Problem 14 · AMC 8 Stretch
Core
Number Theory
Arithmetic & Operations
pattern-recognitiontranslate-text-into-mathematics
Start adding \(1 + 2 + 3 + \cdots\), keeping a running total, until the total is a three-digit number with all three digits the same (like 111, 222, ..., 999). How many numbers do you add?
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Answer: 36 numbers (the total is 666)
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Hint 1 of 4
The running total after adding up to \(n\) is \(\tfrac{n(n+1)}{2}\). It has to be at most 999, so \(n\) can't be too big.
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Hint 2 of 4
A number 'xxx' (all digits equal) is \(x \times 111\), and \(111 = 3 \times 37\).
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Hint 3 of 4
So you need \(\tfrac{n(n+1)}{2}\) to be a multiple of 111, meaning \(n(n+1)\) is a multiple of \(2 \times 3 \times 37\). The prime 37 must divide \(n\) or \(n+1\).
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Approach: Triangular number must be a repdigit multiple of 111
- The running total after \(n\) numbers is \(S = \tfrac{n(n+1)}{2} \le 999\). Since \(\tfrac{44 \times 45}{2} = 990\) and \(\tfrac{45 \times 46}{2} = 1035\), we need \(n \le 44\).
- A repdigit 'xxx' equals \(x \times 111 = x \times 3 \times 37\), so \(n(n+1) = 2 \times 3 \times 37 \times x\). The prime 37 must divide \(n\) or \(n+1\), and with \(n \le 44\) the only options are \(n = 37\) or \(n = 36\).
- \(n = 37\): \(37 \times 38 = 1406\), not divisible by 3 (digit sum 11), so no. \(n = 36\): \(36 \times 37 = 1332 = 222 \times 6\), so the total is \(666\).
- Check: \(1 + 2 + \cdots + 36 = \tfrac{36 \times 37}{2} = 666\). So you add 36 numbers.
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