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Problem #34 · August 30, 2026

Natural numbers

A three-digit number is 1212 times the sum of its digits. Find this number.
  1. A120120
  2. B144144
  3. C216216
  4. D108108
  5. E324324

WORKED EXAMPLE

Solution

Let the required number be N=12SN = 12S, where SS is the sum of its digits. A number and the sum of its digits leave the same remainder when divided by 99, so the difference 12S−S=11S=9S+2S12S - S = 11S = 9S + 2S is divisible by nine; hence SS is also divisible by 99. For the number to have three digits, S∈{9,18,27}S \in \{9, 18, 27\}, that is, N∈{108,216,324}N \in \{108, 216, 324\}; checking shows that only for 108108 does 1+0+8=9=S1 + 0 + 8 = 9 = S hold.
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