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can someone please describe me why just the perfect square has actually odd number of factors.why does other number not has actually odd number of factors? I understand it yet don"t find any sdrta.netmetical proof.Please assist me

For a given number $n$ we can group its divisors in bag $(d,frac nd)$, other than that if $n=m^2$ this would pair $m$ with itself.

You can constantly list the components of a number, N, into pairs $(a_i,b_i)$ where $a_i le sqrt N le b_i$. This means that a number will always have an even variety of factors, uneven the number is a perfect square, in which instance one pair will consists of the exact same two numbers. The two examples listed below should show why.

eginalign extfactors &; extof $36 = 6^2$ \ hline 1 &,, 36 \ 2 &,, 18 \ 3 &,, 12 \ 4 &,, 9 \ 6 &,, 6 & extA total of $9$ factors \ hlineendalign

eginalign extfactors &; extof $12$ \ hline 1 &, 12 \ 2 &, 6 \ 3 &, 4 & extA total of $6$ factors \ hlineendalign

I agree that it is counter-intuitive. The capture is that when "Factors" is written, generally what is actually meant is "unique factors". Therefor if a number is a perfect square, that will have actually an even number of total factors, however odd number of UNIQUE factors, i beg your pardon is what is truly meant.

E.g. 36: 1x36, 2x18, 3x12, 4x9, 6x6 - 10 total numbers yet only 9 unique ones. For funny we simply decide to overlook the 2nd "6".Hope that helps.

Well, without the square

**root**, the square

**no.**would certainly be even. Because the square

**root**is multiplied by

*itself*, climate there is just 1 more factor, not 2.

*Example*: 64. Has actually 6 components (excluding 8 (squared)). V 8, there is currently 7 factors... But not eight because the **root** (in this case, 8) will be repeated, and there is just no allude in repeating it.

See more: What Is The Bond Order Of Hi? Why Bond Strength In This Order

I expect your inquiry is answered! :)

Prove the for $ninBbbZ^+$, $n$ has an odd variety of positive divisors if and also only if $n$ is the square of an integer.

If a perfect square $N$ has actually nine factors less 보다 its square root, how countless factors walk $N^5$ have?

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