1.

For number N=35700, find (i) number of divisors (ii) number of proper divisors (iii) number of even divisors (iv) number of odd divisors (v) sum of all divisors

Answer»

Solution :`N=35700=5^(2)xx2^(2)xx3^(1)xx7^(1)xx17^(1)`
(i) Number of divisors
`=(2+1)xx(2+1)xx(1+1)xx(1+1)xx(1+1)`
`=3xx3xx2xx2xx2`
=72
(II) Number of proper divisors =72-2=70
(III) Number of even divisors `=3xx2xx2xx2xx2` (as 2 MUST occur at least once)
=48
(iv) Number of ODD divisors =72-48=24
(v) Sum of divisors `=(5^(@)+5^(1)+5^(2))(2^(@)+2^(1)+2^(2))(3^(@)+3^(1))(7^(@)+7^(1))(17^(@)+17^(1))`
`=31xx7xx4xx8xx18`
=124992


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