Problem 1 · 1991 AJHSME
Easy
Arithmetic & Operations
subtraction
1,000,000,000,000 − 777,777,777,777 =
(A) 222,222,222,222 (B) 222,222,222,223 (C) 233,333,333,333 (D) 322,222,222,223 (E) 333,333,333,333
Show answer
Answer: B — 222,222,222,223.
Show hints
Hint 1 of 2
Compare with 999,999,999,999 − 777,777,777,777, which is all 2's.
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Hint 2 of 2
Subtracting from one more than that adds 1.
Show solution
Approach: subtract from the all-9's number, then adjust
999,999,999,999 − 777,777,777,777 = 222,222,222,222. Our top number is 1 larger, so the answer is 222,222,222,222 + 1 = 222,222,222,223 .
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Problem 2 · 1991 AJHSME
Easy
Arithmetic & Operations
order-of-operations
(A) 4 (B) 8 (C) 12 (D) 16 (E) 20
Show answer
Answer: C — 12.
Show hint
Hint 1
The fraction bar groups the top and the bottom — finish each before dividing.
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Approach: simplify numerator and denominator
Top: 16 + 8 = 24. Bottom: 4 − 2 = 2. 24 ÷ 2 = 12 .
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Problem 3 · 1991 AJHSME
Easy
Arithmetic & Operations
powers-of-ten
Two hundred thousand times two hundred thousand equals
(A) four hundred thousand (B) four million (C) forty thousand (D) four hundred million (E) forty billion
Show answer
Answer: E — forty billion.
Show hints
Hint 1 of 2
Multiply 2 × 2 and count the zeros.
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Hint 2 of 2
200,000 has five zeros.
Show solution
Approach: multiply, then count zeros
(2 × 10⁵) × (2 × 10⁵) = 4 × 10¹⁰. 4 × 10¹⁰ = 40,000,000,000 = forty billion .
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Problem 4 · 1991 AJHSME
Medium
Algebra & Patterns
sum-near-round-number
If 991 + 993 + 995 + 997 + 999 = 5000 − N , then N =
(A) 5 (B) 10 (C) 15 (D) 20 (E) 25
Show answer
Answer: E — 25.
Show hints
Hint 1 of 2
Each term is a little under 1000; how much under?
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Hint 2 of 2
The five shortfalls add to N.
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Approach: measure each term below 1000
The five numbers fall short of 1000 by 9, 7, 5, 3, 1, totaling 25, so the sum is 5000 − 25. Thus N = 25 .
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Problem 5 · 1991 AJHSME
Medium
Geometry & Measurement
parity tiling
A “domino” is made up of two small squares:
Which of the “checkerboards” illustrated below CANNOT be covered exactly and completely by a whole number of non-overlapping dominoes?
(A) 3 × 4 (B) 3 × 5 (C) 4 × 4 (D) 4 × 5 (E) 6 × 3
Show answer
Answer: B — 3 × 5.
Show hints
Hint 1 of 2
Each domino covers exactly 2 squares, so a board needs an even number of squares.
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Hint 2 of 2
Compute each board's area.
Show solution
Approach: a tileable board must have an even number of squares
The boards have 12, 15, 16, 20, and 18 squares. Only 3 × 5 = 15 is odd. An odd number of squares can't be split into dominoes, so 3 × 5 cannot be covered.
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Problem 6 · 1991 AJHSME
Hard
Logic & Word Problems
array-search
Which number in the array below is both the largest in its column and the smallest in its row? (Columns go up and down, rows go right and left.)
10 6 4 3 2 11 7 14 10 8 8 3 4 5 9 13 4 15 12 1 8 2 5 9 3
(A) 1 (B) 6 (C) 7 (D) 12 (E) 15
Show answer
Answer: C — 7.
Show hints
Hint 1 of 2
You want a number that is both the biggest in its column and the smallest in its row.
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Hint 2 of 2
Scan each column for its max, then check whether that entry is the min of its row.
Show solution
Approach: test column-max entries against their rows
In the second column the largest entry is 7, and across its row (11, 7, 14, 10, 8) it is the smallest. So the number that is both is 7 .
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Problem 7 · 1991 AJHSME
Hard
Arithmetic & Operations
estimation factoring
The value of (487,000)(12,027,300) + (9,621,001)(487,000) (19,367)(.05)
(A) 10,000,000 (B) 100,000,000 (C) 1,000,000,000 (D) 10,000,000,000 (E) 100,000,000,000
Show answer
Answer: D — About 10,000,000,000.
