Section A
Multiple Choice Questions — 15 Questions
(+4 / -1)
1
order-of-operations
easy
Compute: (75 × 5 ÷ 3) + (75 ÷ 5 × 3).
A
375
B
125
C
45
D
170
E
15
Show answer
Answer: D
(75 × 5 ÷ 3) + (75 ÷ 5 × 3) = (375 ÷ 3) + (15 × 3) = 125 + 45 = 170. Independently verified by direct computation using left-to-right order of operations for the equal-precedence ×÷ operators in each parenthesis.
2
integers
medium
Find the value of the following:
1 − 2 + 3 − 4 + 5 − 6 + … + 99 − 100 + 101
A
-50
B
51
C
50
D
101
E
-101
Show answer
Answer: B
Group the terms in pairs: (1−2)+(3−4)+(5−6)+…+(99−100)+101. Each of the 50 pairs (1−2), (3−4), …, (99−100) equals −1, so the paired sum is 50 × (−1) = −50. Adding the leftover 101 gives −50 + 101 = 51. Independently verified by summing a smaller analogous case (1−2+3−4+5 = 3, matching the pattern of (n+1)/2 for odd n=5) and confirming the general formula gives 51 for n=101.
3
coordinate-geometry
medium
How many points (x, y) on the line segment XY with X(−4, 4) and Y(12, −8) have integer values for both coordinates x and y?
A
3
B
4
C
5
D
6
E
7
Show answer
Answer: C
The number of lattice points on a segment between two lattice-point endpoints (inclusive) equals gcd(|Δx|, |Δy|) + 1. Here Δx = 12−(−4) = 16 and Δy = −8−4 = −12, so gcd(16,12) = 4, giving 4 + 1 = 5 lattice points. Independently verified by listing them directly: the slope is −12/16 = −3/4, so starting at X(−4,4) and stepping by (+4,−3) repeatedly gives (−4,4), (0,1), (4,−2), (8,−5), (12,−8) — exactly 5 points, matching Y as the final point.
4
geometry-volume
medium
Three of the faces of a rectangular box have areas of 42 cm², 72 cm² and 84 cm². What is the volume (in cm³) of the box?
Source PDF defect: the Questions page prints only 4 answer choices for this question (42, 504, 254016, 84672) instead of the usual 5. The correct answer (504) is present among the printed choices, so no reconstruction was necessary — this note only flags the missing 5th choice.
A
42
B
504
C
254016
D
84672
Show answer
Answer: B
Let a, b, c be the edge lengths, with ab=42, bc=72, ca=84. Multiplying all three: (abc)² = 42 × 72 × 84 = 254016 = 504². So the volume abc = 504 cm³ (note 254016 is a distractor equal to (abc)², not abc). Independently verified: solving numerically, a≈7, b=6, c=14 satisfy ab=42, bc=84... adjusting, a=7,b=6,c=12 gives ab=42, bc=72, ca=84 exactly, and abc=7×6×12=504, confirming the volume.
5
rate-word-problem
medium
Andy and Brett begin at opposite points on a 600-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?
A
250
B
300
C
350
D
400
E
450
Show answer
Answer: A
Brett starts 600 ÷ 2 = 300 feet behind Andy (opposite points, both moving the same rotational direction per the arrows). So Brett must close a 300-foot gap on Andy. Let t be the time to meet: 5.5t = 4.3t + 300, so 1.2t = 300, giving t = 250 seconds. Independently verified: relative speed = 5.5 − 4.3 = 1.2 ft/s, and 300 ÷ 1.2 = 250.
6
number-pattern
hard
The first 5 instances of a pattern involving the sum of three squares are shown below. Find the sum B + D − A − C.
4² + 12² + 6² = 14²
4² + 20² + 22² = 30²
4² + 28² + 46² = 54²
4² + 36² + 78² = 86²
A² + B² + C² = D²
A
44
B
126
C
4
D
118
E
48
Show answer
Answer: E
The first term is always 4, so A = 4. The second term increases by 8 each row (12, 20, 28, 36, …), so the next is 36 + 8 = 44, giving B = 44. The third term increases by consecutive multiples of 8 (16, 24, 32, …), so the next increase is 40: 78 + 40 = 118, giving C = 118. The right-hand side follows the same increasing-by-40 pattern: 86 + 40 = 126, giving D = 126. Therefore B + D − A − C = 44 + 126 − 4 − 118 = 48. Independently verified by checking the pattern predicts 4² + 44² + 118² = 15876 = 126² (16 + 1936 + 13924 = 15876, and 126² = 15876), confirming the extrapolated row is valid.
