Section A
Multiple Choice Questions — 15 Questions
(+4 / -1)
1
arithmetic
easy
What is the value of
(888 + 666 + 444 + 111) − (555 + 333 + 222 + 777)?
A
111
B
222
C
333
D
444
E
555
Show answer
Answer: B
The sum of the first set of numbers is 888 + 666 + 444 + 111 = 2109.
The sum of the second set of numbers is 555 + 333 + 222 + 777 = 1887.
The difference is 2109 − 1887 = 222.
2
area
easy
Ashley has a rectangle made out of paper that is 8 cm by 12 cm. She folds it in half twice, first vertically and then horizontally. The new rectangle looks just like the original rectangle but smaller. What is the area of the new smaller rectangle in square cm?
A
6
B
12
C
18
D
24
E
30
Show answer
Answer: D
The area of the original rectangle is 8 × 12 = 96 cm². When it is folded in half, the area is cut in half and 96 ÷ 2 = 48 cm². When the paper is again folded in half the area is once again cut in half and 48 ÷ 2 = 24 cm².
3
arithmetic
medium
What is twice the value of 9 + 7 × 7 + 5 × 5 + 3 × 3 + 2?
A
188
B
94
C
93
D
186
E
190
Show answer
Answer: A
2 × (9 + 49 + 25 + 9 + 2) = 2 × 94 = 188.
4
calendar
medium
Summer vacation lasts for 85 days. During summer vacation, what is the greatest number of Fridays that could occur?
A
12
B
13
C
15
D
18
E
20
Show answer
Answer: B
85 days equals 12 weeks and 1 day. To maximize the number of Fridays, make the first day a Friday. The remaining 84 days then contain exactly 12 Fridays, so the greatest number of Fridays that could occur is 12 + 1 = 13.
5
arithmetic-series
medium
Find the sum of the following:
117 + 104 + 91 + 78 + 65 + 52 + 39 + 26 + 13
A
130
B
65
C
150
D
600
E
585
Show answer
Answer: E
117 + 104 + 91 + 78 + 65 + 52 + 39 + 26 + 13
= (117 + 13) + (104 + 26) + (91 + 39) + (78 + 52) + 65
= 130 × 4 + 65 = 585.
6
logic-digits
hard
TOM's PIN code is a 5-digit number with all different digits. The thousands digit is twice the ten thousands digit; the hundreds digit is 50% more than the thousands digit; the tens digit is 1 more than the ten thousands digit; and the ones digit is 2 more than the tens digit. What is the value of Tom's PIN code ÷ 65?
A
379
B
24635
C
6565
D
3790
E
24700
Show answer
Answer: A
Since the thousands digit is twice the ten thousands digit, the ten thousands digit can only be 1, 2, 3 or 4.
Since the hundreds digit is 50% more than the thousands digit, the ten thousands digit cannot be 4 (its hundreds digit would need to be 12, which is impossible).
Since the tens digit is 1 more than the ten thousands digit, the ten thousands digit cannot be 1 (its tens digit would then repeat the thousands digit).
Since the ones digit is 2 more than the tens digit, the ten thousands digit cannot be 3 (its ones digit would then repeat the thousands digit).
This leaves ten thousands digit = 2, thousands digit = 4, hundreds digit = 6, tens digit = 3, ones digit = 5.
Therefore, Tom's PIN code is 24635, and 24635 ÷ 65 = 379.
7
geometry-perimeter
hard
Sixteen 1 cm by 1 cm tiles are arranged in 4 rows and 4 columns to form a single square as shown. The perimeter of the square is 16 cm. If the shape is cut into exactly two pieces along the edges of the tiles, what is the greatest total perimeter of the two pieces, in centimetres, that can be made?
A
12
B
14
C
22
D
24
E
36
Show answer
Answer: E
Maximize the perimeter by cutting along as many tile edges as possible. In order to cut along the most possible tile edges, the cut should not run directly straight across or down — an indirect (staircase) cut maximizes the perimeter. The cut can run along at most 10 cm of tile edges. Since the cut creates a new edge on each of the two pieces, it adds 10 × 2 = 20 cm to the original perimeter. Thus, the greatest combined perimeter of the two pieces is 16 + 20 = 36 cm.
