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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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08 Jul 2013, 00:53



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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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27 Dec 2013, 09:20
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. Remember two properties 'bout perfect squares The number of distinct factors of N is even, of course one will always need pairs and will always have the factor 1 remaining hence always odd, so the answer is NO, N is not a perfect squares The sum of all distinct factors of N is even, of course, same reason, all the pairs will add up to an even number +1 = odd Hence, D is the correct answer choice Hope it helps Cheers! J



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Re: a perfect square [#permalink]
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07 Feb 2014, 00:13
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. Hello Bunuel What if n=1 ? Question says n is a Positive Integer. is 1 considered a perfect square ? Please clarify. Thankyou



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Re: a perfect square [#permalink]
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07 Feb 2014, 05:21
niyantg wrote: 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. Hello Bunuel What if n=1 ? Question says n is a Positive Integer. is 1 considered a perfect square ? Please clarify. Thankyou Yes, 1 is a perfect square: 1 = 1^1.
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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? (1) The number [#permalink]
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05 Aug 2015, 01:31
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. Bunuel Do you have 510 questions to practice on Perfect Square.
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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
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16 Aug 2015, 11:22
honchos wrote: 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. Bunuel Do you have 510 questions to practice on Perfect Square.
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