Problem 445

Solution 1

Since $20\times20\times20=8000$ and $30\times30\times30=27\,000$, then we might guess that the three consecutive odd numbers whose product is 9177 are closer to 20 than they are to 30.

Using trial and error, we determine that $21\times23\times25=12\,075$, which is too large.

The next smallest set of three consecutive odd numbers is $19,21,23$ and the product of these three numbers is $19\times21\times23=9177$, as required.
Thus, the sum of the three consecutive odd numbers whose product is 9177 is $19+21+23=63$.

Solution 2

We begin by determining the prime numbers whose product is 9177.

(This is called the prime factorization of 9177.)
This prime factorization of 9177 is shown in the following factor tree.

[[IMAGE0]]

That is, $9177=3\times3059=3\times7\times437=3\times7\times19\times23$. Since $3\times7=21$, then $9177=21\times19\times23$ and so the three consecutive numbers whose product is 9177 are $19,21,23$.
Thus, the sum of the three consecutive odd numbers whose product is 9177 is $19+21+23=63$.