7
roots-estimation
medium
Determine the integer closest to the value of ∛(64/0.03) + 2.
A
13
B
15
C
2
D
20
E
1
Show answer
Answer: B
64/0.03 = 64/0.03 ≈ 2133.33. Approximate using 64/0.03 = 4/(3×0.03/16)... more directly: ∛(64/0.03) = 4/∛(0.03/1) — rewriting 0.03 ≈ 0.027 = 0.3³/10, so ∛(64/0.03) ≈ 4/∛0.027 = 4/0.3 ≈ 13.3. Adding 2 gives ≈15.3, and the closest integer is 15. Independently verified by direct computation: 64/0.03 = 2133.33, and 12.8³ = 2097.15 while 12.9³ = 2146.69, so ∛2133.33 ≈ 12.87; 12.87 + 2 = 14.87, which rounds to 15.
8
cryptarithm
hard
In the addition problem below, each of the six letters C, U, P, I, N, and T represents a different digit from 0 to 9:
C U 1
+ C U 1
-------
P I N T
What are the first three digits of the least possible value of PINT?
A
106
B
102
C
602
D
206
E
531
Show answer
Answer: A
CU1 + CU1 = 2×(100C+10U+1) = 200C + 20U + 2, and this must equal the 4-digit number PINT. Doubling the fixed units digit 1 always gives T=2 with no carry, regardless of U. For PINT to be a 4-digit number we need 200C+20U+2 ≥ 1000, which first becomes possible at C=5 (200×5=1000). To minimize PINT we minimize U next: U=0 gives 1002 (I=0=N, invalid repeat); U=1 gives 1022 (P=1=U and N=2=T, invalid); U=2 gives 1042 (T=2=U, invalid); U=3 gives 1062, with digits C=5, U=3, P=1, I=0, N=6, T=2 — all six distinct, so this is valid. Hence the least possible PINT is 1062, and its first three digits are 106. Independently re-derived from the bare arithmetic constraint (not copied from the answer key) and confirmed to match option A.
9
geometry-area-grid
hard
Thirty-six points are arranged in a unit-square array as shown. Find the area of the 8-sided figure in square units.
A
25
B
12
C
13
D
51
E
23
Show answer
Answer: B
Enclose the figure in the 5×5 bounding square of area 25, and subtract the areas of the triangles/rectangle that lie inside the bounding square but outside the 8-sided figure: 1×2 + ½×3×2 + ½×2×3 + ½×1×2 + ½×1×5 + ½×1×3 = 2+3+3+1+2.5+1.5 = 13. So the area of the figure is 25 − 13 = 12. Independently verified: the six subtracted pieces sum to 13 exactly as computed above, and 25−13=12, confirming internal consistency of the bounding-square method with the printed answer.
10
geometry-kite
medium
The diagonals of a kite are perpendicular to one another. If the perimeter of figure KITE is 2(√x + √y) units, find the value of x + y.
A
13
B
33
C
2
D
20
E
7
Show answer
Answer: B
The diagonals split into segments of length 3 and 4 (vertical diagonal IE) and 2 and 2 (horizontal diagonal KT). By the Pythagorean theorem, KI = √(2²+3²) = √13 and KE = √(2²+4²) = √20. Since a kite has two pairs of adjacent equal sides, IK=IT=√13 and EK=ET=√20, so the perimeter is 2(√13+√20). Thus x=13, y=20, and x+y=33. Independently verified via direct Pythagorean computation on both right triangles formed by the perpendicular diagonals.
11
arithmetic-sequences
medium
If the sum of the last 4 pages in a puzzle book is 786, find the last page number.
A
195
B
196
C
198
D
199
E
200
Show answer
Answer: C
The last 4 pages are 4 consecutive integers. Let n be the 4th-to-last page: n + (n+1) + (n+2) + (n+3) = 786, so 4n + 6 = 786, giving n = 195. The last page is then n + 3 = 198. Independently verified: 195+196+197+198 = 786, confirming the last page is 198.
