Section A
Multiple Choice Questions — 15 Questions
(+4 / -1)
1
number-theory
easy
How many numbers of the series 1/1, 1/2, 1/3, … 1/98, 1/99, 1/100 have a terminating decimal expansion?
The source PDF prints no A/B/C/D/E letters for these options; labels here follow the exact on-page top-to-bottom order.
A
15
B
16
C
14
D
13
E
18
Show answer
Answer: A
A fraction 1/n (already in lowest terms since the numerator is 1) has a terminating decimal expansion exactly when n's only prime factors are 2 and/or 5. Counting such n from 1 to 100: 1 itself (1/1 = 1); powers of 2 — 2, 4, 8, 16, 32, 64 (6 numbers); powers of 5 — 5, 25 (2 numbers); products 2^a·5 — 10, 20, 40, 80 (4 numbers); products 2^a·25 — 50, 100 (2 numbers). Total = 1 + 6 + 2 + 4 + 2 = 15.
2
number-theory
easy
Solve: √(4³) − ∛(2⁶)
A
2
B
4
C
8
D
12
E
1
Show answer
Answer: B
√(4³) = √64 = 8. ∛(2⁶) = 2^(6/3) = 2² = 4. So the expression equals 8 − 4 = 4.
3
algebra
medium
Solve the following expression: (∛7 − 1) × (∛49 + ∛7 + 1)
A
48
B
50
C
6
D
8
E
7
Show answer
Answer: C
Let a = ∛7, so a³ = 7 and a² = ∛49. The expression becomes (a − 1)(a² + a + 1), which is the standard factorisation of a³ − 1. So the value is a³ − 1 = 7 − 1 = 6.
4
algebra
medium
How many positive integers N satisfy the following equation? 441 − 42N + N² = 64
The source PDF prints only 4 options for this question (no fifth option); labels here are A-D, following the exact on-page top-to-bottom order.
A
2
B
0
C
4
D
3
Show answer
Answer: A
Rearranging, N² − 42N + 441 = 64. The left side is a perfect square: N² − 2(21)N + 21² = (21 − N)², so (21 − N)² = 64, giving 21 − N = ±8. Thus N = 13 or N = 29 — both positive integers, so there are 2 values of N.
5
number-theory
medium
Find the sum of all the unique prime factors of 656656.
A
74
B
1044
C
553
D
473
E
687
Show answer
Answer: A
656656 is even; dividing by 2 repeatedly (5 times) gives 656656 = 2⁴ × 41041. Then 41041 = 41 × 1001, and 1001 = 7 × 11 × 13 (a well-known factorization, also checkable via the divisibility rule for 11). So 656656 = 2⁴ × 7 × 11 × 13 × 41, and the unique prime factors are 2, 7, 11, 13, 41. Their sum is 2 + 7 + 11 + 13 + 41 = 74.
6
geometry
medium
In the adjoining figure, AB is the diameter of the circle and C, D lie on the same side of the diameter. If ∠ABC = 75° and ∠DAC = 33°, then find the measurement of ∠DCA, in degrees.
A
33
B
24
C
72
D
42
E
27
Show answer
Answer: D
A, B, C, D all lie on the circle, so ABCD is a cyclic quadrilateral and opposite angles sum to 180°. So ∠ADC = 180° − ∠ABC = 180° − 75° = 105°. In triangle ADC, the angles sum to 180°, so ∠DCA = 180° − ∠DAC − ∠ADC = 180° − 33° − 105° = 42°.
7
3d-geometry
medium
A metal cylinder is re-casted into a cone with twice the radius. If the height of the cylinder was 12 m, then what is the height of the new cone?
A
24 cm
B
16 cm
C
6 cm
D
9 cm
E
36 cm
Show answer
Answer: D
Recasting preserves volume. Let the cylinder have radius r and height 12, so its volume is πr²(12). The cone has radius 2r and unknown height h, so its volume is (1/3)π(2r)²h = (4/3)πr²h. Setting the volumes equal: (4/3)πr²h = πr²(12), so h = 12 × 3/4 = 9.
8
geometry
easy
In the diagram, ABC is a right-angled triangle, AD = 4 and CD = 6. Find the length of DB if CD is perpendicular to AB.
The source PDF prints only 4 options for this question (no fifth option); labels here are A-D, following the exact on-page top-to-bottom order.
A
9
B
10
C
4
D
12
Show answer
Answer: A
Since CD is the altitude from the right angle at C to the hypotenuse AB, triangle ADC is similar to triangle CDB (both are right triangles sharing the angle at D, with the altitude creating equal corresponding angles). This gives AD/CD = CD/DB, i.e. 4/6 = 6/DB, so DB = 36/4 = 9.
9
geometry
medium
A rhombus has a length of 17 and the length of one of its diagonals is 16. Find the length of the other diagonal.
