Problem 14 · 1988 AJHSME
Hard
Number Theory
factor-pairsmax-sum
◇ and △ are whole numbers and ◇ × △ = 36. The largest possible value of ◇ + △ is
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Answer: E — 37.
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Hint 1 of 2
With the product locked at 36, you only get to choose which factor pair to use. To make the sum as *big* as possible, do you want the two numbers close together or far apart?
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Hint 2 of 2
For a fixed product, the sum grows as the factors spread apart and shrinks as they bunch up. So go straight for the most lopsided pair.
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Approach: for a fixed product, spread the factors apart
- The factor pairs of 36 are (1,36), (2,18), (3,12), (4,9), (6,6), with sums 37, 20, 15, 13, 12. The most spread-out pair, 1 and 36, gives the biggest sum.
- 1 + 36 = 37.
- Why this transfers: when a product is fixed, balanced factors (like 6 × 6) give the *smallest* sum and the most lopsided factors (1 × 36) give the *largest*. Knowing which end you want lets you jump to the answer without testing every pair.
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