A natural number has prime factorization given by n = 2 x 3 y 5 z , where y and z are such that y + z = 5 and y - 1 + z - 1 = 5 6 , y > z . Then the number of odd divisors of n , including 1 , is:

Question

A natural number has prime factorization given by n = 2 x 3 y 5 z , where y and z are such that y + z = 5 and y - 1 + z - 1 = 5 6 , y &#62; z . Then the number of odd divisors of n , including 1 , is:

Aicharya · Accepted Answer

y + z = 5 1 y + 1 z = 5 6 &#160; &#160; y &#62; z &#8658; y = 3 , z = 2 &#8658; n = 2 x . 3 3 . 5 2 = 2 . 2 . 2 &#8230; 3 . 3 . 3 5 . 5 For calculating the odd divisor of n = 2 x 3 y 5 z , x must be 0 . Hence, Number of odd divisors = 3 + 1 &#215; 2 + 1 = 4 &#215; 3 = 12 .

A natural number has prime factorization given by n = 2 x 3 y 5 z , where y and z are such that y + z = 5…

Solution