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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