Problem:
Let w,x,y, and z be whole numbers. If 2wâ‹…3xâ‹…5yâ‹…7z=588, then what does 2w+3x+5y+7z equal?
Answer Choices:
A. 21
B. 25
C. 27
D. 35
E. 56
Solution:
Factor 588 into 22â‹…31â‹…50â‹…72. Thus w=2,x=1,y=0, and z=2, and 2w+3x+5y+7z=21.
Answer: A​.
The problems on this page are the property of the MAA's American Mathematics Competitions