If n-1- the values of the positive integer m for which nm-1 divides a-1-n-n2-n63 is-areA- 8B- 16C- 32D- 64

The correct options are A 8 B 16 C 32a=1+n+n^2+… +n^63 ⇒ a=1(n^64 1)n 1 ⇒ a = (n^32 1)(n^32+1)n 1 ⇒ a = (n^16 1)(n^16+1)(n^32+1)n 1 nbsp; ∴ a=(n+1)(n^2+1)(n^4 ...

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Byju's Answer

Standard XI

Mathematics

Permutation: n Different Things Taken All at a Time When All Are Not Different

If n>1, the v...

Question

If $n > 1$ , the values of the positive integer $m$ for which $n^{m} + 1$ divides $a = 1 + n + n^{2} + \dots \dots + n^{63}$ is/are

8

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16

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32

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64

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Solution

The correct options are
A $8$
B $16$
C $32$
$a = 1 + n + n^{2} + \dots + n^{63}$
$\Rightarrow a = \frac{1 (n^{64} - 1)}{n - 1} \Rightarrow a = \frac{(n^{32} - 1) (n^{32} + 1)}{n - 1} \Rightarrow a = \frac{(n^{16} - 1) (n^{16} + 1) (n^{32} + 1)}{n - 1}$
$∴ a = (n + 1) (n^{2} + 1) (n^{4} + 1) (n^{8} + 1) (n^{16} + 1) (n^{32} + 1)$

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If n>1, the values of the positive integer m for which nm+1 divides a=1+n+n2+……+n63 is/are

The correct options are A 8 B 16 C 32a=1+n+n2+…+n63 ⇒a=1(n64−1)n−1⇒a=(n32−1)(n32+1)n−1⇒a=(n16−1)(n16+1)(n32+1)n−1 ∴a=(n+1)(n2+1)(n4+1)(n8+1)(n16+1)(n32+1)

If $n > 1$ , the values of the positive integer $m$ for which $n^{m} + 1$ divides $a = 1 + n + n^{2} + \dots \dots + n^{63}$ is/are