Problem 4 · 1991 AJHSME
Medium
Algebra & Patterns
sum-near-round-number
If 991 + 993 + 995 + 997 + 999 = 5000 − N, then N =
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Answer: E — 25.
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Hint 1 of 3
Five numbers each just shy of 1000 would total exactly 5000 — and the right side is written as 5000 − N. So N isn't the sum; it's how much the sum FALLS SHORT of 5000. How far below 1000 is each term?
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Hint 2 of 3
Don't add the five big numbers. Add only the small gaps (how far each is under 1000); those gaps are exactly N.
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Hint 3 of 3
The gaps are 9, 7, 5, 3, 1 — consecutive odd numbers. There's a quick way to total those without adding one at a time.
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Approach: compare each term to 1000 — add the tiny shortfalls instead of the big numbers
- The five terms are each near 1000, so pretend they're all 1000: that's 5000. But each is a little less — by 9, 7, 5, 3, 1. The total shortfall is exactly what gets subtracted from 5000.
- Add the gaps: 9 + 7 + 5 + 3 + 1 = 25. So the sum is 5000 − 25, matching 5000 − N, which gives N = 25.
- Speed note: 9 + 7 + 5 + 3 + 1 is the first five odd numbers, and the sum of the first k odd numbers is always k² — here 5² = 25, no adding needed.
- Why this transfers: when numbers cluster near a round value, measure each from that value and add the small differences — far lighter than adding the bulky numbers and far safer than miscounting digits.
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