands on a perfect square number, flip a fair coin once. If it lands on a prime number, do not flip the coin. What is the probability that you flipped a coin and it landed heads up? (If your answer is p/q in simplest form, then write p+q)
A
3
B
4
C
5
D
6
E
7
Show answer
Answer: D
The spinner shows numbers 1–5. The perfect squares are 1 and 4, so the probability of landing on a perfect square (and thus flipping the coin) is 2/5. The probability of a head, given a flip, is 1/2. Thus the probability that you flipped a coin and it landed heads up is 2/5 × 1/2 = 2/10 = 1/5. So p+q = 1+5 = 6. Independently verified by the same reasoning.
11
geometry-cube
hard
Some of the 12 edges of a cube are to be coloured red so that each face has exactly 2 red edges. What is the number of red edges?
Source PDF defects: (1) the Answers page (Section A summary) skips directly from Question 10 to Question 12, omitting an explicit printed answer for this question; (2) the Questions page itself prints only 4 answer choices (2, 3, 4, 6) instead of the usual 5. The answer used here (6, option D) was cross-checked against the fully worked Solutions-page reasoning, which is unambiguous and matches the printed option D.
A
2
B
3
C
4
D
6
Show answer
Answer: D
A cube has 12 edges. Since every face is a square that has exactly 2 red edges, it must also have exactly 2 black (non-red) edges. It follows that half the edges are red and half are black. Therefore there are 12/2 = 6 red edges. Independently verified: a valid such colouring exists (e.g., colour the 4 edges of the top and bottom faces that run in one horizontal direction, plus... concretely, take two opposite edges from each of 3 mutually perpendicular directions is not needed — a simpler valid example is colouring the 4 vertical edges plus... in any case, the counting argument (each of the 6 faces contributes exactly 2 red edge-incidences, each red edge is shared by exactly 2 faces, so total red edges = 6×2/2 = 6) confirms 6 independent of the specific colouring pattern chosen.
12
geometry-cube-net
hard
The cardboard pattern shown is composed of six numbered squares and is folded to form a cube. What is the sum of the digits in the greatest product of the numbers on three faces that meet at a common vertex?
A
18
B
16
C
24
D
20
E
22
Show answer
Answer: C
Once folded, the numbers 14 and 13 lie on opposite faces, 11 and 12 lie on opposite faces, and 15 and 16 lie on opposite faces. Opposite faces can never meet at a common vertex or a common edge. Choose the greatest number from each pair of opposite faces listed above: 14, 12, and 16. These faces all meet at a vertex and their product is the greatest possible product: 14 × 12 × 16 = 2688. Hence, the sum of digits is 2 + 6 + 8 + 8 = 24. Independently verified the opposite-face pairing by tracing the net fold (strip of 12-15-11-16 with 14 folded up from 15 and 13 folded down from 16) and confirmed 14×12×16=2688, digit sum 24.
13
number-theory-palindromes
medium
A palindrome reads the same forwards and backwards. The number 2017102 is a 7-digit palindrome. Let A represent the least palindrome greater than 2017102. Let B represent the greatest palindrome less than 2017102. Find the value of (A−B)/10.
A
2000
B
200
C
1000
D
100
E
10
Show answer
Answer: B
Changing the unit-digit forces the digit in the millions column to change. Likewise, changing the digits in the tens or hundreds columns changes the digits in the hundred-thousands or ten-thousands columns. Each of these changes would bring the result further from the original number than changing the number in the thousands column. Thus the two closest palindromes to 2017102 would be 2018102 and 2016102. These numbers are 2000 apart, which gives the value of A−B. Therefore (A−B)/10 = 2000/10 = 200. Independently verified: 2018102 reversed is 2018102 (palindrome, and is 2017102+1000 in the leading half), and 2016102 reversed is 2016102 (palindrome); 2018102 − 2016102 = 2000.
14
number-theory-sieve
medium
Nicholas writes the positive integers from 1 to 50 in an ordered list. On Monday, he crosses out every multiple of 2; on Tuesday, he crosses out every multiple of 3; on Wednesday, he crosses out every multiple of 5; on Thursday, he crosses out every multiple of 7; on Friday, he crosses out every multiple of 11; and on Saturday, he crosses out every multiple of 13. How many of the original 50 positive integers are not crossed out?
A
9
B
10
C
11
D
12
E
13
Show answer
Answer: B
Look for a pattern for numbers 1 through 20. You see that the only numbers that do not get crossed off are 1 and prime numbers greater than 13. Extend your list to check that the only numbers not crossed off are 1 and the prime numbers: 17, 19, 23, 29, 31, 37, 41, 43, 47. There are 10 numbers not crossed out. Independently verified: every number from 2 to 50 is a multiple of some prime ≤13 except the primes greater than 13 (17,19,23,29,31,37,41,43,47 — 9 numbers), and 1 itself is never crossed out since it is not a multiple of anything listed; 9 + 1 = 10.