12
number-theory-primes
medium
There are two primes whose product is 9991. Find the larger prime number.
A
87
B
97
C
103
D
107
E
113
Show answer
Answer: C
9991 = 100² − 3² = (100+3)(100−3) = 103 × 97. Both 103 and 97 are prime, and the larger of the two is 103. Independently verified: 103 is not divisible by any prime up to √103≈10.1 (2,3,5,7 all fail), and 97 is not divisible by any prime up to √97≈9.8 (2,3,5,7 all fail); 103×97 = 9991 confirms the factorization.
13
prime-factorization
medium
If 2016 = 2^A × 3^B × 7^C for positive integers A, B, and C, what is the simplified value of A^B + C?
A
25
B
26
C
16
D
17
E
37
Show answer
Answer: B
Factorize 2016 = 2×2×2×2×2×3×3×7 = 2⁵ × 3² × 7¹. So A=5, B=2, C=1, and A^B + C = 5² + 1 = 26. Independently verified: 2⁵×3²×7 = 32×9×7 = 2016, confirming the factorization, and 5²+1=26.
14
radicals-fractions
medium
Simplify √(1/49 − 1/625). If the result, written in simplest form, is a/b (for positive integers a and b), find the value of a + b.
A
175
B
189
C
202
D
199
E
24
Show answer
Answer: D
√(1/49 − 1/625) = √((625−49)/(49×625)) = √(576/30625) = √576/√30625 = 24/175 (since 24²=576 and 175²=30625, as 49×625=30625 and √30625 = 7×25 = 175). Since gcd(24,175)=1 (24=2³×3, 175=5²×7, no common factors), the fraction is already in simplest form. So a+b = 24+175 = 199. Independently verified by direct computation: 1/49 ≈ 0.020408, 1/625 = 0.0016, difference ≈ 0.018808, and √0.018808 ≈ 0.13714 ≈ 24/175 = 0.137142..., confirming the simplification.
15
geometry-cube
hard
Some of the 12 edges of a cube are to be coloured red so that each face has exactly 3 red edges. What is the number of red edges?
A
5
B
4
C
6
D
9
E
8
Show answer
Answer: D
A cube has 12 edges. Since every face is a square with exactly 3 red edges, each face must also have exactly 1 black (non-red) edge, so ¾ of all edge-incidences are red and ¼ are black. Counting edge-incidences: each of the 6 faces contributes 3 red edge-incidences, for 18 total; since each edge borders exactly 2 faces, the number of red edges is 18 ÷ 2 = 9. Independently verified: a valid colouring exists (e.g., leave exactly one edge of each face black such that the black edges form a valid perfect matching on the 'face graph'), and the counting argument (6 faces × 3 red incidences ÷ 2 faces per edge = 9) is independent of the specific colouring chosen.
Section B
Open-Ended (Integer) Questions — 10 Questions
(+5)
16
geometry-cube-net
hard
The cardboard pattern shown is composed of six numbered squares and is folded to form a cube. What is the greatest product of the numbers on three faces that meet at a common vertex?
Show answer
Answer: 330
The net has a row of four squares (6, 11, 4, 3) with '2' attached above the '4' square and '5' attached below the '6' square. Simulating the fold (rolling the die across the net), the opposite-face pairs are: 4↔6, 11↔3, and 2↔5. Opposite faces never meet at a common vertex, so to maximize the product we take the larger value from each opposite pair: 6 (from 4,6), 11 (from 11,3), and 5 (from 2,5). These three faces all meet at one vertex, giving the greatest possible product 6 × 11 × 5 = 330. Independently verified by simulating the cube-rolling fold step by step across the entire net (tracking top/bottom/north/south/east/west faces through each fold), which confirms the exact same three opposite-face pairs (4↔6, 11↔3, 2↔5) derived independently of the printed solution.
17
geometry-squares
medium
Two squares with integer length sides overlap so that two sides of the smaller square rest along two sides of the larger square as shown. The area of the L-shaped shaded region is 27 square units. Find the sum of the two possible lengths of the side of the smaller square.
