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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 |
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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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