Problem 19 · 1993 AJHSME
Hard
Algebra & Patterns
pair-terms
(1901 + 1902 + 1903 + … + 1993) − (101 + 102 + 103 + … + 193) =
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Answer: A — 167,400.
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Hint 1 of 2
Don't add up either giant sum. Pair them up instead: 1901 with 101, 1902 with 102, … Every top number is the same distance above its partner. What's that distance?
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Hint 2 of 2
Each pair differs by exactly 1901 − 101 = 1800. If you know how many pairs there are, the whole subtraction is one multiplication.
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Approach: subtract term-by-term, then multiply
- Match each top term with the bottom term directly below it: 1901−101, 1902−102, …, 1993−193. Every single pair differs by 1800.
- Count the pairs: from 1901 to 1993 is 1993 − 1901 + 1 = 93 terms (and likewise 101 to 193). So the answer is 93 × 1800 = 167,400.
- Why this transfers: subtracting two long sums term-by-term beats computing each sum separately — the messy totals never appear. Just watch the count: 'how many numbers from a to b' is b − a + 1, the classic off-by-one to nail down.
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