Section A
Multiple Choice Questions — 15 Questions
(+4 / -1)
1
number-theory
easy
Simplify: ⁵√(9¹⁰) + √((−2)⁶)
A
81
B
89
C
8
D
90
E
9
Show answer
Answer: B
⁵√(9¹⁰) = 9^(10/5) = 9² = 81, since taking the fifth root is the same as raising to the power 1/5, and (9¹⁰)^(1/5) = 9^2. Also √((−2)⁶) = √64 = 8, since (−2)⁶ = 64 (an even power of a negative number is positive) and the principal square root of 64 is 8. So the expression equals 81 + 8 = 89.
2
number-theory
easy
Find the number of multiples of 3 or 5 from 1 to 50.
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
16
B
26
C
23
D
24
Show answer
Answer: C
By inclusion-exclusion, the count of multiples of 3 or 5 up to 50 equals (multiples of 3) + (multiples of 5) − (multiples of both, i.e. multiples of 15). Multiples of 3 up to 50: ⌊50/3⌋ = 16. Multiples of 5 up to 50: ⌊50/5⌋ = 10. Multiples of 15 up to 50: ⌊50/15⌋ = 3. So the total is 16 + 10 − 3 = 23.
3
algebra
easy
Find the value of k, if k = 5.6(5.6 + 0.4) + 0.4(5.6 + 0.4)
A
6
B
18
C
24
D
36
E
42
Show answer
Answer: D
Both terms share the common factor (5.6 + 0.4) = 6. Factoring it out: k = (5.6 + 0.4)(5.6 + 0.4) = 6 × 6 = 36.
4
coordinate-geometry
medium
The points (0, 1) and (1, 2) are the coordinates of the adjacent corners of a square. Find the sum of the y-coordinates of the points diagonally opposite to (0, 1) in a square.
A
2
B
3
C
4
D
5
E
6
Show answer
Answer: C
Let A = (0,1) and B = (1,2), so side AB has direction vector (1,1). Since two different squares can be built on either side of segment AB, rotate this vector by ±90° to get the other two vertices. Rotating (1,1) by −90° gives (1,−1): the square A(0,1), B(1,2), C(2,1), D(1,0) has C diagonally opposite A, with y-coordinate 1. Rotating (1,1) by +90° gives (−1,1): the square A(0,1), B(1,2), C(0,3), D(−1,2) has C diagonally opposite A, with y-coordinate 3. Summing the y-coordinates of the two possible diagonally-opposite points: 1 + 3 = 4.
5
algebra
medium
Emily and Hannah had the same number of tickets at first. Emily gave 5 tickets to Hannah. Hannah then bought 2 more tickets. The number of tickets Hannah had, in the end, was four times as many as the number of tickets Emily had. How many tickets did Emily have at first?
A
9
B
11
C
18
D
10
E
7
Show answer
Answer: A
Let x be the number of tickets each had at first. After Emily gives 5 to Hannah: Emily has x − 5, Hannah has x + 5. After Hannah buys 2 more: Hannah has x + 7. Since Hannah's final count is 4 times Emily's: x + 7 = 4(x − 5), so x + 7 = 4x − 20, giving 3x = 27 and x = 9.
6
algebra
easy
Find the value of the following: 101 − 102 + 103 − 104 + 105 − 106 + … + 199 − 200 + 201
A
201
B
50
C
-50
D
-201
E
151
Show answer
Answer: E
Group the terms from 101 to 200 into consecutive pairs (101−102), (103−104), …, (199−200). Each pair equals −1, and there are (200−101+1)/2 = 50 such pairs, contributing 50 × (−1) = −50. Adding the leftover +201 gives −50 + 201 = 151.
7
number-theory
medium
The whole numbers are arranged five at a time, in every row, as shown (1st row: 1,2,3,4,5; 2nd row: 10,9,8,7,6; 3rd row: 11,12,13,14,15; 4th row: 20,19,18,17,16; 5th row: 21,22,23,24,25; and so on, alternating direction each row). What is the sum of the row number and the column number for 2019?
