Problem:
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?
Answer Choices:
A. 3
B. 5
C. 6
D. 8
E. 10
Solution:
To find A and B, it is sufficient to consider only 303⋅505, because 0 is in the thousands place in both factors.
⋯ 303× ⋯ 505⋯ 1515⋯ 1500⋯ 3015
So A=3 and B=5, and the sum is A+B=3+5=8.
Answer: D.
The problems on this page are the property of the MAA's American Mathematics Competitions