A natural number has prime factorization given by n = 2x3y5z, where y and z are such that …

Question

A natural number has prime factorization given by n = 2x3y5z, where y and z are such
that y + z = 5 and y-1 + z-1 = $${5 \over 6}$$, y > z. Then the number of odd divisions of n, including 1, is :

Accepted Answer

y + z = 5 ....... (1)

$${1 \over y} + {1 \over z} = {5 \over 6}$$

$$ \Rightarrow {{y + z} \over {yz}} = {5 \over 6}$$

$$ \Rightarrow {5 \over {yz}} = {5 \over 6}$$

$$ \Rightarrow $$ yz = 6

Also, (y - z)2 = (y + z)2 - 4yz

$$ \Rightarrow $$ (y - z)2 = (y + z)2 - 4yz

$$ \Rightarrow $$ (y - z)2 = 25 - 4(6) = 1

$$ \Rightarrow $$ y - z = 1 ..... (2)

from (1) and (2), y = 3 and z = 2

for calculating odd divisor of p = 2x . 3y . 5z

x must be zero

P = 20 . 33 . 52

$$ \Rightarrow $$ Total possible cases = (3050 + 3150 + 3250 + 3350 + .... + 3352)

$$ \therefore $$ Total odd divisors must be (3 + 1) ( 2 + 1) = 12

Solutions · 1