Question

N is the smallest number such that N/2 is a perfect square and N/3 is a perfect cube. Then, the number of divisors of N is?

Answer 1

Answer:

(#M40014045) MBA QUESTION number systems, progressions KEEP AN EYE Keep an eye puzzle N is the smallest no. such that N/2 is a perfect square and N/3 is a perfect cube then no. of divisors of N are

1) 2

2) 3

3) 4

4) 5

5) None.....

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Answer 2

$Let \: \frac{N}{2} = x^{2} \: --(1)\:and$

$\frac{N}{3} = y^{3} \: --(2)$

/* Multiplying equations (1) and (2), we get */

$\frac{N}{2} \times \frac{N}{3} = x^{2}y^{3}$

$\implies N^{2} = 6x^{2}y^{3}$

$\implies N = \sqrt{6x^{2}y^{3}}$

$\implies N = xy \times \sqrt{6y}$

$But , \: N \: is \:a \: natural \: number$

$6y\:must \:be\:a \:perfect\: square$

$\implies y = 6$

/*Put y = 6 equation (2), we get */

$\frac{N}{3} = 6^{3}$

$\implies N = 3 \times (2 \times 3)^{3}$

$=2^{3}\times 3^{4}$

$Number \:of \: Factors \:of \: N$

$= (3+1)(4+1)$

$= 4 \times 5$

$= 20$

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