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by vipulgoyal » Thu Sep 12, 2013 12:17 am
what is the sum of the digits of the positive integer n where n<99 ??
1. n is divisible by the squre of Y
2. y^4 is a two digit positive odd integer
later
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by ganeshrkamath » Thu Sep 12, 2013 12:37 am
vipulgoyal wrote:what is the sum of the digits of the positive integer n where n<99 ??
1. n is divisible by the square of Y
2. y^4 is a two digit positive odd integer
later
1. Not sufficient.

2. y^4 is a 2 digit odd number
Since nothing is mentioned about n, this is not sufficient.

Both:
n = y^2 * Q (Q is a positive integer)
y^4 is a positive odd 2-digit integer.
Case 1: y^2 = 7 => y^4 = 49 and n = multiple of 7
Case 2: y^2 = 9 => y^4 = 81 and n = multiple of 9
We cannot find the sum of the digits of n.
Not sufficient.

Choose E

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by vipulgoyal » Thu Sep 12, 2013 12:57 am
If y would have been int then the ans is C and the sum of digits is 9
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by vinay1983 » Thu Sep 12, 2013 1:40 am
Statement 1

Some number less than 99 that is a square of some othet number....Insuf

Statement 2

I could think of 16 and 81. 2 answers so Insuff

Combining

16 is divisible by 2^2 and 81 is divisible by 3^2. 2 answers In sufficient

Should be E
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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by vipulgoyal » Thu Sep 12, 2013 2:04 am
Vinay, reconsider you ans, statement 2 says "y^4 is a two digit positive odd integer"

y could be root5 or root 7 or 3 but not 2
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by vinay1983 » Thu Sep 12, 2013 2:26 am
vipulgoyal wrote:Vinay, reconsider you ans, statement 2 says "y^4 is a two digit positive odd integer"

y could be root5 or root 7 or 3 but not 2
Yep sorry for that.My bad.So 2 can be rreplaced by √5 or √7 as you said. Thanks man.

For "√" hold Alt tab and then type "251" on number pad simultaeonously
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