8
magic-square
hard
In the figure, the whole numbers from 1 through 7 are to be placed, one per square. The sum of the numbers in the left column, the sum of the numbers in the right column, and the sum of the numbers in each diagonal are the same. What is the largest possible product of the numbers across the grey row?
A
18
B
30
C
210
D
42
E
120
Show answer
Answer: C
To get the largest possible product for the three numbers in the grey row, try using 5, 6 and 7. Since the sum of the numbers in the first column must equal the sum of the numbers in the second column, the sum of the two remaining numbers in the first column must be 2 more than the sum of the two missing numbers in the second column. This forces the missing numbers in the first column to be 2 and 4, while the two missing numbers in the second column must be 1 and 3. The diagonals also work out to the same sum (2 + 6 + 3 = 11) with one such arrangement of the four remaining numbers. The product of the numbers in the grey row is then 5 × 6 × 7 = 210.
9
work-rate
medium
Tom can paint a classroom in 12 hours. Jerry can paint the same classroom in 4 hours. If they work together, each at their own rate, to paint this classroom, how many hours would they take?
A
3
B
4
C
5
D
6
E
8
Show answer
Answer: A
Tom can paint 1/12 of the classroom in one hour, and Jerry can paint 1/4 of the classroom in one hour. Together, they can paint 1/4 + 1/12 = 1/3 of the classroom in one hour. This means the entire classroom takes 3 × 1 = 3 hours to paint.
10
multiples
medium
What's the greatest 3-digit number that is divisible by both 32 and 36?
A
988
B
888
C
864
D
998
E
846
Show answer
Answer: C
Factor 32 and 36: 32 = 2⁵ and 36 = 2² × 3², so their lowest common multiple is 2⁵ × 3² = 288. The greatest 3-digit multiple of 288 is 288 × 3 = 864.
11
multiples
hard
Ari, Barry and Carrie chose a number. Ari said the number was a multiple of 3, 5 and 11. Barry said the number was a multiple of 2 and 7. Carrie said the number had 4 digits, one of which was a 9, but none of which were 6. Find the last 3 digits of the number they chose.
Answer back-derived from the Solutions page: the Answers page for Section A skips directly from Question 10 to Question 12, omitting an explicit printed answer for this question (the same generator defect the grade4 sibling paper shows at the same question number).
A
924
B
310
C
240
D
620
E
930
Show answer
Answer: C
The common multiple of these numbers is their product: (3 × 5 × 11) × (2 × 7). Regroup this as (3 × 7 × 11) × (2 × 5) = 231 × 10 = 2310. This is the smallest multiple of all 5 numbers. Checking multiples of 2310 against Carrie's rule that the number has a 9 digit but no 6 digit: 2310, 4620, 6930, and 9240. Hence, the number they chose is 9240, and its last 3 digits are 240.
12
counting-cubes
hard
There are 165 unit cubes arranged in 5 square layers with no space between the cubes as shown. The layers are 1 by 1, 3 by 3, 5 by 5, 7 by 7, and 9 by 9. If a sixth layer is placed on the bottom of the arrangement following the pattern, how many unit cubes are completely surrounded by six other unit cubes?
A
83
B
74
C
84
D
49
E
35
Show answer
Answer: C
Separate the layers of the structure. None of the cubes in the very top layer or the new bottom layer is completely surrounded by other cubes. A cube in one of the other layers is completely surrounded only where it is covered above by the next-smaller layer, so the top face of the layer above shows exactly which cubes below it are fully enclosed. Counting these covered cubes across all the layers gives 1×1 + 3×3 + 5×5 + 7×7 = 1 + 9 + 25 + 49 = 84 completely surrounded cubes.
13
clock-arithmetic
medium
Tiffany looked at a clock and realized that the number of hours that had passed since midnight was five times the number of hours remaining until noon. What time did the clock show 1 hour ago?
A
7.00
B
8.00
C
9.00
D
10.00
E
11.00
Show answer
Answer: C
There are 12 hours from midnight to noon. Since the number of hours that had passed was five times the number of hours remaining until noon, divide the total 12 hours into six equal units of 2 hours each. The time passed is then 5 × 2 = 10 hours, so the clock currently shows 10:00. One hour ago it showed 9:00.
