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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: Number of divisers of the form 4n+2(n>=0)of the int. 240 is a)4 b)8 c)10 d)3
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pramod6990 (973)

Blazing goIITian

Olaaa!! Perrrfect answer. 177  [221 rates]

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here goes the technical solution...


240 = 2431 51


we want factors of the form 2(2m+1) so we want odd multiples of 2....


so out of four powers we want only one 2.....rest to be choosen frm 3 and 5....


i have two objects 3 and five...total no of selections that i can make....


so total no of ways = (exp of 3 +1)(exp of 5 +1)= 2*2 =4


rate if useful.....


"Logic is the systematic way of reaching the wrong conclusion with confidence" lol.....
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