Problem 437

Solution 1

When $N$ is divided by 10, 11 or 12, the remainder is 7.

This means that $M=N-7$ is divisible by each of 10, 11 and 12.

Since $M$ is divisible by each of 10, 11 and 12, then $M$ is divisible by the least common multiple of 10, 11 and 12.

Since $10 = 2\times 5$, $12 = 2\times 2 \times 3$, and $11$ is prime, then the least common multiple of 10, 11 and 12 is $2 \times 2 \times 3 \times 5 \times 11 = 660$. (To find the least common multiple, we compute the product of the highest powers of each of the prime factors that occur in the given numbers.)

Since $M$ is divisible by 660 and $N=M+7$ is a three-digit positive integer, then $M$ must equal 660. (The next largest multiple of 660 is 1320.)

Therefore, $N=M+7=667$, and so the sum of the digits of $N$ is $6+6+7=19$.

Solution 2

When $N$ is divided by 10, 11 or 12, the remainder is 7.

This means that $M=N-7$ is divisible by each of 10, 11 and 12.

Since $M$ is divisible by each of 10 and 11, then $M$ must be divisible by 110.

We test the first few multiples of 110 until we obtain one that is divisible by 12.

The integers 110, 220, 330, 440, and 550 are not divisible by 12, but 660 is.

Therefore, $M$ could be 660. (This means that $M$ must be 660.)

Finally, $N = M+7=667$, and so the sum of the digits of $N$ is $6+6+7=19$.