MCQ
For all $n \in N, 3^{2n} + 7$ is divisible by:
- ✓$8$
- B$3$
- C$11$
- Dnon of these
✓
Answer
Correct option: A.
$8$
Given number $= 32n + 7$
Let $n = 1, 2, 3, 4, ……..$
$3^{2 n}+7=3^2+7=9+7=16$
$3^{2 n}+7=3^4+7=81+7=88$
$3^{2 n}+7=3^6+7=729+7=736$
Since, all these numbers are divisible by $8$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $8$
Let $n = 1, 2, 3, 4, ……..$
$3^{2 n}+7=3^2+7=9+7=16$
$3^{2 n}+7=3^4+7=81+7=88$
$3^{2 n}+7=3^6+7=729+7=736$
Since, all these numbers are divisible by $8$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $8$
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