Problem:
If 991+993+995+997+999=5000βN, then N=
Answer Choices:
A. 5
B. 10
C. 15
D. 20
E. 25
Solution:
Each of the five numbers on the left side of the equation is approximately equal to 1,000 . Thus N can be found by computing the difference between 1,000 and each number, so N=9+7+5+3+1=25.
OR
Since
991+993+995+997+999β=(1000β9)+(1000β7)+(1000β5)+(1000β3)+(1000β1)=5000β(9+7+5+3+1)=5000β25β
it follows that N=25.
Answer: 25β.
The problems on this page are the property of the MAA's American Mathematics Competitions