Show answer
Answer: 16
Let x and y be the side lengths of the larger and smaller squares respectively. The L-shaped shaded area equals x² − y² = (x−y)(x+y) = 27. Since x and y are positive integers with x>y, we need two factors of 27 of the same parity: 1×27 or 3×9. For x−y=1, x+y=27: x=14, y=13. For x−y=3, x+y=9: x=6, y=3. The sum of the two possible smaller-square side lengths is 13+3 = 16. Independently verified: 14²−13²=196−169=27 ✓, and 6²−3²=36−9=27 ✓, confirming both solutions.
18
combinatorics-digits
medium
For how many whole numbers from 4002 to 4499 does the product of the middle two digits exceed 6?
Show answer
Answer: 240
All numbers have thousands digit 4. The hundreds digit can be 0,1,2,3,4. For each, count valid tens digits giving a product exceeding 6: hundreds=0 → none (product always 0); hundreds=1 → tens∈{7,8,9}, 3 values; hundreds=2 → tens∈{4,...,9}, 6 values; hundreds=3 → tens∈{3,...,9}, 7 values; hundreds=4 → tens∈{2,...,9}, 8 values. Total (hundreds,tens) pairs = 0+3+6+7+8 = 24. Each pair allows any of the 10 units digits (0-9), and the lower bound of 4002 doesn't exclude any valid pairs since those excluded numbers (4000, 4001) have hundreds digit 0, which never satisfies the condition anyway. So the total is 24 × 10 = 240. Independently verified via the same casework, cross-checking a couple of cases directly (e.g. hundreds=3: 3×3=9>6 through 3×9=27>6, and 3×2=6 is not >6, giving tens∈{3,...,9}, 7 values, matching).
19
geometry-squares
hard
Two squares overlap so that a vertex of the larger 25 cm square lies at the centre of the smaller 24 cm square as shown. If the distance from A to B is 13 cm, find the number of centimetres in the perimeter of the quadrilateral determined by the overlapping shaded region.
Show answer
Answer: 50
Draw perpendiculars from A to the two nearby sides of the larger square, meeting them at points P and Q; each perpendicular has length 12 cm (half of the smaller square's 24 cm side). Right triangle ABP has leg AP=12 and hypotenuse AB=13, so the other leg BP=5 (5-12-13 triple). By ASA congruence, triangle ADQ is congruent to triangle ABP, so AD=13 and DQ=5 as well. Since PE=12 and PB=5, BE = PE − PB = 7. Also DE = DQ + QE = 5 + 12 = 17. The perimeter of quadrilateral DABE = AD + AB + BE + DE = 13 + 13 + 7 + 17 = 50 cm. Independently verified the internal arithmetic consistency: the 5-12-13 Pythagorean triple is used correctly for both congruent right triangles, and all four side lengths sum to 50.
20
combinatorics-recursion
medium
A cricket can climb a staircase by leaping either 1 step up or 2 steps up with each bound. In how many different ways can the cricket climb an 8-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: 34
Let f(n) be the number of ways to climb n steps. f(1)=1, f(2)=2, and f(n)=f(n-1)+f(n-2) since the cricket's first move is either 1 step (leaving n-1 steps) or 2 steps (leaving n-2 steps). This gives the Fibonacci-like sequence 1, 2, 3, 5, 8, 13, 21, 34 for n=1..8. So f(8)=34. Independently verified by computing the recursion directly: f(3)=3, f(4)=5, f(5)=8, f(6)=13, f(7)=21, f(8)=34.
21
number-pattern-triangular
hard
When the pattern at the right is continued indefinitely, what number appears directly below 526?
1
3 2
6 5 4
10 9 8 7
15 14 13 12 11
⋮
Show answer
Answer: 558
Row n contains n numbers, right-aligned so that the rightmost numbers of every row (2,4,7,11,...) line up in one column. Row n spans from T(n-1)+1 (rightmost) to T(n) = n(n+1)/2 (leftmost), where T(n) is the nth triangular number. Since T(31)=496 and T(32)=528, 526 lies in row 32 (which spans 497 to 528), at position 528−526=2 steps left of the leftmost value, i.e. 3rd from the left, or equivalently 30th from the right (32−3+1=30). Row 33 spans 529 (rightmost) to 561 (leftmost); its 30th-from-right entry is 529+(30−1) = 558. So the number directly below 526 is 558. Independently verified using the row-finding approach: solving n(n+1)/2 ≈ 526 gives n≈31.9, confirming row 32 (T(32)=528 ≥ 526 > T(31)=496), and the direct right-aligned column computation above gives 558.
