If there are 3n+1 numbers a,b,c,... be all are different and each is prime number then the number of different factors (excluding) of $a^{m} b c . . .$ is

2^{3 n} - m

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(m + 1) 2^{3 n} - 1

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(m + 1) 2^{3 n}

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m 2^{3 n} - 1

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Solution

The correct option is B $(m + 1) 2^{3 n}$
Given, $a^{m} b c . . . . . .$
and all $a, b, c, . . . . . . . . (3 n + 1)$ numbers are prime numbers
$∴$ number of factors $= (m + 1) (1 + 1) (1 + 1)$ ......... (3n times)
$= (m + 1) 2^{3 n}$

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If there are 3n+1 numbers a,b,c,... be all are different and each is prime number then the number of different factors (excluding) of ambc... is

The correct option is B (m+1)23nGiven,ambc......and all a,b,c,........(3n+1) numbers are prime numbers ∴ number of factors =(m+1)(1+1)(1+1)......... (3n times) =(m+1)23n

If there are 3n+1 numbers a,b,c,... be all are different and each is prime number then the number of different factors (excluding) of $a^{m} b c . . .$ is

The correct option is B $(m + 1) 2^{3 n}$
Given, $a^{m} b c . . . . . .$
and all $a, b, c, . . . . . . . . (3 n + 1)$ numbers are prime numbers
$∴$ number of factors $= (m + 1) (1 + 1) (1 + 1)$ ......... (3n times)
$= (m + 1) 2^{3 n}$