Show hints
Hint 1 of 2
The numerator has 487,000 as a common factor — pull it out.
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Hint 2 of 2
Then round each piece to a power of ten.
Show solution
Approach: factor, then estimate with powers of ten
The numerator is 487,000 × (12,027,300 + 9,621,001) ≈ (5 × 10⁵)(2 × 10⁷) = 10¹³. The denominator is about (2 × 10⁴)(0.05) = 10³, so the value ≈ 10¹³ ÷ 10³ = 10¹⁰ .
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Problem 8 · 1991 AJHSME
Medium
Arithmetic & Operations
max-quotient signs
What is the largest quotient that can be formed using two numbers chosen from the set {−24, −3, −2, 1, 2, 8}?
(A) −24 (B) −3 (C) 8 (D) 12 (E) 24
Show answer
Answer: D — 12.
Show hints
Hint 1 of 2
For a big positive quotient, divide a large-magnitude number by a small one.
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Hint 2 of 2
Two negatives divide to a positive.
Show solution
Approach: divide the largest magnitude by the smallest
Dividing −24 by −2 gives a positive quotient of 12, the biggest possible. So the largest quotient is 12 .
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Problem 9 · 1991 AJHSME
Hard
Number Theory
inclusion-exclusion
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
(A) 18 (B) 21 (C) 24 (D) 25 (E) 27
Show answer
Answer: B — 21.
Show hints
Hint 1 of 2
Count multiples of 3 and of 5 separately.
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Hint 2 of 2
Subtract the multiples of 15, which got counted twice.
Show solution
Approach: inclusion-exclusion
Multiples of 3: ⌊46/3⌋ = 15; of 5: ⌊46/5⌋ = 9; of 15: ⌊46/15⌋ = 3. So 15 + 9 − 3 = 21 .
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Problem 10 · 1991 AJHSME
Medium
Geometry & Measurement
base-height
(A) (B) (C) (D) (E)
Show answer
Answer: B — 8.
Show hints
Hint 1 of 2
Use the horizontal side BC as the base.
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Hint 2 of 2
The height is the vertical gap between the top and bottom sides.
Show solution
Approach: area = base × height
The top side BC runs from (0,2) to (4,2), a base of 4, and the parallelogram's height (top row at y=2 down to the bottom at y=0) is 2. Area = 4 × 2 = 8 .
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Problem 11 · 1991 AJHSME
Medium
Counting & Probability
count-pairs
There are several sets of three different numbers whose sum is 15 which can be chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. How many of these sets contain a 5?
(A) 3 (B) 4 (C) 5 (D) 6 (E) 7
Show answer
Answer: B — 4.
Show hints
Hint 1 of 2
If 5 is in the set, the other two numbers must add to 10.
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Hint 2 of 2
Count distinct pairs (not using 5) that sum to 10.
Show solution
Approach: fix the 5, then pair the rest
The other two numbers must sum to 15 − 5 = 10, both different and not 5. Those pairs are (1,9), (2,8), (3,7), (4,6) — 4 sets.
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Problem 12 · 1991 AJHSME
Medium
Arithmetic & Operations
average
If 2 + 3 + 4 3 1990 + 1991 + 1992 N N =
(A) 3 (B) 6 (C) 1990 (D) 1991 (E) 1992
Show answer
Answer: D — 1991.
Show hints
Hint 1 of 2
The left side is the average of 2, 3, 4, which is 3.
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Hint 2 of 2
For the right side to equal 3, N must be the count that makes its average 3 — i.e. the sum divided by 3.
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Approach: both sides are an average
The left side equals 3. The right side is 5973/N, so 5973/N = 3 gives N = 5973 ÷ 3. That is N = 1991 (the middle of 1990, 1991, 1992).
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Problem 13 · 1991 AJHSME
Hard
Number Theory
trailing-zeros factor-2-and-5
How many zeros are at the end of the product 25 × 25 × 25 × 25 × 25 × 25 × 25 × 8 × 8 × 8?
(A) 3 (B) 6 (C) 9 (D) 10 (E) 12
Show answer
Answer: C — 9.
Show hints
Hint 1 of 2
Each trailing zero comes from a pair of factors 2 and 5.
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Hint 2 of 2
Count the total 2's and 5's; the smaller count wins.
Show solution
Approach: match 2's with 5's
The product is 25⁷ · 8³ = 5¹⁴ · 2⁹, giving fourteen 5's and nine 2's. Trailing zeros = the smaller count = 9 .
