Problem 16 · AMC 8 Stretch
Stretch
Number Theory
logical-reasoningaccount-for-all-possibilities
Find the smallest whole number \(n\) that leaves remainder \(1\) when divided by \(5\), and remainder \(3\) when divided by \(7\). (Hint: you can find it by smart listing — no fancy formula needed!)
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Answer: 31
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Hint 1 of 4
Start with the harder condition. List numbers that leave remainder \(3\) when divided by \(7\): \(3, 10, 17, 24, 31, \dots\)
Still stuck? Show hint 2 →
Hint 2 of 4
Now go down that list and check which ones also leave remainder \(1\) when divided by \(5\) (the ones ending in \(1\) or \(6\)).
Still stuck? Show hint 3 →
Hint 3 of 4
Check each: does \(3\) leave remainder \(1\) mod \(5\)? Does \(10\)? Does \(17\)? Keep going.
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Approach: Smart listing and checking against the second condition
- Write the numbers that leave remainder \(3\) when divided by \(7\): \(3, 10, 17, 24, 31, 38, \dots\)
- Check each for 'remainder \(1\) when divided by \(5\)': \(3 \to 3\) (no), \(10 \to 0\) (no), \(17 \to 2\) (no), \(24 \to 4\) (no), \(31 \to 1\) (yes!).
- So \(n = 31\).
- Double-check: \(31 = 6 \times 5 + 1\) (remainder \(1\)) and \(31 = 4 \times 7 + 3\) (remainder \(3\)). Both work.
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