A
404
B
406
C
403
D
405
E
408
Show answer
Answer: B
Each row holds 5 consecutive numbers, so row n covers the numbers from 5(n−1)+1 to 5n. Dividing 2019 by 5 gives 2019 = 5×403 + 4, so 2019 falls in row 404 (the row covering 2016 to 2020). Odd-numbered rows increase left to right; even-numbered rows (like row 404) decrease left to right, so column 1 holds 2020, column 2 holds 2019, column 3 holds 2018, and so on. Thus 2019 is in row 404, column 2, and the required sum is 404 + 2 = 406.
8
3d-geometry
medium
A wire is folded to form a square with an area of 324 cm². The wire is then straightened, cut into pieces of equal length, and those pieces fastened together to form the frame of a cube. If no wire is wasted, find the volume enclosed by the cube (in cm³).
A
125
B
216
C
64
D
36
E
81
Show answer
Answer: B
A square of area 324 cm² has side length √324 = 18 cm, so the total wire length (its perimeter) is 4 × 18 = 72 cm. A cube's frame has 12 equal edges, so each edge of the new cube has length 72 ÷ 12 = 6 cm. The enclosed volume is 6³ = 216 cm³.
9
number-theory
easy
1001 is the product of 3 different prime numbers. How many factors of 1001 are not prime?
A
3
B
4
C
5
D
6
E
7
Show answer
Answer: C
1001 = 7 × 11 × 13, a product of 3 distinct primes. The number of divisors of a product of 3 distinct primes (each to the first power) is 2³ = 8: namely 1, 7, 11, 13, 77, 91, 143, 1001. Of these 8 divisors, exactly 3 (7, 11, 13) are prime. So the number of divisors that are not prime is 8 − 3 = 5.
10
number-theory
medium
What is the total number of different ways that the blanks below can be filled in so that the resulting four-digit number is a multiple of 11? 9 _ 5 _
A
9
B
10
C
18
D
11
E
8
Show answer
Answer: A
Write the number as 9A5B, where A and B are digits (0-9, not necessarily distinct). A number is divisible by 11 exactly when the alternating digit sum (sum of digits in odd positions minus sum of digits in even positions) is a multiple of 11 — here that is (9+5) − (A+B) = 14 − (A+B). For this to be a multiple of 11 with A+B between 0 and 18, we need A+B = 3 or A+B = 14. A+B = 3 gives 4 pairs: (0,3),(1,2),(2,1),(3,0). A+B = 14 gives 5 pairs: (5,9),(6,8),(7,7),(8,6),(9,5). In total there are 4 + 5 = 9 ways.
11
number-theory
hard
Find the positive difference between the sum of all possible integers x and the sum of all possible integers y that satisfy the equation x^y = 256.
A
16
B
227
C
32
D
29
E
256
Show answer
Answer: B
Since 256 = 2⁸ is a positive integer greater than 1, y must be a positive integer (y=0 gives 1, and negative y would force |x|<1, impossible for integer x giving 256). Listing every integer pair (x,y) with x^y = 256: since y is odd only for y=1 (x=256, as 256 is not a perfect cube, 5th power, or 7th power of an integer), and for even y we can also use negative bases: y=1 → x=256; y=2 → x=±16; y=4 → x=±4; y=8 → x=±2. That is 7 ordered solution pairs in total: (256,1), (16,2), (−16,2), (4,4), (−4,4), (2,8), (−2,8). Summing x over all 7 pairs: 256 + 16 − 16 + 4 − 4 + 2 − 2 = 256 (the ± pairs cancel). Summing y over all 7 pairs (with repetition, since y=2,4,8 each occur twice for the ± bases): 1 + 2 + 2 + 4 + 4 + 8 + 8 = 29. The positive difference is |256 − 29| = 227.
12
combinatorics
hard
Six boys of differing heights are arranged in 2 rows of 3 each. Each boy is taller than the boys in front of him and also taller than the boys to his left. How many arrangements for the six boys are possible?