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Problem 14 · 1991 AJHSME
Stretch
Counting & Probability
worst-case parity
Several students are competing in a series of three races. A student earns 5 points for winning a race, 3 points for finishing second, and 1 point for finishing third. There are no ties. What is the smallest number of points a student must earn in the three races to be guaranteed of earning more points than any other student?
(A) 9 (B) 10 (C) 11 (D) 13 (E) 15
Show answer
Answer: D — 13.
Show hints
Hint 1 of 2
Each race awards 5, 3, 1 — all odd — so a 3-race total is always odd.
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Hint 2 of 2
Find the lowest odd total that no rival can match.
Show solution
Approach: find the lowest total that beats every possible rival
Race totals are sums of three odd numbers, hence always odd. With 13 points (say 5 + 5 + 3), the best a rival can scrape together from the leftovers is 5 + 3 + 3 = 11. Any total of 11 could be tied, so the guaranteed-winning minimum is 13 .
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Problem 15 · 1991 AJHSME
Hard
Geometry & Measurement
surface-area invariance
(A) (B) (C) (D) (E)
Show answer
Answer: C — the same.
Show hints
Hint 1 of 2
Count the faces of the cut-out cube that were on the surface, and the new faces it uncovers.
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Hint 2 of 2
Removing a corner cube trades the same number of squares away as it reveals.
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Approach: faces removed equal faces uncovered
Cutting the unit cube out of the corner removes three exposed unit squares but uncovers three new ones inside the notch. The two amounts cancel, so the surface area stays the same .
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Problem 16 · 1991 AJHSME
Stretch
Geometry & Measurement
folding spatial
(A) (B) (C) (D) (E)
Show answer
Answer: B — 9.
Show hints
Hint 1 of 2
Track one square's position through each of the four folds.
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Hint 2 of 2
Each fold maps the grid onto half its size; see which number ends up on top.
Show solution
Approach: follow the folds step by step
Folding top-over-bottom, then bottom-over-top, then right-over-left, then left-over-right collapses the 4 × 4 grid to a single stack. Carefully tracking which square lands on top, it is the square numbered 9 .
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Problem 17 · 1991 AJHSME
Stretch
Counting & Probability
max-independent sum
An auditorium with 20 rows of seats has 10 seats in the first row. Each successive row has one more seat than the previous row. If students taking an exam are permitted to sit in any row, but not next to another student in that row, then the maximum number of students that can be seated for an exam is
(A) 150 (B) 180 (C) 200 (D) 400 (E) 460
Show answer
Answer: C — 200.
Show hints
Hint 1 of 2
In a row of n seats, the most students with gaps between them is ⌈n/2⌉.
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Hint 2 of 2
Add ⌈n/2⌉ for n = 10, 11, …, 29.
Show solution
Approach: max per row, then pair rows from the ends
Rows 10 through 29 fit ⌈n/2⌉ students each: 5, 6, 6, 7, 7, …, 14, 14, 15. Pair the smallest with the largest, next-smallest with next-largest: each pair sums to 20 (e.g. 5+15, 6+14, …). Ten pairs × 20 = 200 .
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Problem 18 · 1991 AJHSME
Hard
Fractions, Decimals & Percents
pictograph percent
(A) (B) (C) (D) (E)
Show answer
Answer: C — 30%.
Show hints
Hint 1 of 2
Even with no scale, each X stands for the same number of employees, so count X's.
Still stuck? Show hint 2 →
Hint 2 of 2
Add the X's for 5 years or more and divide by the total number of X's.
Show solution
Approach: count symbols, then take the fraction
Each X represents the same number of employees, so the percent is just (X's for 5+ years) ÷ (total X's). Counting the columns, that fraction comes out to 30% .
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Problem 19 · 1991 AJHSME
Hard
Algebra & Patterns
minimize-rest
The average (arithmetic mean) of 10 different positive whole numbers is 10. The largest possible value of any of these numbers is
(A) 10 (B) 50 (C) 55 (D) 90 (E) 91
Show answer
Answer: C — 55.
Show hints
Hint 1 of 2
The ten numbers add to 10 × 10 = 100.
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Hint 2 of 2
To make one as big as possible, make the other nine as small as possible (and different).
Show solution
Approach: shrink the other nine numbers
The numbers total 100. The nine smallest different positive numbers are 1, 2, …, 9, summing to 45. So the largest can be 100 − 45 = 55 .
