Problem 15 · 1995 AJHSME
Hard
Number Theory
repeating-decimalcyclicity
What is the 100th digit to the right of the decimal point in the decimal form of 4/37?
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Answer: B — 1.
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Hint 1 of 2
A repeating decimal loops forever in a fixed block, so the digits go in cycles. You don't write out 100 digits — you find the block, then figure out where the 100th digit lands WITHIN the cycle.
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Hint 2 of 2
Find the repeating block of 4/37 (it's short). Then use the leftover after dividing 100 by the block length to pick the right digit in the block.
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Approach: find the repeating block, then locate position 100 inside the cycle
- Do the division: 4 ÷ 37 = 0.108108108… The block '108' repeats, length 3. So the digit at any position depends only on where it sits inside that 3-long cycle.
- Position 100: how many whole blocks fit, with what leftover? 100 = 3 × 33 + 1, leftover 1. After 33 complete '108' blocks you're at the 1st digit of the next block.
- The 1st digit of '108' is 1.
- Why this transfers: for any repeating decimal, divide the position by the block length and read the leftover — leftover 1 → 1st digit, leftover 2 → 2nd, and a leftover of 0 means you've landed on the LAST digit of the block. This 'cycle position' trick handles powers, units digits, and calendars too.
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