A
6
B
12
C
5
D
10
E
8
Show answer
Answer: C
Label the boys 1 (shortest) through 6 (tallest) by height and place them into a 2-row × 3-column grid so that values increase left-to-right along each row and increase from front row to back row down each column — this is exactly the count of standard Young tableaux of a 2×3 rectangular shape. Boy 1 must sit at the front-left (nothing can be smaller than it to its left or in front), and boy 6 must sit at the back-right. Systematically placing boys 2 through 5 into the remaining valid positions (each new boy must go in the leftmost/frontmost still-available slot consistent with the ordering) yields exactly 5 valid completions, matching the well-known count of standard Young tableaux for a 2×n rectangle, which is the nth Catalan number: for n=3, C₃ = 5.
13
number-theory
medium
The sequence 1, 2, 3, …, 7, 8, 10, … consists of all positive integers except integers which contain the digit 9. What is the 300th term in the sequence?
A
309
B
339
C
363
D
243
E
359
Show answer
Answer: C
Numbers with no digit '9' behave like base-9 numerals using the digit set {0,1,2,...,8} in place of base-9 digits {0,...,8}: the nth term of the sequence (n ≥ 1) is obtained by writing n in base 9 and reading those digits as an ordinary base-10 number. For n = 300: dividing repeatedly by 9, 300 = 3×81 + 6×9 + 3×1, so 300 in base 9 is '363'. Reading '363' as a base-10 number gives the 300th term: 363. (Check: the 8th term is 8 itself since base-9 of 8 is '8', and the 9th term is 10 since base-9 of 9 is '10' — matching the given sequence.)
14
3d-geometry
medium
A box has a volume of 70 cm³. If the length, width, and height of the box are all different prime numbers, what is the surface area (in cm²) of the box?
A
118
B
104
C
90
D
55
E
20
Show answer
Answer: A
70 = 2 × 5 × 7, and this is the only way to write 70 as a product of 3 distinct primes, so the box has dimensions 2, 5, and 7. The surface area of a box with dimensions a, b, c is 2(ab + bc + ca) = 2(2×5 + 5×7 + 7×2) = 2(10 + 35 + 14) = 2(59) = 118 cm².
15
rate-problems
medium
Andy and Brett begin at opposite points on a 900-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
450
B
400
C
375
D
350
E
300
Show answer
Answer: C
The figure shows both arrows curving in the same rotational sense around the track (both clockwise), so Andy and Brett jog in the same direction rather than toward each other. Starting at opposite points, Brett (behind, since he is faster) must close a 900/2 = 450-foot gap on Andy. Since they move the same way, the gap closes at their relative speed, 5.5 − 4.3 = 1.2 feet/sec. The time to close the 450-foot gap is 450 ÷ 1.2 = 375 seconds.
Section B
Open-Ended (Integer) Questions — 10 Questions
(+5)
16
algebra
medium
The sum of the number of pieces of candy that Joe has and the number of pieces of candy that Tracy has is 26. Joe gives Tracy n pieces of candy so that she will have twice the number of pieces of candy as Joe initially had. How many different values are there for n?
Show answer
Answer: 5
Let a = Joe's initial candy count and b = Tracy's, so a + b = 26. After the transfer, Tracy has b + n pieces, and this must equal twice Joe's original amount: b + n = 2a. Substituting b = 26 − a: 26 − a + n = 2a, so n = 3a − 26. Since Joe gives away n pieces, n must be between 0 and a (he can't give more candy than he has, and n ≥ 0), i.e. 0 ≤ 3a − 26 ≤ a. From 3a − 26 ≥ 0: a ≥ 26/3 ≈ 8.67, so a ≥ 9 (a is a whole number of candies). From 3a − 26 ≤ a: 2a ≤ 26, so a ≤ 13. So a ranges over 9, 10, 11, 12, 13, giving n = 3(9)−26=1, 3(10)−26=4, 3(11)−26=7, 3(12)−26=10, 3(13)−26=13. That is 5 distinct values of n.