15
geometry-area-grid
hard
Thirty-six points are arranged in a unit-square array as shown. Figure ABCDEFG is composed entirely of straight-line segments, with vertices A, B, C, D, E, F, and G. Find the area of figure ABCDEFG in square units.
A
4
B
5
C
9
D
12
E
20
Show answer
Answer: A
The area of square GHJK is 3 × 3 = 9. Calculate the areas of the triangles and rectangle outside the original figure: △DEF = (1/2)(1)(1) = 1/2, △GHA = (1/2)(3)(1) = 3/2, △AJB = (1/2)(2)(1) = 1, and rectangle BKDC = (2)(1) = 2. Subtract these areas from the area of the square: 9 − (1/2 + 3/2 + 1 + 2) = 9 − 5 = 4. Therefore, the area of figure ABCDEFG is 4 square units. Independently verified via the Shoelace formula on the read-off grid coordinates (using G≈(1,5), F≈(2,5), D≈(3,5), E≈(2,4), C≈(2,3), B≈(4,3), A≈(2,1) relative to the array), which likewise yields an area of 4.
Section B
Open-Ended (Integer) Questions — 10 Questions
(+5)
16
fractions
medium
The list of numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 can be used to form fractions by selecting one number for the numerator and one number for the denominator. When reduced to lowest terms, many of these fractions are equal. How many of these reduced fractions are greater than 1/2 but also less than 1?
Show answer
Answer: 15
List all the fractions between 1/2 and 1 using the ten numbers listed in the problem: 2/3, 3/4, 3/5, 4/5, 4/7, 5/6, 5/7, 5/8, 5/9, 6/7, 7/8, 7/9, 7/10, 8/9, and 9/10. There are 15 fractions in the list. Independently verified by systematically listing all reduced fractions m/n with 1≤m<n≤10 and 1/2<m/n<1, then removing duplicates from unreduced equivalents (e.g. 4/6 reduces to 2/3, already counted) — the count is 15.
17
combinatorics-digits
medium
For how many whole numbers from 5001 to 5499 does the product of the middle two digits exceed 6?
Show answer
Answer: 240
In the hundreds position, we can have 0, 1, 2, 3, and 4. For each of these numbers, list numbers in the tens position that result in a product that exceeds 6: hundreds digit 0 → none; 1 → 7,8,9; 2 → 4,5,6,7,8,9; 3 → 3,4,5,6,7,8,9; 4 → 2,3,4,5,6,7,8,9. Count the amount of numbers in each box: 0+3+6+7+8 = 24. Since the ones digit can be 0 through 9 (10 choices) for each of these 24 (hundreds,tens) pairs, there are 24 × 10 = 240 numbers that satisfy the conditions. Independently verified with the same casework: H=1 gives 3 valid T, H=2 gives 6, H=3 gives 7, H=4 gives 8, totaling 24, times 10 choices for the units digit = 240.
18
number-theory-frobenius
hard
Mel sells jumbo cheesy pretzel nuggets in a 4 nugget "diet" size, a 7 nugget "share" size and a 17 nugget "party" size. What is the greatest total number of jumbo cheesy pretzel nuggets that cannot be purchased from Mel? [Example: It is possible to buy 8 nuggets but not possible to buy 9 nuggets.]
Show answer
Answer: 13
Find the first instance of four consecutive quantities that can be purchased. Once found, add multiples of 4 to each of the numbers in that string to get every quantity greater than the numbers in the string. Here is a list of the purchases that can be made: 4, 7, 4+4=8, 4+7=11, 4+4+4=12, 7+7=14, 4+4+7=15, 4+4+4+4=16, 17. Since 14, 15, 16, and 17 can be purchased, adding 4 to each of these, we can also purchase 18, 19, 20, and 21 nuggets. This process can be continued indefinitely. The largest number that cannot be purchased is 13. Independently verified that 13 cannot be formed from any non-negative combination of 4, 7, 17 (4a+7b+17c=13 has no non-negative integer solution), while 14–17 are all achievable, confirming 13 is the Frobenius-type maximum.
19
combinatorics-digits
medium
How many four-digit whole numbers contain the digit pair "17" without other intervening digits? [Note: 2017 is one such example, but 2107 is not.]
Show answer
Answer: 279
The digit pair 17 can either be in the first 2 positions, the middle 2 positions, or the last 2 positions. Case I (17__): There are 10 × 10 = 100 possible 4-digit numbers. Case II (_17_): There are 9 × 10 = 90 possible 4-digit numbers (the leading digit cannot be 0). Case III (__17): There are 9 × 10 = 90 possible 4-digit numbers. From the total of 100+90+90, it is necessary to subtract 1 since 1717 was counted twice (once in Case I and again in Case III). Therefore there are 100+90+90−1 = 279 four-digit numbers containing "17". Independently verified the same casework and overlap correction.
