The natural number has prime factorization given by N-23-5x-7y- where x-y-5 and 1x-1y-56 x-y- Then the number of even divisors of N excluding N- is equal to

N=2^3·5^x·7^y Given, 1x+1y=56 ⇒ xy=6 (∵ x+y=5 ) ∴ x=3,y=2 (∵ x gt;y) ∴ N=2^3·5^3·7^2 nbsp; Number of even divisors of N=3(3+1)(2+1)=36 Hence, nbsp;Nu ...

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The natural number has prime factorization given by $N = 2^{3} \cdot 5^{x} \cdot 7^{y},$ where $x + y = 5$ and $\frac{1}{x} + \frac{1}{y} = \frac{5}{6} (x > y) .$ Then the number of even divisors of $N ($ excluding $N),$ is equal to

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Solution

The correct option is D $35$
$N = 2^{3} \cdot 5^{x} \cdot 7^{y}$
Given, $\frac{1}{x} + \frac{1}{y} = \frac{5}{6}$
$\Rightarrow x y = 6 (∵ x + y = 5)$
$∴ x = 3, y = 2 (∵ x > y)$
$∴ N = 2^{3} \cdot 5^{3} \cdot 7^{2}$
Number of even divisors of $N = 3 (3 + 1) (2 + 1) = 36$
Hence, Number of even divisors of $N ($ excluding $N)$ is $35.$

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The natural number has prime factorization given by N=23⋅5x⋅7y, where x+y=5 and 1x+1y=56 (x>y). Then the number of even divisors of N (excluding N), is equal to

The correct option is D 35N=23⋅5x⋅7y Given, 1x+1y=56 ⇒xy=6 (∵x+y=5) ∴x=3,y=2 (∵x>y) ∴N=23⋅53⋅72 Number of even divisors of N=3(3+1)(2+1)=36 Hence, Number of even divisors of N (excluding N) is 35.

The natural number has prime factorization given by $N = 2^{3} \cdot 5^{x} \cdot 7^{y},$ where $x + y = 5$ and $\frac{1}{x} + \frac{1}{y} = \frac{5}{6} (x > y) .$ Then the number of even divisors of $N ($ excluding $N),$ is equal to