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Problem 20 · 1991 AJHSME
Stretch
Number Theory
cryptarithm place-value
(A) (B) (C) (D) (E)
Show answer
Answer: A — 1.
Show hints
Hint 1 of 2
Add the column-place values: ABC + AB + A = 111·A + 11·B + C.
Still stuck? Show hint 2 →
Hint 2 of 2
Set that equal to 300 and find digits A, B, C.
Show solution
Approach: expand by place value and solve
111A + 11B + C = 300. Taking A = 2 gives 11B + C = 78, so B = 7 and C = 1 (271 + 27 + 2 = 300). All digits differ, so C = 1 .
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Problem 21 · 1991 AJHSME
Medium
Algebra & Patterns
linear-rate
For every 3° rise in temperature, the volume of a certain gas expands by 4 cubic centimeters. If the volume of the gas is 24 cubic centimeters when the temperature is 32°, what was the volume in cubic centimeters when the temperature was 20°?
(A) 8 (B) 12 (C) 15 (D) 16 (E) 40
Show answer
Answer: A — 8.
Show hints
Hint 1 of 2
From 32° down to 20° is a 12° drop — how many 3° steps is that?
Still stuck? Show hint 2 →
Hint 2 of 2
Each step removes 4 cm³.
Show solution
Approach: count the 3° steps and subtract
A 12° drop is four 3° steps, each shrinking the gas by 4 cm³, for 16 cm³ less. So the volume was 24 − 16 = 8 cm³.
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Problem 22 · 1991 AJHSME
Hard
Counting & Probability
complementary-counting parity
(A) (B) (C) (D) (E)
Show answer
Answer: D — 7/9.
Show hints
Hint 1 of 2
A product is even unless BOTH spins are odd.
Still stuck? Show hint 2 →
Hint 2 of 2
Find each spinner's chance of landing odd, then multiply for 'both odd'.
Show solution
Approach: 1 minus the chance both are odd
Spinner 1 is odd (1 or 3) with probability 2/3; spinner 2 is odd (5) with probability 1/3. Both odd: 2/3 · 1/3 = 2/9, so an even product has probability 1 − 2/9 = 7/9 .
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Problem 23 · 1991 AJHSME
Stretch
Counting & Probability
inclusion-exclusion venn
The Pythagoras High School band has 100 female and 80 male members. The orchestra has 80 female and 100 male members. There are 60 females who are in both band and orchestra. Altogether there are 230 students who are in either band or orchestra or both. The number of males in the band who are NOT in the orchestra is
(A) 10 (B) 20 (C) 30 (D) 50 (E) 70
Show answer
Answer: A — 10.
Show hints
Hint 1 of 2
First find how many distinct females there are, then subtract to get the distinct males.
Still stuck? Show hint 2 →
Hint 2 of 2
Use inclusion-exclusion on the males to find those in both groups.
Show solution
Approach: count distinct males, then split out the overlap
Distinct females = 100 + 80 − 60 = 120, so distinct males = 230 − 120 = 110. Then males in both = 80 + 100 − 110 = 70. Males in band but not orchestra = 80 − 70 = 10 .
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Problem 24 · 1991 AJHSME
Stretch
Geometry & Measurement
volume-decomposition
A cube of edge 3 cm is cut into N smaller cubes, not all the same size. If the edge of each smaller cube is a whole number of centimeters, then N =
(A) 4 (B) 8 (C) 12 (D) 16 (E) 20
Show answer
Answer: E — 20.
Show hints
Hint 1 of 2
The total volume is 27, and pieces must be 1×1×1 or 2×2×2.
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Hint 2 of 2
Only one 2-cube fits in a 3-cube; the rest are unit cubes.
Show solution
Approach: fit one 2-cube, fill the rest with unit cubes
One 2 × 2 × 2 cube uses 8 of the 27 cubic cm, leaving 19 unit cubes (two 2-cubes won't fit in an edge of 3). That's 1 + 19 = 20 cubes, and they aren't all the same size.
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Problem 25 · 1991 AJHSME
Stretch
Geometry & Measurement
geometric-fraction
(A) (B) (C) (D) (E)
Show answer
Answer: C — 243/1024.
Show hints
Hint 1 of 2
Each change turns the middle fourth of every black triangle white.
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Hint 2 of 2
So each change keeps 3/4 of the black area; apply that five times.
Show solution
Approach: multiply by 3/4 once per change
Turning the middle fourth white leaves 3/4 of each black triangle black, so after 5 changes the black fraction is (3/4)⁵. (3/4)⁵ = 243/1024.
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