Problem:
Let n be a positive integer such that 21β+31β+71β+n1β is an integer. Which of the following statements is not true:
Answer Choices:
A. 2 divides n
B. 3 divides n
C. 6 divides n
D. 7 divides n
E. n>84
Solution:
The number 21β+31β+71β+n1β is greater than 0 and less than 21β+31β+71β+11β<2. Hence,
21β+31β+71β+n1β=4241β+n1β
is an integer precisely when it is equal to 1. This implies that n=42, so the answer is (E).
Answer: Eβ.
The problems on this page are the property of the MAA's American Mathematics Competitions