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Re: Is the positive integer N a perfect square? [#permalink]
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31 Jul 2012, 01:26
ashish8 wrote: How is B sufficient? Sum of distinct factors of a perfect square is odd, but if n is 2, then the sum is also odd. (2) states: The sum of all distinct factors of N is even. Since the sum of distinct factors of a perfect square must be odd, we can conclude that N is not a perfect square. So, the answer to the question "Is N a perfect square?" is a definite NO. Therefore, (2) sufficient. Not only perfect squares have the sum of their distinct factors odd. As you mentioned, for 2, the sum of its factors is odd, and it is not a perfect square. So, if a number is a perfect square, then the sum of its factors is necessarily odd, but the reciprocal is not true. Meaning, if the sum of the factors is odd, the number is not necessarily a perfect square, it might be or not. But if the sum of the distinct factors is even, then certainly the number cannot be a perfect square.
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Re: Is the positive integer N a perfect square? [#permalink]
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31 Jul 2012, 01:32
ashish8 wrote: How is B sufficient? Sum of distinct factors of a perfect square is odd, but if n is 2, then the sum is also odd. Also check this: Tips about the perfect square: 1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square; 2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50; 3. A perfect square ALWAYS has an ODD number of Oddfactors, and EVEN number of Evenfactors. The reverse is also true: if a number has an ODD number of Oddfactors, and EVEN number of Evenfactors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors); 4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even. Hope it helps.
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Re: perfect square [#permalink]
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31 Jul 2012, 23:19
cipher wrote: Pkit wrote: Please help me. This is the question of 700 level from MGMAT CAT1 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. I have got this question wrong , but I would argue with the OA provided by MGMAT. . Please provide your reasons and explanations. Thank you. 1) The number of distinct factors of N is even. Suppose N = 4. It has 3 distinct factors: 1, 2 and 4. Suppose N = 9. It has 3 distinct factors: 1, 3 and 9. Suppose N = 16. It has 5 distinct factors: 1, 2, 4, 8, and 16. Suppose N = 64. It has 7 distinct factors: 1, 2, 4, 8, 16, 32, and 64. But that not the case. In fact, the case is opposite. So it is sufficient because N is not a square. (2) The sum of all distinct factors of N is even. If you follow the above pattern, you see 1 is always there. The sum of all distinct factors except 1 of N is even. If you add 1 on the even sum, that odd. So N is not a square. But that not the case. In fact, the case is opposite. So it is sufficient because N is again not a square. So D. PS: A perfect square always have odd number of factors, for e.g a integer \(n\) and its square \(n^2\) Now, \(n\) will have always have even number of factors, (take any number and you will realise that factors come in pairs), now \(n^2\) will have all factors which \(n\) has + one more which is \(n^2\) Hi there, I have a problem with this method. I think it is flawed but luckily works here. We can see that the two statements should be true for perfect squares, but in no way have we proved that it is not true for nonperfect square. For instance, getting a few examples of perfect squares and seeing that the sum of the factors is always odd, doesn't imply that summing the factors of a nonperfect square would not be odd... The only way to properly answer is to know the properties given by Bunuel IMO.



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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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21 Aug 2014, 01:26
Bunuel wrote: tingle15 wrote: I have a doubt...
Consider N=18, Its factors are: 1, 2, 3, 6, 9, 18. The sum of factors is 39 which is odd... Am i missing something? Tips about the perfect square: 1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square; 2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50; 3. A perfect square ALWAYS has an ODD number of Oddfactors, and EVEN number of Evenfactors. The reverse is also true: if a number has an ODD number of Oddfactors, and EVEN number of Evenfactors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors); 4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even. NEXT:There is a formula for Finding the Number of Factors of an Integer: First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers. The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\) Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors. Back to the original question:Is the positive integer N a perfect square? (1) The number of distinct factors of N is even > let's say \(n=a^p*b^q*c^r\), given that the number of factors of \(n\) is even > \((p+1)(q+1)(r+1)=even\). But as we concluded if \(n\) is a perfect square then powers of its primes \(p\), \(q\), and \(r\) must be even, and in this case number of factors would be \((p+1)(q+1)(r+1)=(even+1)(even+1)(even+1)=odd*odd*odd=odd\neq{even}\). Hence \(n\) can not be a perfect square. Sufficient. (2) The sum of all distinct factors of N is even > if \(n\) is a perfect square then (according to 3) sum of odd factors would be odd and sum of even factors would be even, so sum of all factors of perfect square would be \(odd+even=odd\neq{even}\). Hence \(n\) can not be a perfect square. Sufficient. Answer: D. Hope it helps. Hi, could you explain why " A perfect square ALWAYS has an ODD number of Oddfactors, and EVEN number of Evenfactors" is true? Thanks
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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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24 Aug 2014, 21:49
vad3tha wrote: Hi, could you explain why " A perfect square ALWAYS has an ODD number of Oddfactors, and EVEN number of Evenfactors" is true? Thanks
Here is a post that explains this: http://www.veritasprep.com/blog/2010/12 ... tsquares/
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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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17 Feb 2015, 08:58
Hi @veritasprepkarishma/@Bunuel,
For 2nd statement if we take 1)4perfect squaresum of distinct factors is 2 or 4(2*2 or 4*1) Condition satisfied
2)8not a perfect squaresum of distinct factors is 2 or 8(2*2*2 or 8*1) Condition satisfied still not perfect square
Then how can D be the answer?



