Two different numbers are divided by 7

Question

Can anyone tell me whether this problem is wrong? Should it be "the smallest possible three-digit product" or "the greatest possible three-digit product"? If it is greatest possible three-digit product, can you show me how to solve the problem. Thank you.

When two different numbers are divided by 7, remainders of 2 and 3, respectively, are left. What is the greatest possible three-digit product of these two numbers?

Answer 1

The problem works either way, regardless of whether the smallest or greatest 3-digit product is asked for.

In any case, we can say that the product is $6\bmod7$, as the two original numbers are $2\bmod7$ and $3\bmod7$ respectively, the product of residues being $2×3=6$.

To find the greatest possible 3-digit product, look at the numbers from 999 downwards that are $6\bmod7$ and see if they split into two factors with residues $2,3\bmod7$. But the first number we try fits perfectly: $$993=141×7+6=3×331=(0×7+3)(47×7+2)$$ Hence 993 is the answer to the question as originally stated.

If we are looking for the smallest 3-digit product, we do the same thing, only searching from 100 upwards: $$104=14×7+6=52×2=(7×7+3)(0×7+2)$$ Hence 104 is the answer to this modified question.

Answer 2

Clearly such a number would be of the form $7k+6$, the largest such number under $1000$ is $993$.

If we take $b=3$ then we can make $a=331=7\times 47+2$. So the maximum is $993$, obtained when $a=331$ and $b=3$.