AMC 10 · 2025 · #22

Grade 7 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

divisibility-rulesstars-and-barsconditional-probability complementary-counting ↑ Prerequisites: divisibility-rules

📏 Long solution 💡 4 insights

Problem

A seven-digit positive integer is chosen at random from all seven-digit numbers whose digits add up to 61. What is the probability that the chosen number is divisible by 11?

Pick an answer.

(A)

$\frac{3}{14}$

(B)

$\frac{3}{11}$

(C)

$\frac{2}{7}$

(D)

$\frac{4}{11}$

(E)

$\frac{3}{7}$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

The digit sum 61 is only 2 short of the maximum 63, so instead of counting big digits directly, count how much each digit falls short of 9. That complement swap d_i = 9 - a_i turns a hard 'sum to 61' count into an easy 'sum to 2' count, and it works again for the divisibility case. Split the job into a denominator (all numbers with digit sum 61) and a numerator (those also divisible by 11).

1STEP 1

Split into numerator and denominator

Every seven-digit number with digit sum 61 is equally likely, so the probability is (those divisible by 11) over (all of them).

P = (#{digit sum 61 and divisible by 11})/(#{digit sum 61})

2STEP 2

Flip digits to their shortfall

Digits crowd near 9, so let d_i = 9 - a_i be each shortfall; the seven 9's give 63, leaving d₁+…+d₇ = 2.

Σ_i=1⁷(9-d_i)=61 ⟹ d₁+d₂+…+d₇ = 2

3STEP 3

Count all numbers with digit sum 61

Count nonnegative solutions of d₁+…+d₇ = 2; the cap 9 never bites and every digit stays ≥ 7, so stars and bars gives 28 numbers.

C(2+7-1, 7-1)=C(8, 2)=28

4STEP 4

Write the divisibility-by-11 condition

The rule for 11 alternates positions: with O = a₁+a₃+a₅+a₇ and E = a₂+a₄+a₆, we need O - E = 11k, while O + E = 61.

O-E = 11k, O+E = 61

5STEP 5

Pin down k with the extreme principle

Adding gives 2O = 61 + 11k, so k is odd; the caps E ≤ 27 and O ≤ 36 kill every odd k but k = 1, leaving O = 36, E = 25.

k=1 → O = 36, E = 25

6STEP 6

Count the divisible cases

O = 36 forces all four odd digits to 9; for E = 25 the same shortfall trick gives d₂+d₄+d₆ = 2, so stars and bars counts 6 ways.

C(2+3-1, 3-1)=C(4, 2)=6

7STEP 7

Form the probability

Just 6 of the 28 numbers work, so the probability is $\frac{6}{28}$ , which reduces to $\frac{3}{14}$ , choice (A).

P = 6/28 = 3/14

Answer

3/14

The probability 3/14 ≈ 0.214 is between 0 and 1, as any probability must be, and it is below 1/4, which fits the fact that only a special alternating-sum alignment (all four odd-position digits equal to 9) lets a near-maximal number be divisible by 11. Both counts are cross-checkable by listing digit patterns: digit sum 61 means either five 9's and two 8's (C(7, 2)=21 ways) or six 9's and one 7 (7 ways), totaling 28; of these, exactly the ones with the 7 or both 8's in even positions work, giving 6.

💡Key takeaway

When digits are packed near 9, count how far each falls short of 9 instead of counting the digits themselves.

Split into numerator and denominator
Flip digits to their shortfall
Count all numbers with digit sum 61
Write the divisibility-by-11 condition
Pin down k with the extreme principle
Count the divisible cases
Form the probability