A
15
B
30
C
32
D
34
E
√545
Show answer
Answer: B
The diagonals of a rhombus bisect each other at right angles, so each side satisfies side² = (d1/2)² + (d2/2)². With side = 17 and d1 = 16: 17² = 8² + (d2/2)², so (d2/2)² = 289 − 64 = 225, giving d2/2 = 15 and d2 = 30.
10
rate-problems
medium
Tom travelled the first 25 km at a speed of 60 km/h. Then due to the conditions of the remaining road, he had to slow down and travel the rest of the distance with a speed of x km/h. If the average speed over 50 km was found to be 40 km/h, find the value of x.
A
36
B
20
C
30
D
15
E
None of these
Show answer
Answer: C
Time for the first 25 km = 25/60 h. Total time for 50 km at an average of 40 km/h = 50/40 = 5/4 h. Time for the remaining 25 km = 5/4 − 25/60 = 15/12 − 5/12 = 10/12 = 5/6 h. So x = distance/time = 25 ÷ (5/6) = 25 × 6/5 = 30 km/h.
11
number-theory
medium
Count the number of square numbers between (6⁴ + 1) and (4⁶ − 1).
A
28
B
17
C
29
D
16
E
27
Show answer
Answer: E
6⁴ = 1296, so 6⁴ + 1 = 1297; since 36² = 1296, the first perfect square strictly greater than 1297 is 37² = 1369. 4⁶ = 4096, so 4⁶ − 1 = 4095; since 64² = 4096 (which is not less than 4095), the largest perfect square strictly less than 4095 is 63² = 3969. So the perfect squares in range are 37² through 63², which is 63 − 37 + 1 = 27 squares.
12
number-theory
hard
Find the 1000th root of 10^(10^10).
A
10^(10^7)
B
10^(7^10)
C
√(10^√(10^√10))
D
7^(10^10)
E
10^(1/10)
Show answer
Answer: A
Taking the 1000th root is the same as raising to the power 1/1000: (10^(10^10))^(1/1000) = 10^(10^10 / 1000) = 10^(10^10 / 10^3) = 10^(10^7).
13
3d-geometry
hard
A cuboid has faces of areas of 24, 15 and 10 square units. If the lengths of the sides of the cuboid are a, b and c, then find abc + 2(a + b + c).
A
72.5
B
49
C
90
D
85
E
Insufficient information
Show answer
Answer: D
Let ab = 24, bc = 15, ca = 10 (the three face areas). Multiplying all three: (abc)² = 24 × 15 × 10 = 3600 = 60², so abc = 60. Then a = abc/bc = 60/15 = 4, b = abc/ca = 60/10 = 6, c = abc/ab = 60/24 = 2.5. So abc + 2(a+b+c) = 60 + 2(4 + 6 + 2.5) = 60 + 25 = 85. (This result is independent of which face area is paired with which product, since the final expression is symmetric in a, b, c.)
14
combinatorics
medium
What is the greatest number of regions that 7 lines can divide a circle into?
The source PDF prints only 4 options for this question (no fifth option); labels here are A-D, following the exact on-page top-to-bottom order.
A
14
B
29
C
15
D
35
Show answer
Answer: B
The maximum number of regions n lines can divide a circle into (each pair of lines intersecting inside the circle at a distinct point) follows the formula n(n+1)/2 + 1. For n = 7: 7×8/2 + 1 = 28 + 1 = 29.
15
combinatorics
hard
How many unique necklaces can be made such that they contain 7 equally spaced beads of 7 different colours?
A
5040
B
720
C
2520
D
360
E
840
Show answer
Answer: D
Arranging 7 distinct beads in a line gives 7! = 5040 orderings. Since the arrangement is a necklace, we must divide out its symmetries: rotational symmetry (7 equivalent rotations give the same necklace) divides by 7, and mirror (flip) symmetry divides by a further 2. So the count is 7!/(7×2) = 5040/14 = 360.
Section B
Open-Ended (Integer) Questions — 10 Questions
(+5)
16
coordinate-geometry
medium
In the figure, the difference between any two neighbouring horizontal and vertical dots is 1. Find the area of the triangle.
Show answer
Answer: 15
Placing the dot grid on a coordinate system with the bottom-left dot at the origin, the triangle's vertices are at (0,3), (7,6), and (3,0). Using the shoelace formula: Area = ½|x1(y2−y3) + x2(y3−y1) + x3(y1−y2)| = ½|0(6−0) + 7(0−3) + 3(3−6)| = ½|0 − 21 − 9| = ½(30) = 15. (Equivalently, subtracting the three corner right triangles from the 7×6 bounding rectangle: 42 − 10.5 − 4.5 − 12 = 15.)
17
algebra
medium
If n! = n × (n−1)! × … × 2 × 1, find the simplest value of the following: (20! + 21!) / (2×19! + 4×19! + 16×19!)
