can someone please describe me why only the perfect square has odd number of factors.why does other number not has odd numbers of factors? I understand it but don't find any mathmetical proof.Please help me

Suppose we have a factor \(n\) of the number \(N\). Then \(N/n\) is also a factor of the number \(N\).

These factors are different, unless \(n = N/n\), or \(N=n^2\).

So unless \(N\) is a square, every factor \(n\) can be paired with \(N/n\) and thus there is an even number of factors.

If \(N=n^2\), then every factor \(m\) (\(m \neq n\)) can be paired with \(N/m\). Adding the factor \(n\) that can't be paired with a different factor, we have an odd number of factors.

Then the prime factorization of N is a product of primes with even powers. Thus the total number of factors is (even+1)(even+1)....(even+1), which is odd.

( I dont know how to format power terms here, otherwise i would have given you a nice complete solution. Hope this works.)

I did factorizing this way so that the number on the left side (or right side) of the multiplication sign is a factor of the number. It can be easily seen that 24 has even number of factors.
You see, at one step of factorizing 36, we come across a term with same numbers on both sides of the multiplication sign (\(6 \times 6\)) which when reversed makes no difference. Hence, you can see that 36, a square number, has odd number of factors.

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TopNewestSuppose we have a factor \(n\) of the number \(N\). Then \(N/n\) is also a factor of the number \(N\).

These factors are different, unless \(n = N/n\), or \(N=n^2\).

So unless \(N\) is a square, every factor \(n\) can be paired with \(N/n\) and thus there is an even number of factors.

If \(N=n^2\), then every factor \(m\) (\(m \neq n\)) can be paired with \(N/m\). Adding the factor \(n\) that can't be paired with a different factor, we have an odd number of factors.

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Hai Mashrur.

Let N be a perfect square.

Then the prime factorization of N is a product of primes with even powers. Thus the total number of factors is (even+1)

(even+1)....(even+1), which is odd.( I dont know how to format power terms here, otherwise i would have given you a nice complete solution. Hope this works.)

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Just look at the following examples: \(24\) is not a square number but \(36\) is a perfect square number.

\(24=1 \times 24\)

\(24=2 \times 12\)

\(24=3 \times 8\)

\(24=4 \times 6\)

\(24=6 \times 4\)

\(24=8 \times 3\)

\(24=12 \times 2\)

\(24=24 \times 1\)

and.........................................................

\(36=1 \times 36\)

\(36=2 \times 18\)

\(36=3 \times 12\)

\(36=4 \times 9\)

\(36=6 \times 6\)

\(36=9 \times 4\)

\(36=12 \times 3\)

\(36=18 \times 2\)

\(36=36 \times 1\)

I did factorizing this way so that the number on the left side (or right side) of the multiplication sign is a factor of the number. It can be easily seen that 24 has even number of factors. You see, at one step of factorizing 36, we come across a term with same numbers on both sides of the multiplication sign (\(6 \times 6\)) which when reversed makes no difference. Hence, you can see that 36, a square number, has odd number of factors.

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Gaussian pairing solves this problem before one can spell 'brilliant'

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Lets have an example,

9=3x3 if you are doing a factor tree, you cant write the same number twice.

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