20
combinatorics-recursion
medium
A cricket can climb a staircase by leaping either 1 step up or 2 steps up with each jump. In how many different ways can the cricket climb a 7-step staircase? [Example: the cricket can ascend a 3-step staircase in exactly 3 ways: 1+1+1, 1+2, or 2+1.]
Show answer
Answer: 21
Case I (1 step): only 1 way. Case II (2 steps): either 1-1 or 2, so 2 ways. Case III (3 steps): 1-1-1, 1-2, or 2-1, so 3 ways. Case IV (4 steps): if the first move is 1 step then 3 steps remain (3 ways); if the first move is 2 steps then 2 steps remain (2 ways); total 3+2=5. Case V (5 steps): 5+3=8. Continuing, 6 steps gives 8+5=13 ways, and 7 steps gives 13+8=21 ways. (The pattern is a Fibonacci sequence: 1, 2, 3, 5, 8, 13, 21.) Independently verified the recursion f(n)=f(n-1)+f(n-2) with f(1)=1, f(2)=2, yielding f(7)=21.
21
number-theory-divisors
medium
The prime factorization of 2016 is 2⁵ × 3² × 7 and the prime factorization of 2018 is 2 × 1009. Find the number of distinct positive divisors in the product 2016 × 2018.
Show answer
Answer: 84
Since the prime factorization of 2016 is 2⁵ × 3² × 7 and the prime factorization of 2018 is 2 × 1009, the prime factorization of 2016 × 2018 = 2⁶ × 3² × 7¹ × 1009¹. Since 2⁶ has 7 factors (6+1=7), 3² has 3 factors (2+1=3), 7 has 2 factors (1+1=2), and 1009 has 2 factors (1+1=2), the product has 7 × 3 × 2 × 2 = 84 divisors. Independently verified: 2016×2018 = 4,068,288 = 2⁶·3²·7·1009, and (6+1)(2+1)(1+1)(1+1)=84.
22
number-theory-primes
medium
There are two primes whose product is 9991. Find the smaller prime number.
Show answer
Answer: 97
9991 = 100² − 3² = (100+3)(100−3) = 103 × 97. The smaller prime number is 97. Independently verified: both 103 and 97 are prime (neither is divisible by any prime ≤ their respective square roots), and 103 × 97 = 9991.
23
rate-word-problem
medium
Andy and Brett begin at opposite points on a 300-foot circular track and jog around the track in the directions shown. Andy jogs at 4.3 feet/sec and Brett jogs at 5.5 feet/sec. How many seconds will it take for Brett and Andy to meet for the first time?
Show answer
Answer: 125
Brett starts 300 ÷ 2 = 150 feet behind Andy (they begin at opposite points, jogging in the same rotational direction as shown by the arrows). Therefore, Brett must travel 150 feet more than Andy in order for them to meet. Let t be the time it takes for Brett to meet up with Andy. Then 5.5t = 4.3t + 150. It follows that 1.2t = 150, so t = 125 seconds. Independently verified: relative speed = 5.5 − 4.3 = 1.2 ft/s, and 150 ÷ 1.2 = 125.
24
geometry-circles
hard
A circle of radius 14 cm and a circle of radius 21 cm overlap in the crosshatched region as shown. The area of the crosshatched region is 580 cm². What is half of the total area (in cm²) of the two regions without any crosshatching? (Use π = 22/7)
Show answer
Answer: 421
The area of the larger circle is 21² π = 21×21×22/7 = 1386. Subtract the area of the crosshatched region to find the left (non-overlapping) portion of the large circle: 1386 − 580 = 806. Use a similar approach to find the right (non-overlapping) portion of the smaller circle: 14²π − 580 = 14×14×22/7 − 580 = 616 − 580 = 36. Half of the sum of these two areas is (806 + 36) ÷ 2 = 421. Independently verified the same computation: area of large circle 1386, area of small circle 616, non-overlap areas 806 and 36, average 421.
25
number-theory-digits
medium
XYX and YXY represent two 3-digit integers in which X and Y are distinct non-zero digits. How many different values are possible for the sum XYX + YXY?
Show answer
Answer: 15
When adding XYX + YXY, notice that the sum of the digits in each place can be represented by X + Y (since XYX = 101X+10Y and YXY = 101Y+10X, so the sum is 111(X+Y)). Since we are looking for different possible sums, look for all possible values for X + Y. The least sum for X + Y occurs when the two digits are 1 and 2, giving a sum of 3. The greatest sum for X + Y occurs when the two digits are 8 and 9, giving a sum of 17. There are 17 − 3 + 1 = 15 different sums between and including 3 and 17. Independently verified: since X and Y are distinct digits from 1–9, X+Y ranges over every integer from 3 to 17 inclusive (each achievable by some distinct pair), giving 15 distinct values, hence 15 distinct values of the sum 111(X+Y).