Show answer
Answer: 20
Factor 19! out of every term. The numerator: 20! + 21! = 20×19! + 21×20×19! = 19!×20×(1+21) = 19!×20×22. The denominator: 19!×(2+4+16) = 19!×22. So the ratio is (20×22)/22 = 20.
18
combinatorics
easy
What is the number of unique 4-letter words that can be formed with the word MATHS?
Show answer
Answer: 120
MATHS has 5 distinct letters. The number of ways to choose and arrange 4 of them in order is the permutation ⁵P₄ = 5!/(5−4)! = 5×4×3×2 = 120.
19
number-theory
hard
In a 3-digit number XYZ, X, Y and Z are non-zero digits in an arithmetic progression. If XYZ − ZXY = M, find the largest value of a 3-digit number M.
Show answer
Answer: 756
Let d be the common difference, so X − Y = d and Y − Z = d (X = Y+d, Z = Y−d). Then XYZ − ZXY = (100X+10Y+Z) − (100Z+10X+Y) = 90X + 9Y − 99Z. Substituting X = Y+d and Z = Y−d: 90(Y+d) + 9Y − 99(Y−d) = 189d. Since X, Y, Z must all be non-zero digits (1-9), d can be at most 4 (e.g. Y=5, X=9, Z=1 gives d=4, and indeed 951 − 195 = 756). So the largest value of M is 189×4 = 756.
20
sets
easy
In a class of 50 students if 15 like maths, 27 like English and 13 like none then how many like both Maths and English?
Show answer
Answer: 5
The number who like at least one of Maths or English is 50 − 13 = 37. By inclusion-exclusion, |Maths ∪ English| = |Maths| + |English| − |Maths ∩ English|, so 37 = 15 + 27 − Both, giving Both = 15 + 27 − 37 = 5.
21
number-theory
medium
The numbers 411, 666 and 870 give the same remainder when divided by a positive integer N. Find the greatest possible value of N.
Show answer
Answer: 51
If 411, 666, 870 all leave the same remainder when divided by N, then N must divide every pairwise difference: 666−411=255, 870−411=459, 870−666=204. The greatest such N is the GCD of these differences. 255 = 3×5×17, 459 = 3³×17, 204 = 2²×3×17. The common factors are 3 and 17, so GCD = 51.
22
number-theory
medium
In a sequence of numbers, each subsequent term is the sum of cubes of digits of the previous term. If one such sequence starts with the number 244, find the 2021st term in it.
Show answer
Answer: 244
Starting from 244: 2³+4³+4³ = 8+64+64 = 136. From 136: 1³+3³+6³ = 1+27+216 = 244. So the sequence is 244, 136, 244, 136, … repeating with period 2 from the very first term. Odd-numbered terms (1st, 3rd, 5th, …) are all 244, and 2021 is odd, so the 2021st term is 244.
23
combinatorics
hard
In how many ways can you reach any circle on the 10th row from the circle on the top if you can only move down left or right but cannot pass through the black circle?
Show answer
Answer: 326
Each step moves down-left or down-right (2 choices), so the number of paths from the top to any node on the 10th row (9 steps down) is 2⁹ = 512. The blocked (black) node sits 4 steps below the top with 2 choices to its left and 2 to its right among nodes at that depth, giving C(4,2) = 6 paths from the top to that node. From the blocked node, reaching the 10th row takes 5 more down-steps, giving 2⁵ = 32 paths. So paths that pass through the blocked node number 6 × 32 = 192, and paths avoiding it number 512 − 192 = 320.
24
algebra
hard
Find the maximum value of the following expression: √(781 − X) + √(X − 59)
Show answer
Answer: 38
Let a = √(781−X) and b = √(X−59), so a² + b² = (781−X) + (X−59) = 722, a constant independent of X. By the QM-AM inequality, (a+b)² ≤ 2(a²+b²), so a+b ≤ √(2×722) = √1444 = 38, with equality when a = b, i.e. 781−X = X−59, giving X = 420 (which is within the valid domain 59 ≤ X ≤ 781). Checking: at X=420, √361 + √361 = 19+19 = 38. So the maximum value is 38.
25
combinatorics
hard
The factorisation of a semi-prime number is p1 × p2 where p1 and p2 are prime numbers. Let the probability that the sum of faces of 3 standard dice is a semi-prime number be a/b in simplest form. Find the value of a + b.
Show answer
Answer: 139
There are 6³ = 216 equally likely outcomes for 3 standard dice. The source's solution identifies 6, 10, 14, and 15 as the semi-prime sums in range and counts the number of ordered outcomes giving each: sum 6 has 10 ways, sum 10 has 27 ways, sum 14 has 15 ways, and sum 15 has 10 ways, totalling 10+27+15+10 = 62 favourable outcomes. So the probability is 62/216, which simplifies (dividing by 2) to 31/108. With a=31 and b=108, a+b = 139.