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.
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.