If \(n \in N\), then \(11^{n+2}+12^{2 n+1}\) is divisible by (a) 113 (b) 123 (c) 133 (d) None of these

We have the expression \(11^{n+2}+12^{2n+1}\). We will analyze it step by step.

To determine the divisibility, we need to find the remainders of the expression when divided by 113, 123 and 133. We can use the Remainder Theorem, which states that if a number \(x\) leaves a remainder \(r\) when divided by \(d\), then \(x^n\) will leave a remainder of \(r^n\) when divided by \(d\).

Let's find the remainders of \(11^{n+2}\) when divided by 113, 123, and 133: \(11^{n+2} \equiv 11^2 * 11^n\) When dividing by 113: \((11^2) \equiv 121 \equiv 8 \pmod{113}\) So, \(11^{n+2} \equiv 8 * 11^n \pmod{113}\) When dividing by 123: \((11^2) \equiv 121 \equiv -2 \pmod{123}\) So, \(11^{n+2} \equiv (-2) * 11^n \pmod{123}\) When dividing by 133: \((11^2) \equiv 121 \equiv -12 \pmod{133}\) So, \(11^{n+2} \equiv (-12) * 11^n \pmod{133}\)

Now let's find the remainders of \(12^{2n+1}\) when divided by 113, 123, and 133: \(12^{2n+1} \equiv 12 * (12^2)^n\) When dividing by 113: \((12^2) \equiv 144 \equiv -1 \pmod{113}\) So, \(12^{2n+1} \equiv 12*(-1)^n \pmod{113}\) When dividing by 123: \((12^2) \equiv 144 \equiv 4 \pmod{123}\) So, \(12^{2n+1} \equiv 12 * 4^n \pmod{123}\) When dividing by 133: \((12^2) \equiv 144 \equiv 11 \pmod{133}\) So, \(12^{2n+1} \equiv 12 * 11^n \pmod{133}\)

Now let's combine the two parts of the expressions: For 113: \(11^{n+2} + 12^{2n+1} \equiv 8 * 11^n + 12*(-1)^n \pmod{113}\) For 123: \(11^{n+2} + 12^{2n+1} \equiv (-2) * 11^n + 12 * 4^n \pmod{123}\) For 133: \(11^{n+2} + 12^{2n+1} \equiv (-12) * 11^n + 12 * 11^n \pmod{133}\) Here, we see that 11^n is a common term for both parts of the expression, from which we can conclude that the expression is divisible by a number that divides both of them. \(11^{n+2} + 12^{2n+1} = (12 - 11) * 11^n + 12 * 11^n\) \(11^{n+2} + 12^{2n+1} = 11^n(12 - 11 + 12) = 11^n(13)\) So, the given expression is divisible by 11 and 13. This gives the factors as 11, 13, and 11*13 = 143. Since 143 is not an option, we will check the given numbers: For 113 and 123, we can't find \(n\) such that \(11^{n+2} + 12^{2n+1}\) is divisible by those. However, for 133: \(11^{n+2} + 12^{2n+1}\equiv 0 \pmod{133}\) As 133 is divisible by 11, so we can say that our answer is 133 and the answer is (c).