Problem 19 · AMC 8 Stretch
Core
Number Theory
specification-without-loss-of-generalitylogical-reasoning
The sum of any \(5\) numbers in a row (like \(8, 9, 10, 11, 12\)) is always divisible by \(5\), because it equals \(5\) times the middle number. What is the sum \(8 + 9 + 10 + 11 + 12\)?
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Answer: 50 (= 5 x 10)
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Hint 1 of 4
For \(5\) numbers in a row, there is a middle number. How does each number compare to the middle one?
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Hint 2 of 4
Pair up the numbers around the middle: the one just below and the one just above add to twice the middle. What does the whole sum equal in terms of the middle number?
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Hint 3 of 4
If the sum equals \(5\) times the middle number, it's automatically divisible by \(5\). The middle of \(8, 9, 10, 11, 12\) is \(10\).
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Approach: Pair around the middle term
- Take \(5\) consecutive numbers and call the middle one \(m\): they are \(m-2, m-1, m, m+1, m+2\).
- Add them: the \(-2\) and \(+2\) cancel, the \(-1\) and \(+1\) cancel, leaving \(5m\) — five times a whole number, so always divisible by \(5\).
- For \(8, 9, 10, 11, 12\) the middle is \(m = 10\), so the sum is \(5 \times 10 = 50\).
- (Contrast: \(4\) in a row, like \(1 + 2 + 3 + 4 = 10\), has no single middle and \(10\) is not divisible by \(4\).)
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