Question

Find the number of integers \(n\) such that \(1 \leq n \leq 100\) and \(n^2 + 3n + 2\) is divisible by 5.

• A. 20
• B. 21
• C. 19
• D. 18

Hint:

For divisibility of polynomial expressions modulo \(m\), factorize if possible and check residues satisfying divisibility conditions.

Answer 1

The Correct Option is A

The objective is to determine the count of integers \(n\), within the range \(1 \leq n \leq 100\), for which the expression \(n^2 + 3n + 2\) is divisible by 5. This condition is equivalent to solving \(n^2 + 3n + 2 \equiv 0 \pmod{5}\). We analyze the congruence based on the remainder of \(n\) when divided by 5:

• If \(n \equiv 0 \pmod{5}\), \(n^2 + 3n + 2 \equiv 0^2 + 3(0) + 2 \equiv 2 \pmod{5}\).
• If \(n \equiv 1 \pmod{5}\), \(n^2 + 3n + 2 \equiv 1^2 + 3(1) + 2 \equiv 1 + 3 + 2 \equiv 6 \equiv 1 \pmod{5}\).
• If \(n \equiv 2 \pmod{5}\), \(n^2 + 3n + 2 \equiv 2^2 + 3(2) + 2 \equiv 4 + 6 + 2 \equiv 12 \equiv 2 \pmod{5}\).
• If \(n \equiv 3 \pmod{5}\), \(n^2 + 3n + 2 \equiv 3^2 + 3(3) + 2 \equiv 9 + 9 + 2 \equiv 20 \equiv 0 \pmod{5}\).
• If \(n \equiv 4 \pmod{5}\), \(n^2 + 3n + 2 \equiv 4^2 + 3(4) + 2 \equiv 16 + 12 + 2 \equiv 30 \equiv 0 \pmod{5}\).

Thus, the expression is divisible by 5 when \(n \equiv 3 \pmod{5}\) or \(n \equiv 4 \pmod{5}\). We now count the integers in the range [1, 100] that satisfy these conditions:

• For \(n \equiv 3 \pmod{5}\): These are numbers of the form \(5k + 3\). For \(k=0\), \(n=3\). For \(k=19\), \(n=98\). Thus, there are \(19 - 0 + 1 = 20\) such numbers.
• For \(n \equiv 4 \pmod{5}\): These are numbers of the form \(5k + 4\). For \(k=0\), \(n=4\). For \(k=19\), \(n=99\). Thus, there are \(19 - 0 + 1 = 20\) such numbers.

The total count of integers \(n\) in the range \(1 \leq n \leq 100\) for which \(n^2 + 3n + 2\) is divisible by 5 is the sum of the counts from these two cases: \(20 + 20 = 40\).