Problem:
For how many integers N between 1 and 1990 is the improper fraction N+4N2+7 not in lowest terms?
Answer Choices:
A. 0
B. 86
C. 90
D. 104
E. 105
Solution:
Since N+4N2+7=N+4(N−4)(N+4)+23, the numerator and denominator will have a nontrivial common factor exactly when N+4 and 23 have a factor in common. Because 23 is a prime, N+4 is a multiple of 23 when N=−4+23k for some integer k. Solving 1<−4+23k<1990 yields 235<k<862316, or k=1,2,…,86.