The sum of the digits of a number N is 23. The remainder when N is divided by 11 is 7. What is the remainder when N is divided by 33?
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16
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7
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29
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13
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5
Since the sum of the digits of N is 23, N is congruent to 23, which is congruent to 2 modulo 3. We are also given that N is congruent to 7 modulo 11. By testing the options, only 29 leaves a remainder of 2 when divided by 3 and a remainder of 7 when divided by 11.
A number divided by 11 with a remainder of 7 can be written as 11k plus 7. Testing multiples of 11 shows that the number 119 satisfies both the remainder condition for 11 and has digits that sum to 11, which is less than 23, but we must consider if another number works; adding 99 to 119 gives 218, where the digit sum is exactly 11, so we add 99 again to get 317 with a digit sum of 11, but adding 99 to 416 gives 515 with a digit sum of 11. The only numbers leaving a remainder of 7 when divided by 11 and having a digit sum of 23 are extremely large, so we evaluate the general form modulo 33; since 11k plus 7 modulo 33 yields remainders dependent on k, we find the maximum possible remainder is 29. Thus, the remainder when N is divided by 33 is 29.