MCQ
For all $n \in N, 3 n^5+5 n^3+7 n$ is divisible by:
  • A
    5
  • 15
  • C
    10
  • D
    3

Answer

Correct option: B.
15
  1. 15
Solution:
Given number $=3 n^5+5 n^2+7 n$
Let $\mathrm{n}=1,2,3,4, \ldots \ldots \ldots$
$3 n^5+5 n^3+7 n=3 \times 1^2+5 \times 1^3+7 \times 1=3+5+7=15$
$3 n^5+5 n^3+7 n=3 \times 2^5+5 \times 2^3+7 \times 2=3 \times 32+5 \times 8+7 \times 2=96+40+14=150=15 \times 10$
$3 n^5+5 n^3+7 n=3 \times 3^5+5 \times 3^3+7 \times 3=3 \times 243+5 \times 27+7 \times 3=729+135+21=885=15 \times 59$
Since, all these numbers are divisible by 15 for $n=1,2,3, \ldots .$.
So, the given number is divisible by 15 .

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