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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 \(n^2\)
Please provide your reasons and explanations. Thank you.
Probably the best way of solving would be making the chart of perfect squares and its factors to check both statements, but below is the algebraic approach if needed.
Couple of things: 1. Note that if \(n\) is a perfect square powers of its prime factors must be even, for instance: \(36=2^2*3^2\), powers of prime factors of 2 and 3 are even.
2. 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.
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. For instance 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).
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.
There are some tips about the perfect square: • The number of distinct factors of a perfect square is ALWAYS ODD. • The sum of distinct factors of a perfect square is ALWAYS ODD. • A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. • Perfect square always has even number of powers of prime factors.
Re: Is the positive integer N a perfect square? [#permalink]
26 Jul 2012, 12:49
Agree with that the solutions presented above could probably be the way to do it in the test but have a few considerations regarding the problem.
1. It says N is positive. Why would we not consider 1 as a test value too? It is positive and is a perfect square. 2. Why would 'distinct factors' not include the negative factors as well? (Which will make 1 have even number of distinct factors 1 and -1)
Please provide your reasons and explanations. Thank you.
Probably the best way of solving would be making the chart of perfect squares and its factors to check both statements, but below is the algebraic approach if needed.
Couple of things: 1. Note that if \(n\) is a perfect square powers of its prime factors must be even, for instance: \(36=2^2*3^2\), powers of prime factors of 2 and 3 are even.
2. 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.
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. For instance 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).
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.
There are some tips about the perfect square: • The number of distinct factors of a perfect square is ALWAYS ODD. • The sum of distinct factors of a perfect square is ALWAYS ODD. • A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. • Perfect square always has even number of powers of prime factors.
Hope it helps.
Thanks Bunuel for reminding us the very useful properties mentioned above.
For those interested, we can easily justify why the sum of the distinct divisors of a perfect square is odd.
If the number N is an odd perfect square, then all its divisors are odd. They come in pairs, \((1,N), (d_1, N/d_1), (d_2,N/d_2)...\), except \(\sqrt{N}\) (we count it only once, its pair being itself). So, we have an odd number of odd divisors, whose sum will certainly be odd.
If N is an even perfect square, then N must be of the form \(N=2^{2n}M\), where M is an odd perfect square. All the odd divisors of N are the divisors of M, and as we have seen above, their sum (and number) is odd. All the even divisors of N are obviously even, so again, the sum of all the divisors is odd. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: Is the positive integer N a perfect square? [#permalink]
26 Jul 2012, 13:39
But in this particular question, it has not been specified if N is not equal to one. Hence, N could be 1, perfect square, and have even number of factors namely 1 and -1.
Re: Is the positive integer N a perfect square? [#permalink]
27 Jul 2012, 07:37
Expert's post
ShalabhAr wrote:
But in this particular question, it has not been specified if N is not equal to one. Hence, N could be 1, perfect square, and have even number of factors namely 1 and -1.
Posted from my mobile device
Factor is a "positive divisor" (at least on the GMAT). For example the factors of 4 are 1, 2, and 4 ONLY. _________________
Re: Is the positive integer N a perfect square? [#permalink]
30 Jul 2012, 23:49
Expert's post
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.
The sum of distinct factors of a perfect square is ALWAYS ODD. (2) says that "the sum of all distinct factors of N is even", hence N is not a perfect square. _________________
Re: Is the positive integer N a perfect square? [#permalink]
31 Jul 2012, 00:26
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. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: Is the positive integer N a perfect square? [#permalink]
31 Jul 2012, 00:32
Expert's post
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 Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors 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.
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 non-perfect 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 non-perfect square would not be odd...
The only way to properly answer is to know the properties given by Bunuel IMO.
Re: Is the positive integer N a perfect square? [#permalink]
23 Nov 2014, 17:24
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Re: Is the positive integer N a perfect square? [#permalink]
02 Nov 2015, 11:12
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!
Re: Is the positive integer N a perfect square? [#permalink]
02 Nov 2015, 11:18
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.
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