For how many integers N between 1 and 1990 is the improper fraction (N^2+7)/(N+4) not in lowest terms?

1990 AMC 12 Problem 19

Attempt Problem 19 of the 1990 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1990 AMC 12 solutions, or check the answer key.

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19.

For how many integers

N

between

1

and

1990

is the improper fraction

\frac{N^2+7}{N+4}

not in lowest terms?

0

86

90

104

105

Reduce

N^2+7

modulo

N+4

The only possible common prime factor is a divisor of

23

Modulo

N+4,

we have

N\equiv-4,

N^2+7\equiv16+7=23.

Thus the fraction is reducible exactly when

N+4

is divisible by

23,

N\equiv19\pmod{23}.

The values are

19,42,\ldots,1974,

a total of

86.

Thus the correct answer is B.