2007 AMC 8 Problem 18

Attempt Problem 18 of the 2007 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

The product of the two 9999-digit numbers 303,030,303,…,030,303303,030,303,\ldots,030,303 and 505,050,505,…,050,505505,050,505,\ldots,050,505 has thousands digit AA and units digit B.B. What is A+B?A+B?
33
55
66
88
1010
Answer: D
Concepts:modular arithmeticdigits
Small hint:
Only the last four digits can affect the thousands and units digits.
Big hint:
Reduce each 9999-digit number to its last four digits before multiplying.
Solution:
Only the last four digits can affect the thousands digit and units digit of the product. The two numbers end in 03030303 and 05050505, so it is enough to compute 303⋅505303\cdot 505.
Since 303⋅505=153015303\cdot 505=153015, the last four digits of the full product are 30153015.
This gives A=3A=3 and B=5B=5, so A+B=8A+B=8.
Thus, D is the correct answer.
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