2025 AMC 10B Problem 22

Attempt Problem 22 of the 2025 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10B solutions, or check the answer key.

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22.

A seven-digit positive integer is chosen at random. What is the probability that the number is divisible by 11,11, given that the sum of its digits is 61?61?

314\dfrac{3}{14}

311\dfrac{3}{11}

27\dfrac{2}{7}

411\dfrac{4}{11}

37\dfrac{3}{7}

Answer: A
Concepts:conditional probabilitydivisibilitystars and bars
Difficulty rating: 2040
Solution:

A digit sum of 61=63261 = 63 - 2 means all seven digits are 99 except for a total deficit of 2,2, which gives (2+66)=28\binom{2 + 6}{6} = 28 numbers. For divisibility by 1111 we need OE0(mod11),O - E \equiv 0 \pmod{11}, where OO sums the 44 odd-position digits and EE the 33 even ones. Write the deficits as dO+dE=2.d_O + d_E = 2. Then OE=9dOO - E = 9 - d_O +dE=112dO,+ d_E = 11 - 2 d_O, a multiple of 1111 only when dO=0.d_O = 0. So all of the deficit 22 falls on the 33 even positions, giving (2+22)=6\binom{2 + 2}{2} = 6 ways. The probability is 628=314.\tfrac{6}{28} = \tfrac{3}{14}. Therefore, the answer is A.

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