Middle-school contest problems: counting, geometry, and number sense at real AMC 8 difficulty. Try each problem first, then open the strategy analysis and worked solution.
Problem 1AMC 8 · Counting with Restrictions
How many three-digit positive integers have a hundreds digit that is twice the units digit, an odd tens digit, and all three digits different?
A. 12 · B. 16 · C. 18 · D. 20 · E. 24
Strategy analysis
Anchor on the most restrictive digit first: the units digit forces the hundreds digit, so split into cases by the units digit and count the tens choices left in each case.
Worked solution and answer
Let the units digit be u. Then the hundreds digit is 2u, so u can only be 1, 2, 3, or 4. The tens digit must be odd and different from both u and 2u. If u = 1, the hundreds digit is 2, and the odd tens digits different from 1 are 3, 5, 7, 9: 4 choices. If u = 2, the hundreds digit is 4, and the odd tens digits are 1, 3, 5, 7, 9: 5 choices. If u = 3, the hundreds digit is 6, and the odd tens digits different from 3 are 1, 5, 7, 9: 4 choices. If u = 4, the hundreds digit is 8, and the odd tens digits are 1, 3, 5, 7, 9: 5 choices. The total is 4 + 5 + 4 + 5 = 18.
Answer: C
Problem 2AMC 8 · Geometry: Area by Subtraction
Each side of a 12 by 12 square is divided into segments of lengths 4 and 8, going around the square in the same direction. The four division points are joined to form an inner quadrilateral. What is the area of the inner quadrilateral?
A. 64 · B. 72 · C. 76 · D. 80 · E. 96
Strategy analysis
Don't find the inner figure directly — subtract the four corner triangles from the whole square. The trick is noticing all four cut-off triangles are congruent.
Worked solution and answer
The whole square has area 12² = 144. At each corner, the division points cut off a right triangle with legs 4 and 8, so each has area (4 × 8) ÷ 2 = 16. There are four such triangles, with total area 64. The inner quadrilateral therefore has area 144 − 64 = 80. (In fact it is a square of side √80.)
Answer: D
Problem 3AMC 8 · Number Theory: Remainders
What is the remainder when 2²⁰²⁵ + 3²⁰²⁵ is divided by 5?
A. 0 · B. 1 · C. 2 · D. 3 · E. 4
Strategy analysis
Cycle hunting: powers modulo 5 repeat in a short cycle, so only the exponent's remainder modulo the cycle length matters — never compute the giant power.
Worked solution and answer
Modulo 5, the powers of 2 cycle 2, 4, 3, 1 with period 4. Since 2025 ÷ 4 leaves remainder 1, 2²⁰²⁵ ≡ 2 (mod 5). The powers of 3 cycle 3, 4, 2, 1, also with period 4, so 3²⁰²⁵ ≡ 3 (mod 5). Adding gives 2 + 3 = 5 ≡ 0 (mod 5).
Answer: A
Problem 4AMC 8 · Algebra: Word Problem with a Twist
A rectangle's length is 3 more than twice its width. If both the length and the width are increased by 4, the area increases by 88. What is the perimeter of the original rectangle?
A. 28 · B. 32 · C. 36 · D. 40 · E. 44
Strategy analysis
Write the area change as an equation and watch the squared terms cancel — the twist is that the w² terms vanish, leaving a simple linear equation.
Worked solution and answer
Let the width be w, so the length is 2w + 3. The original area is w(2w + 3) = 2w² + 3w. After increasing both dimensions by 4, the area is (w + 4)(2w + 7) = 2w² + 15w + 28. The increase is (2w² + 15w + 28) − (2w² + 3w) = 12w + 28. Setting this equal to 88 gives 12w = 60, so w = 5 and the length is 13. The perimeter is 2(5 + 13) = 36.
Answer: C
Problem 5AMC 8 · Logic: Truth-Tellers and Liars
On an island, each person either always tells the truth or always lies. Pam says, "Quinn always lies." Quinn says, "Pam and I both always lie." Which of the following must be true?
A. Both tell the truth · B. Pam tells the truth, Quinn lies · C. Pam lies, Quinn tells the truth · D. Both lie · E. Cannot be determined
Strategy analysis
Assume one person's type and chase the contradiction: start with "Pam lies" and show it breaks Quinn's statement, which pins down both types at once.
Worked solution and answer
Suppose Pam lies. Then her statement "Quinn always lies" is false, so Quinn tells the truth. But Quinn says "Pam and I both always lie," which would then have to be true — yet Quinn herself tells the truth, a contradiction. So Pam tells the truth, meaning Quinn does always lie. Check: Quinn's statement "both always lie" is indeed false, consistent with Quinn lying. Therefore Pam tells the truth and Quinn lies.
