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$(1)::$ How Many ways the number $18900$ can be split into $2$ factors which are relatively prime.

$(2)::$ The no. of ways in which the number $94864$ can be resolve as a product of $2$ factors .

$\bf{Solution\; (1)::}$ Given in book:: Given $18900 = 2^2\times 3^3 \times 5^2 \times 7.$

So no. of Different prime factors in $18900$ is $4,$ i.e $2,3,5,7$

So no. of ways in which $18900$ can be resolve $18900$

into product of $2$ relatively prime factors is $\displaystyle = \frac{2^4}{2} = 8$.

$\bf{Solution \; (2)::}$ Given in Book :: Given $94864 = 2^4\times 7^2\times 11^2.$

no. of ways in which the number $94864$ can be resolve as a product of $2$ factors

$\displaystyle = \frac{[(4+1).(2+1).(2+1)+1]}{2} = 23.$

I did not understand the above solutions, Please explain it to me in Detail.

Thanks

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    $\begingroup$ Try $12=2^2\times 3^1$ instead. Find (b) the all pairs of factors and (a) which of these are coprime. Write each number in these pairs as products of primes, and see how the formulae you have been given explain how to count them. $\endgroup$ – Henry Oct 8 '14 at 3:20
  • $\begingroup$ Where in what book? $\endgroup$ – Jonas Meyer Oct 8 '14 at 4:02
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For 1, all the factors of each prime have to stay together, otherwise the two factors will not be coprime. You can consider it to be $18900=4 \times 27 \times 25 \times 7$ One factor will be the product of some subset of $\{4,27,25,7\}$ and the other factor will be all the rest. Each factorization gets counted twice, once when each factor is the first one selected.

For 2, we don't have the requirement of coprime. For the factors of $2$, you can have any number of $2$'s from $0$ through $4$, so there are five choices. This is where the $(4+1)$ comes from. If $94864$ were not a square, the product of the number of factors plus one would be even. Again we would divide by $2$ because you can choose each factorization in two ways. The +1 comes because it is a square, so you can only choose the factorization of two square roots in one way.

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  • $\begingroup$ Thanks Ross Millikan, But I did not understand the line..If $94864$ were not a square, the product of the number of factors plus one would be even. Again we would divide by 2 because you can choose each factorization in two ways., plz explain me , Thanks $\endgroup$ – juantheron Oct 11 '14 at 3:14
  • $\begingroup$ Think of 12 and 16. 12 is not a square, it factors as 2^2\cdot 3^1, so has $(2+1)(1+1)=6$ factors: $1,2,3,4,6,12$. We can therefore split it three ways: pick any of the six for the first factor, that forces the second factor, but you have counted each pair twice. $16$ is a square, it factors as $2^4$, so has $4+1=5$ factors: $1,2,4,8,16$. This is only odd for squares, because there is one factorization that uses the square root (here $4$) twice. So you have five choices for the first factor, but two of the pairs are counted twice. $\endgroup$ – Ross Millikan Oct 12 '14 at 23:58

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