17
logic
medium
The integers from 1 through 10 inclusive are placed in the 10 circles shown (arranged in a triangular pyramid of 4 rows: the bottom row reads A, 10, B, 8; the row above it has an unlabeled circle, 7, and an unlabeled circle; the row above that has two unlabeled circles; and the top row is a single unlabeled circle), with one integer in each circle. Each number in the upper 3 rows is the positive difference between the numbers in the 2 circles immediately below it. Find the sum of A and B.
Show answer
Answer: 12
The row above the bottom row consists of the pairwise positive differences of adjacent bottom entries: |A−10|, |10−B|, |B−8|. The middle entry of that row is given as 7, so |10−B| = 7, giving B = 3 (B = 17 is not a valid single digit). That leaves the bottom-row-to-third-row values as |A−10|, 7, |3−8| = |A−10|, 7, 5. The remaining unused digits from 1-10 (after 10, 3, 8, 7, 5 are placed) are {1, 2, 4, 6, 9}, and one of these is 2 (visible in the figure's second row alongside an unknown D). Testing A = 9: the row above the bottom becomes |9−10|=1, 7, 5. The next row up is |1−7|=6 and |7−5|=2 — matching the figure's given 2. The top circle is then |6−2| = 4. So {A, and the three unknown upper values} = {9, 1, 6, 4}, which together with the already-placed {10, 3, 8, 7, 5, 2} exactly uses every integer 1 through 10 once — confirming A = 9. Therefore A + B = 9 + 3 = 12.
18
number-theory
medium
How many positive 4-digit numbers have exactly 3 factors?
Show answer
Answer: 14
A positive integer has exactly 3 divisors if and only if it is the square of a prime number p (its divisors are then 1, p, and p²). We need 1000 ≤ p² ≤ 9999, i.e. √1000 ≈ 31.6 ≤ p ≤ √9999 ≈ 99.995, so p ranges over primes from 32 to 99 inclusive. These primes are 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 — a count of 14 primes, giving 14 such 4-digit numbers.
19
number-theory
medium
Find the greatest number that divides 298, 348 and 873 with the same remainder in each case.
Show answer
Answer: 25
If N divides 298, 348, and 873 leaving the same remainder each time, then N must divide every pairwise difference of these numbers: 348 − 298 = 50, 873 − 348 = 525, 873 − 298 = 575. The greatest such N is GCD(50, 525, 575). 50 = 2 × 5², 525 = 3 × 5² × 7, 575 = 5² × 23. The only common factor is 5², so GCD = 25.
20
logic
hard
Jessica lies on Wednesdays, Thursdays and Fridays while she tells the truth on all other days. Tim lies on Sundays, Mondays and Tuesdays but he is truthful on all other days. On what day of the week could they both say, "Tomorrow, I will lie"? (Shade the following for your answer: Monday=001, Tuesday=002, …, Saturday=006, Sunday=007)
Show answer
Answer: 2
A person can only say "Tomorrow, I will lie" on a day when their truth-status changes overnight: either they are truthful today and will lie tomorrow (a true statement, consistent with speaking truthfully today), or they are lying today and will be truthful tomorrow (a false statement, consistent with lying today). Any day where today's and tomorrow's status are the same makes the statement impossible to say consistently. For Jessica (lies Wed/Thu/Fri, truthful Sat/Sun/Mon/Tue), the status flips overnight only between Tuesday(truthful)→Wednesday(lying) and between Friday(lying)→Saturday(truthful); so Jessica can say it on Tuesday or Friday. For Tim (lies Sun/Mon/Tue, truthful Wed/Thu/Fri/Sat), the status flips only between Tuesday(lying)→Wednesday(truthful) and Saturday(truthful)→Sunday(lying); so Tim can say it on Tuesday or Saturday. The only day both can say it is Tuesday, coded as 002.
21
number-theory
medium
In Mathematics, n! = n × (n−1) × … × 1. For example, 4! = 4 × 3 × 2 × 1. Find the greatest prime factor of the sum 16! + 18!