14
cryptarithm
hard
In the multiplication AB × BA = A45B, A and B represent different digits, AB and BA are 2-digit numbers and A45B is a 4-digit number. If AB is less than BA, what is the 2-digit number AB?
A
18
B
81
C
14
D
58
E
50
Show answer
Answer: A
The product of the ones digits in AB and BA results in the ones digit of A45B, so B × A must end in B, which means A must be 1. The product 1B × B1 = 145B must be greater than 1450, so B must be 8 or 9. Only B = 8 satisfies the multiplication, since 18 × 81 = 1458. So AB = 18.
15
geometry-area
hard
If the area of the shaded part of the picture on the right is 18 cm², find the area (in cm²) of the triangle CAB.
A
18
B
6
C
3
D
27
E
9
Show answer
Answer: D
Find the area of each shaded triangle using Area = (1/2) × base × height, where the base and height are the legs of each right triangle. Let x be the length of a small square in the diagram. The total area of the three shaded triangles is (2x·x)/2 + (2x·2x)/2 + (2x·3x)/2 = 6x² = 18, so x² = 18 ÷ 6 = 3. The area of triangle CAB = (6x · 3x)/2 = 9x² = 9 × 3 = 27 cm².
Section B
Open-Ended (Integer) Questions — 10 Questions
(+5)
16
cryptarithm
hard
In the following cryptarithm, each different letter represents a different digit in the 6-digit numbers. If B is not 0, find the last three digits of the smallest possible sum. (For example, if your answer is 12345, then write 345.)
Source PDF defect: the Questions page prints only this instructional paragraph and omits the actual cryptarithm diagram/word-equation itself. The underlying addends were recovered from the worked Solutions page (see solution below).
Show answer
Answer: 24
Strategy: use place value and number sense. Since the problem asks for the least sum possible, assign the leading (hundred-thousands) letter the least possible value, which is 1, since it cannot be 0. Assign the ten-thousands-place letters with the least remaining unused digits. Since the tens-place letter repeats across all four addends, assign it 0, then assign 2, 3 and 4 to the varying ten-thousands digits in some order. The next least available digits for the two remaining constant letters (thousands/hundreds place, and ones place) are 5 and 6 respectively. This produces the four addends 105506, 125506, 135506 and 145506, which sum to 512024. Therefore, the last three digits of the least possible sum 512024 are 024.
17
age-problems
medium
Jethro is now 3 times as old as his son Hezekiah. Six years ago, Jethro was 4 times as old as his son Hezekiah was then. Find the sum of Jethro's and Hezekiah's present ages.
Show answer
Answer: 72
Six years ago, let Hezekiah's age be 1 unit, so Jethro's age was 4 units. Now, Hezekiah is (1 unit + 6) and Jethro is (4 units + 6). Since Jethro is now 3 times as old as Hezekiah, 4 units + 6 = 3 × (1 unit + 6), so 4 units + 6 = 3 units + 18, giving 1 unit = 12. Thus, Hezekiah is 12 + 6 = 18 years old and Jethro is 4 × 12 + 6 = 54 years old. The sum of their present ages is 54 + 18 = 72.
18
number-pattern
medium
Gerald likes to create patterns with dots. He created the following figures below. The first four figures are shown below. How many dots will the 16th figure contain?
Show answer
Answer: 481
The number of dots in the figures is 1, 5, 13, 25, 41, 61, ..., and the differences between consecutive terms are consecutive multiples of 4 (+4, +8, +12, +16, +20, ...). Continuing this pattern of increasing differences up to the 16th figure gives 481 dots.
19
number-theory
medium
The product of k and 1980 is a square number. If k is a whole number, what is the least possible value of k?
Show answer
Answer: 55
1980 = 2 × 2 × 3 × 3 × 5 × 11. The factors 2² and 3² already form square pairs, but 5 and 11 each appear only once, so neither forms a square pair on its own. The least possible value of k must supply one more factor of 5 and one more factor of 11 so that every prime appears an even number of times. Thus, k = 5 × 11 = 55.
