2025 AMC 10B Problem 22

Question

A seven-digit positive integer is chosen at random. What is the probability that the number is divisible by

11,

given that the sum of its digits is

61?

\dfrac{3}{14}

\dfrac{3}{11}

\dfrac{2}{7}

\dfrac{4}{11}

\dfrac{3}{7}

A digit sum of

61

is

2

below the maximum

63,

so the digits are all

9

except a total shortfall of

2

For divisibility by

11

use the alternating digit sum; express it through the shortfalls on the odd- versus even-position digits

A digit sum of

61 = 63 - 2

means all seven digits are

9

except for a total deficit of

2,

which gives

\binom{2 + 6}{6} = 28

numbers. For divisibility by

11

we need

O - E \equiv 0 \pmod{11},

where

O

sums the

4

odd-position digits and

E

the

3

even ones. Write the deficits as

d_O + d_E = 2.

Then

O - E = 9 - d_O

+ d_E = 11 - 2 d_O,

a multiple of

11

only when

d_O = 0.

So all of the deficit

2

falls on the

3

even positions, giving

\binom{2 + 2}{2} = 6

ways. The probability is

\tfrac{6}{28} = \tfrac{3}{14}.

Therefore, the answer is A.

Problem 22 in other years