Show answer
Answer: 307
Factor out 16!: 16! + 18! = 16! + 16!×17×18 = 16!×(1 + 17×18) = 16!×(1 + 306) = 16!×307. The prime factors of 16! are all primes up to 16 (namely 2, 3, 5, 7, 11, 13). Checking 307 for primality: it is not divisible by 2, 3, 5, 7, 11, 13, or 17 (the only primes up to √307 ≈ 17.5), so 307 is prime. Since 307 far exceeds the largest prime factor of 16! (which is 13), the greatest prime factor of the whole sum is 307.
22
geometry
medium
Find the angle, in degrees, between the hour hand and the minute hand of a clock at 4:20.
Show answer
Answer: 10
At H hours and M minutes, the minute hand is at 6M degrees from 12, and the hour hand is at 30H + 0.5M degrees from 12. At 4:20 (H=4, M=20): minute hand position = 6×20 = 120°. Hour hand position = 30×4 + 0.5×20 = 120 + 10 = 130°. The angle between them is |130 − 120| = 10°, which is already at most 180°, so the answer is 10 degrees.
23
3d-geometry
hard
Three cubes with volumes of 1 cm³, 8 cm³ and 125 cm³ are glued together to form a solid figure. What is the least possible surface area of the solid figure?
Show answer
Answer: 168
The three cubes have side lengths 1, 2, and 5 (since 1³=1, 2³=8, 5³=125), with total surface area (if kept separate) of 6×1² + 6×2² + 6×5² = 6 + 24 + 150 = 180 cm². Gluing two cubes face-to-face hides an area equal to twice the contact area (once on each cube). To minimize the final surface area, maximize the total hidden contact area. Placing the side-2 cube centered on top of the side-5 cube hides a 2×2 = 4 area on each, removing 2×4 = 8. Then placing the side-1 cube in the corner where it can touch both the top of the side-5 cube (a 1×1 contact, hiding 1 on each, removing 2×1 = 2) and the side of the side-2 cube simultaneously (another 1×1 contact, hiding 1 on each, removing another 2×1 = 2) hides a total of 4 for the small cube — more than if it were simply stacked on top of the medium cube alone (which would hide only 2). This corner placement is the arrangement that minimizes surface area, removing 8 + 4 = 12 in total: 180 − 12 = 168 cm².
24
geometry
hard
The diagram shows a circle with radius 1 that rolls without slipping around the inside of a right-angled triangle with sides of lengths 9, 12 and 15. The circle rolls once around the triangle, returning to its starting point. What distance does the centre of the circle travel?
Show answer
Answer: 24
As the circle of radius r=1 rolls around the inside of the triangle, its center traces a smaller triangle similar to the original, offset inward by a distance r from each side. This inner triangle's inradius is exactly r less than the original triangle's inradius. The original 9-12-15 right triangle has area = (9×12)/2 = 54 and semi-perimeter s = (9+12+15)/2 = 18, so its inradius is R = Area/s = 54/18 = 3. The inner (center-locus) triangle therefore has inradius R − r = 3 − 1 = 2, and since it is similar to the original triangle with inradius scaled by (R−r)/R = 2/3, its perimeter is also scaled by 2/3: perimeter = (2/3) × (9+12+15) = (2/3) × 36 = 24. So the center of the circle travels a distance of 24.
25
number-theory
hard
Let S be the smallest integer such that 7S has 2019 digits. Find the tens digit of S.
Show answer
Answer: 1
A number has 2019 digits exactly when it is at least 10^2018. So we need the smallest S with 7S ≥ 10^2018, i.e. S ≥ 10^2018/7. Since 10^2018 is not a multiple of 7, S = ⌊10^2018/7⌋ + 1. Now 10^2018/7 = (1/7)×10^2018, and 1/7 = 0.142857142857… repeating with period 6 (the block '142857'). Since 2018 = 6×336 + 2, shifting the decimal point 2018 places means the integer part of (1/7)×10^2018 consists of the 6-digit block '142857' repeated 336 times, followed by the first 2 digits of the next block, '14'. So ⌊10^2018/7⌋ ends in '...142857' + '14', and adding 1 gives S ending in '...142857' + '15' (the trailing '14' becomes '15' after adding 1, with no further carry since 14+1=15 causes no overflow). Thus S's last two digits are '15', so its tens digit is 1.