Problem 18 · 2007 AMC 8
Medium
Number Theory
last-digit
The product of the two 99-digit numbers 303,030,303,…,030,303 and 505,050,505,…,050,505 has thousands digit A and units digit B. What is the sum of A and B?
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Answer: D — 8.
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Hint 1 of 2
Those 99-digit monsters are a scare tactic — the last few digits of a product depend only on the last few digits of the factors. Lop off everything but the tails.
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Hint 2 of 2
Work mod a power of 10: to get the last 4 digits of a product, multiply the last 4 digits of each number; the front digits can't reach back that far.
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Approach: keep only the tails of each factor
- We need the last 4 digits, so keep just the last 4 of each number: 0303 and 0505. Everything to the left lands far past the thousands place.
- Multiply the short versions: 303 · 505 = 153015.
- Read off the last four digits, 3015: thousands digit A = 3, units digit B = 5.
- A + B = 3 + 5 = 8.
- Why this is safe: any digit of the product is built only from digit-pairs at or below its place value, so chopping the high digits leaves the last 4 untouched — the standard 'last digits' shortcut.
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