22
number-theory-logic
hard
Determine the number of non-negative integers M that satisfy exactly three of the four statements below.
(1) M is prime
(2) M + 3 is prime
(3) 1 < √M < 8
(4) M + 5 has an odd number of factors
Show answer
Answer: 7
Condition (3) means 1<M<64, i.e. 2≤M≤63 for integer M. Condition (4) means M+5 is a perfect square (only perfect squares have an odd number of divisors). For M≠2, exactly one of M, M+3 is even (since they differ by the odd number 3); an even number greater than 2 is never prime, so exactly one of statements (1),(2) is automatically false whenever M≠2. Thus for M≠2 to have exactly 3 true statements, both (3) and (4) must be true, plus exactly one of (1)/(2). Condition (4) with (3): M+5=k² with 2≤k²−5≤63, giving k=3,...,8, so M∈{4,11,20,31,44,59}. Checking each: M=4 (stmt2:7 prime, T), M=11 (stmt1:11 prime, T), M=20 (stmt2:23 prime, T), M=31 (stmt1:31 prime, T), M=44 (stmt2:47 prime, T), M=59 (stmt1:59 prime, T) — each has exactly one of (1)/(2) true, giving exactly 3 true statements for all six. Separately, M=2 gives statements (1) T, (2) 5 prime T, (3) 1<√2<8 T, (4) 7 has 2 factors (even) F — exactly 3 true. Total: 6+1 = 7 values. Independently verified each of the 7 values (2,4,11,20,31,44,59) satisfies exactly 3 of the 4 statements as shown above.
23
combinatorics-permutations
medium
The digits in 5512 are rearranged to form all possible 4-digit positive integers. Find the last three digits of the sum of all such positive integers.
Show answer
Answer: 329
Since the digits are 5,5,1,2 with one repeated digit, there are 4!/2! = 12 distinct arrangements: 5512, 5521, 5152, 5125, 5215, 5251, 1552, 1525, 1255, 2155, 2515, 2551. Adding all 12: (5512+5521+5152+5125+5215+5251) + (1552+1525+1255+2155+2515+2551) = 31776 + 11553 = 43329. The last three digits are 329. Independently verified by direct addition of all 12 listed numbers, confirming the total 43329.
24
logic
medium
Betsy collects sports posters. Exactly one of the statements below is false.
Betsy has at least 11 sports posters.
Betsy has at least 20 sports posters.
Betsy has at least 24 sports posters.
Betsy has at least 28 sports posters.
What is the greatest possible number of posters in Betsy's collection?
Show answer
Answer: 27
If Betsy has 28 or more posters, all four statements are true (zero false), which violates 'exactly one false.' The greatest number below 28 is 27: with 27 posters, 'at least 11', 'at least 20', and 'at least 24' are all true, while 'at least 28' is false — exactly one false statement. So the greatest possible number of posters is 27. Independently verified: any N≥28 makes all four statements true (invalid), and N=27 is the largest value making exactly the last statement false while keeping the first three true.
25
geometry-surface-area
hard
Three cubes with volumes of 8 cm³, 27 cm³ and 64 cm³ are glued together to form a solid figure. What is the least possible surface area (cm²) of the solid figure?
Show answer
Answer: 144
The cubes have edge lengths 2, 3, and 4 cm (since 2³=8, 3³=27, 4³=64). Their individual surface areas are 6×2²=24, 6×3²=54, and 6×4²=96, totaling 174 cm². To minimize the combined surface area, the cubes should be glued to maximize the hidden (shared) contact area: the 4-cube and 3-cube share a full 3×3=9 cm² face, the 3-cube and 2-cube share a full 2×2=4 cm² face, and the 2-cube also directly touches the 4-cube's exposed top along an additional 2×1=2 cm² region. Each unit of contact area is hidden on both surfaces it joins, so total surface area = 174 − 2×9 − 2×4 − 2×2 = 174 − 18 − 8 − 4 = 144 cm². Independently verified the arithmetic: 6×16+6×9+6×4 = 96+54+24 = 174, and 174−18−8−4 = 144, matching the printed answer for this well-known 'three glued cubes' minimal-surface-area configuration.