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Source: — Data Sufficiency |

by cans » Thu Sep 15, 2011 9:02 pm
N-> perfect square??
A) Not square. (perfect square has odd number of distinct factors)
Sufficient
B) sum of all distinct=even.
3:1,3 (even)
36:1,2,3,6,12,18,36 (even) insufficient
IMO A
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Cans!!
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by jogi1984 » Thu Sep 15, 2011 9:20 pm
The answer mentioned was D :(
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by sl750 » Fri Sep 16, 2011 9:10 am
36 1*36
2*18
3*12
4*9
6*6
The sum of distinct factors is odd, not even. D is correct
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by GMATGuruNY » Fri Sep 16, 2011 9:32 am
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by HeintzC2 » Fri Sep 16, 2011 12:03 pm
3 is not a perfect square.
cans wrote:N-> perfect square??
A) Not square. (perfect square has odd number of distinct factors)
Sufficient
B) sum of all distinct=even.
3:1,3 (even)
36:1,2,3,6,12,18,36 (even) insufficient
IMO A
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by [email protected] » Fri Sep 16, 2011 3:35 pm
cans wrote:N-> perfect square??
A) Not square. (perfect square has odd number of distinct factors)
Sufficient
B) sum of all distinct=even.
3:1,3 (even)
36:1,2,3,6,12,18,36 (even) insufficient
IMO A
CANS, I THINK THE ANSWER SHOULD BE D
IN YOUR OPTION B YOU FORGET TO COUNT NUMBER 4 AND 9. THEY ARE ALSO THE FACTORS OF 36. THE SUM OF 4+9 IS ODD. THUS AS PER YOUR EXPLANATION. THIS WILL BE THE SUM ODD
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by saketk » Sun Sep 18, 2011 5:25 am
jogi1984 wrote:Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.
two basic things to remember --

1) the numbers of distinct factors of 'PERFECT SQUARES' are ALWAYS Odd..

consider this -- 4 -- 2^2 , number of factor = (2+1) = 3
36 = 2^2*3^2 , number of factors = (2+1) (2+1) =9

2) the sum of all distinct factors of a number is calculated this way

I always use my favorite number - 28 (this is the 2nd PERFECT NUMBER, first being 6)
28 = (2^2)(7^1)
sum of factors = (1 + 2 + 4)(1 + 7) = (7)(8) = 56 [EVEN]

So, using these 2 points you can solve this question. and yeah the answer should be D
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by rohit_gmat » Sun Sep 18, 2011 7:34 am
cans wrote:36:1,2,3,6,12,18,36 (even)
thts 7 numbers thr cans... thts odd :P
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