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?

2007 AMC 8 Problem 18 — Solution

2007 AMC 8 Problem 18

Below is the professionally curated solution for Problem 18 of the 2007 AMC 8, 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.

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Concepts:modular arithmetic digits

Difficulty rating: 1400

18.

The product of the two $99$ -digit numbers $303,030,303,\ldots,030,303$ and $505,050,505,\ldots,050,505$ has thousands digit $A$ and units digit $B.$ What is the sum of $A$ and $B?$

$3$

$5$

$6$

$8$

$10$

Solution:

Only the last four digits can affect the thousands digit and units digit of the product. The two numbers end in $0303$ and $0505$ , so it is enough to compute $303\cdot 505$ .

Since $303\cdot 505=153015$ , the last four digits of the full product are $3015$ .

This gives $A=3$ and $B=5$ , so $A+B=8$ .

Thus, D is the correct answer.

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