Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.0 AIME Find the answer North Macedonia

Determine the sum of all odd 3-digit numbers whose product of digits is equal to $140$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Because $140 = 2 \cdot 2 \cdot 5 \cdot 7 = 4 \cdot 5 \cdot 7$ , it follows that the 3-digit numbers are formed with the digits $4$ , $5$ and $7$ . So the odd 3-digit numbers, with product of digits equal to $140$ , are $457$ , $475$ , $547$ and $745$ . Their sum is $2224$ .

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