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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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17 Feb 2015, 09:39
ssriva2 wrote: Hi @veritasprepkarishma/@Bunuel,
For 2nd statement if we take 1)4perfect squaresum of distinct factors is 2 or 4(2*2 or 4*1) Condition satisfied
2)8not a perfect squaresum of distinct factors is 2 or 8(2*2*2 or 8*1) Condition satisfied still not perfect square
Then how can D be the answer? (2) says that the sum of all distinct factors of N is even. If N = 4, then its factors are 1, 2, and 4 > the sum = 1 + 2 + 4 = 7 = odd. If N = 8, then its factors are 1, 2, 4 and 8 > the sum = 1 + 2 + 4 +8 = 15 = even.
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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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18 Feb 2015, 11:21
goldeneagle94 wrote: Interesting Question !!!
A few facts to review:
A perfect sqaure ALWAYS has an ODD number of factors, whose sum is ALWAYS ODD.
A perfect sqaure ALWAYS has an ODD number of Oddfactors, and EVEN number of Evenfactors.
Using the above facts, you can conclude that both statements are sufficient to answer the question. Perfect Square 36: (6 x 6) (3 X 3 X 2 X 2) 4 total factors, 2 distinct factors, and sum is even…?



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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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18 Feb 2015, 21:33
Hi Kitzrow, You have to note the difference between "factors" and "prime factors" 36 has the following FACTORS: 1 and 36 2 and 18 3 and 12 4 and 9 6 So, there are 9 factors and the sum of those factors is 91. This example matches the prior statements  36 has an ODD number of factors and the sum of those factors is ODD. GMAT assassins aren't born, they're made, Rich
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Re: Is the positive integer N a perfect square? [#permalink]
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02 Nov 2015, 12:12
Question here:
I thought perfect squares always have an even sum of powers of prime factors? For instance, in the example, 36 = 2^2 * 3^2, powers of 2 + 2 = 4. Yet when you look at the total number of factors (2+1) * (2+1) you get 9...and the explanation nulls statement 1 because there are an odd number of factors... Can anyone elaborate on this? Thank you!



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Re: Is the positive integer N a perfect square? [#permalink]
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02 Nov 2015, 12:18
HunterJ wrote: Question here:
I thought perfect squares always have an even sum of powers of prime factors? For instance, in the example, 36 = 2^2 * 3^2, powers of 2 + 2 = 4. Yet when you look at the total number of factors (2+1) * (2+1) you get 9...and the explanation nulls statement 1 because there are an odd number of factors... Can anyone elaborate on this? Thank you! You are missing 1 important point. When you look at number of factors of a perfect square you do ALL factors including 1 and the number itself. Example, 25 = 5^2 > total number of factors = (2+1) =3 , (1,5,25). You can not just add the powers of perfect squares to get the total number of factors. Statement 1 is sufficient as it gives a straight "no" to the question" is n a perfect square" as all perfect squares will have odd number of total factors. Hope this helps.



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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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26 Sep 2017, 03:42
mbaMission 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. Nice explanation Bunuel we could also arrive at those rules by testing square roots and knowing the distinct factor equation Statement 1What Bunuel is demonstrating is that this condition cannot allow a perfect square the number of distinct factors of a number is found by taking the prime factorization of the number and then adding 1 to all the exponents and then multiplying the product being the number of distinct factors. Notice the numbers 9, 49, 100 3^2 = 9 3^(2+1) = 3 distinct factors: 9, 3 ,1 49= 7^2 7^(2+1) = 3 distinct factors :1,7,49 10^2= 100 5^2 2^2= 100 5^(2+1) 2^(2+1) = 9 distinct factors: 100, 50, 25, 20, 10 , 5, 4 ,2 ,1 So we can clearly see any number with an even number of distinct factors cannot be a perfect square all perfect squares will have an odd number of distinct factors suff Statement 2We can just try out a few small values such as 25, 9 , 49  clearly the sum of all distinct factors of any perfect square will be odd suff D




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