Problem 15 · 1996 AJHSME
Hard
Number Theory
units-digitmod-arithmetic
The remainder when the product 1492 · 1776 · 1812 · 1996 is divided by 5 is
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Answer: E — 4.
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Hint 1 of 2
Never multiply those huge four-digit numbers! When you divide by 5, only the LAST digit decides the leftover (because 5 divides evenly into every 10, 100, 1000…). So just watch the units digits.
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Hint 2 of 2
The units digit of the whole product comes only from multiplying the units digits together. Find that final units digit, then ask how much is left over after taking out 5s.
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Approach: only the units digit matters
- Dividing by 5 only cares about the last digit, since 5 splits evenly into 10, 100, 1000…. The four factors end in 2, 6, 2, 6, and the product's last digit is the last digit of 2·6·2·6 = 144, which is 4.
- A number ending in 4 has leftover 4 after pulling out as many 5s as possible (every number ending in 0 or 5 is a clean multiple of 5, so an ending of 4 sits 4 past the last one).
- Why this transfers: for 'remainder when divided by 5 (or 10, or 2),' chop everything down to units digits first. The last digit of a product depends only on the last digits of the factors — a giant computation shrinks to one tiny multiplication.
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