Answer: B
Problem 6AMC 8 · Probability: Dice Products
Two fair six-sided dice are rolled. What is the probability that the product of the two numbers shown is a perfect square?
A. 1/6 · B. 2/9 · C. 1/4 · D. 5/18 · E. 1/3
Strategy analysis
Group the numbers by their squarefree part: a product is a square exactly when the two numbers share the same squarefree part, turning a messy count into small groups.
Worked solution and answer
Write each die value by its squarefree part: 1 and 4 have squarefree part 1; 2 has part 2; 3 has part 3; 5 has part 5; 6 has part 6. A product is a perfect square exactly when both numbers have the same squarefree part. From part 1 we get 2 × 2 = 4 ordered pairs: (1,1), (1,4), (4,1), (4,4). Parts 2, 3, 5, and 6 each contribute one pair: (2,2), (3,3), (5,5), (6,6). That is 8 favorable pairs out of 36, so the probability is 8/36 = 2/9.
Answer: B
Problem 7AMC 8 · Geometry: Shaded Area
ABCD is a square of side 8. Quarter-circles of radius 8 centered at A and C are drawn inside the square, and the overlapping lens-shaped region is shaded. What is the area of the shaded region?
A. 16π − 32 · B. 32π − 32 · C. 16π − 64 · D. 32π − 64 · E. 64π − 64
Strategy analysis
Use inclusion–exclusion on areas: the two quarter-circles together cover the whole square, so the overlap equals their total area minus the square's area.
Worked solution and answer
Each quarter-circle has radius 8, so each has area (1/4)π(8²) = 16π, and the two together have area 32π. The two quarter-circles exactly cover the square, whose area is 64, with their overlap counted twice. So 32π − (overlap) = 64, giving overlap = 32π − 64.
Answer: D
Problem 8AMC 8 · Sequences: Growing Differences
The sequence 3, 7, 12, 18, 25, … continues so that each difference between consecutive terms is one more than the previous difference. What is the 20th term?
A. 238 · B. 244 · C. 250 · D. 256 · E. 262
Strategy analysis
When differences grow by a constant amount, the terms are quadratic — sum the arithmetic progression of differences instead of listing 20 terms.
Worked solution and answer
The differences are 4, 5, 6, 7, …, an arithmetic progression starting at 4. The nth term equals 3 plus the sum of the first n − 1 differences: aₙ = 3 + (n − 1)(4 + (n + 2))/2 = 3 + (n − 1)(n + 6)/2. For n = 20, this is 3 + (19 × 26)/2 = 3 + 247 = 250.
Answer: C
Problem 9AMC 8 · Counting: Grid Placement
Three identical coins are placed on distinct squares of a 4 × 4 grid so that no two coins share a row or a column. In how many ways can this be done?
A. 72 · B. 96 · C. 120 · D. 144 · E. 192
Strategy analysis
Build the placement in three independent stages — choose the rows, choose the columns, then match them — and multiply. The matching step is the one most solvers forget.
Worked solution and answer
Choose which 3 of the 4 rows contain coins: C(4,3) = 4 ways. Choose which 3 of the 4 columns contain coins: C(4,3) = 4 ways. Then match the chosen rows to the chosen columns in 3! = 6 ways. Since the coins are identical, each matching gives a distinct placement. The total is 4 × 4 × 6 = 96.
Answer: B
Problem 10AMC 8 · Number Theory: Divisibility Counting
How many positive integers less than 200 are divisible by 3 or by 5, but not by 7?
A. 66 · B. 71 · C. 76 · D. 78 · E. 79
Strategy analysis
Count with inclusion–exclusion twice: first the numbers divisible by 3 or 5, then subtract the ones among them that are also divisible by 7 (counted via lcms, not by listing).
Worked solution and answer
Numbers below 200 divisible by 3: ⌊199/3⌋ = 66. Divisible by 5: ⌊199/5⌋ = 39. Divisible by both, i.e. by 15: ⌊199/15⌋ = 13. So 66 + 39 − 13 = 92 numbers are divisible by 3 or 5. Among these, those also divisible by 7 are divisible by 21 or 35: ⌊199/21⌋ = 9 and ⌊199/35⌋ = 5, minus those divisible by 105: ⌊199/105⌋ = 1, giving 9 + 5 − 1 = 13. The answer is 92 − 13 = 79.