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p^3 / 3920

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by amitabhprasad » Wed Oct 29, 2008 11:12 am
The way I solved this question was
Take factor of 3920 = 5*7*7*4*4
all I needed was another 5*5*7*4 to make the number a perfect cube
thus
p^3 = (5*7*7*4*4)(5*5*7*4)
==> p = 5*7*4 = 140
hence "5"
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by earth@work » Wed Oct 29, 2008 2:24 pm
agree with parallel_chase, OA =5 is correct
p^3/3920 = P^3/2^4 * 5 * 7^2
now, p^3 shud be s.t it is divided by 3920 = 2^4 * 5 * 7^2 completely.
for which we need to have 2^4 * 5 * 7^2 in numerator p^3
this imples MINIMUM p must have atleast
no. of 2's = 2x2
no.of 5's= 5
no. of 7's = 7
total p =2x2x5x7=140
sum =1+4+0 = 5

another way.....
p^3 to be completely divided by 2^4 * 5 * 7^2
i.e p to be completely divided by 2^(4/3)*5^(1/3)*7^(2/3)
this will give us 2*2*5*7 ..... for 2^(4/3) means more than one 2, while for 5 & 7 just one of each works.... i this method we will need more of thinking.
so p=2^2*5*7=140
so OA =5 is correct IMO
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