Matematické Fórum

stuart clark · 11. 10. 2014 15:06

The no. of ways of factorize $1400$ as a product of $2$ factors, are

Brano · 21. 10. 2014 13:19

it is the same as the number of divisors of $1400$

stuart clark · 22. 10. 2014 07:28

Yes Brano.

Like for $18 = 2\times 3^2 = (1\times 18) = (6\times 3) = (9\times 2)$

Thanks

Brano · 22. 10. 2014 09:46

↑ stuart clark:
Yes you're right, order shouldn't matter so if $d(n)$ is the number of divisors, then $d(n)/2$ rounded up is the answer. Note that you will round up only in the case when $d(n)$ is odd, i.e. $n=k^2$ and you are supposed to include $k\times k$ .

stuart clark · 22. 10. 2014 10:01

Thanks ↑ Brano: Got it.

$1400 = 2^3\times 5^2 \times 7$ . So no. of ways in which we can factorise into product of $2$ factors is $= \frac{(3+1)\times (2+1)\times (1+1)}{2} = 12$

Matematické Fórum

#1 11. 10. 2014 15:06

ways to factorize 100 as a product of 2 factors

#2 21. 10. 2014 13:19

Re: ways to factorize 100 as a product of 2 factors

#3 22. 10. 2014 07:28

Re: ways to factorize 100 as a product of 2 factors

#4 22. 10. 2014 09:46

Re: ways to factorize 100 as a product of 2 factors

#5 22. 10. 2014 10:01

Re: ways to factorize 100 as a product of 2 factors

Zápatí