20
cryptarithm
hard
In the cryptarithm shown, each letter represents a different digit. What is the least possible value of the three-digit number SUM? A cannot equal 0.
ADD
ADD
ADD
+ADD
————
SUM
Show answer
Answer: 576
Strategy: find multiples of 4 that are "small" 3-digit numbers, since SUM = 4 × ADD. The only possible value for A is 1 or 2. Let A = 1. If ADD = 122, then 4 × 122 = 488, which would require U to equal M, but they must be different, so ADD ≠ 122. If ADD = 133, then 4 × 133 = 532, which is impossible since it would require U to equal D. If ADD = 144, then 4 × 144 = 576, which works and uses all different digits. Since we want the least possible number, SUM = 576.
21
volume
hard
A rectangular brick has a volume of 80 cubic units, and each of its dimensions is a whole number. If the brick's total surface area is to be as small as possible, what is the area of its smallest face, in square units?
Source PDF defect: this question's text is entirely missing from the Questions section (page 10 ends at Question 20 and page 11 begins directly at Question 22 — there is no page or content anywhere in the 20-page document for Question 21). The text above was reconstructed from the fully worked Solutions-page reasoning (a whole-number brick of volume 80 with minimized surface area) and cross-checked against the printed Answer Key value of 16; treat it as a reconstruction, not a verbatim transcription.
Show answer
Answer: 16
Factor 80 and apply the principle that the least surface area occurs when the brick is most regular (most cube-like). The prime factors of 80 are 2⁴ × 5, so the dimensions of the most cube-like solid are 4 × 4 × 5. The faces with the smallest area have dimensions 4 × 4, and their area is each 16 square units. [Note: the total surface area of the brick is 2 × (4×4 + 4×5 + 4×5) = 112 square units.]
22
logic-puzzle
hard
The numbers 1 through 12 are placed in the diagram, one in each circle, so that the sum of the numbers along each line is the same. What is the largest possible value of this sum per line?
Show answer
Answer: 37
Let the sum of the numbers along each line be S, and let the numbers on the three vertices (corner circles) of the triangle be a, b and c. Adding the numbers along all 3 lines, every number from 1 through 12 except a, b and c is added exactly once, while a, b and c are each added twice. Thus 3S = (1+2+...+12) + a + b + c = 78 + a + b + c, so S = (78 + a + b + c) / 3. The largest possible value of S occurs when a + b + c is as large as possible, i.e. using the three largest available numbers 10, 11 and 12. Hence S = (78 + 10 + 11 + 12) / 3 = 111 / 3 = 37.
23
number-pattern
hard
The region inside the circle can be cut into two parts by drawing 1 line through it, as shown. If the circle is cut into 232 parts, then what is the least number of lines that must be drawn through the region?
Show answer
Answer: 21
Build a table of the number of lines drawn and the largest number of parts obtained: 1 line → 2 parts, 2 lines → 4 parts (+2), 3 lines → 7 parts (+3), 4 lines → 11 parts (+4), with the differences increasing by 1 each time. Continuing the pattern, 5 lines give 11 + 5 = 16 parts, and so on. Continuing this pattern of increasing differences, a circle divided by 21 lines gives the largest number of parts as 232.
24
multiples
hard
How many digits are there in the least multiple of 41 whose only digit is 1?
Show answer
Answer: 5
The least number greater than 41 whose only digit is 1 is 111. However, 111 is not divisible by 41, and neither is 1111. When 11111 is divided by 41 the result is exactly 271, which means 11111 is the least multiple of 41 whose only digit is 1, and it has 5 digits.
25
cryptarithm
hard
In the cryptarithm shown, different letters represent different digits. If two letters are the same, they represent the same digit. What is the greatest value that GOO could be?
DUCK
+DUCK
————
GOOSE
Show answer
Answer: 188
List the possible digits and cross them off once used: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since DUCK + DUCK = GOOSE is a 5-digit number, the only possible value for G is 1, so cross off 1. To make GOO as great as possible, let D = 9, which makes O = 8. This means U + U = 2U must be less than 10 (no further carry into O), so U = 4. Thus, O = 8, DUCK = 9435 or 9436, and GOOSE = 18870 or 18872. The greatest value of GOO is 188.