Counting Factors

How many natural number factors are there of 560?


The answer is 20.

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5 solutions

Yan Yau Cheng
Apr 14, 2014

The Prime Factorisation of 560 is: 560 = 2 4 × 5 × 7 560 = 2^4\times5\times7

Each factor of 560 can either have no factors of 2, 1 factor of 2, 2 factors of 2, 3 factors of 2, or 4 factors of 2, so there are 5 possibilities. There could either be 1 factor of 5 or no factors of 5 so there are 2 possibilities. There is either 1 factor of 7 or no factors of 7, so there are 2 possibilities. So there are 5 × 2 × 2 = 20 5\times2\times2 = \boxed{20} factors of 560

We could write all the natural factors of n as part of the function d(n). To find the number of factors we would simply add one onto all the powers of the prime factorization of 560. This works out as (4+1)(1+1)(1+1) = 20

Elliott Macneil - 7 years, 1 month ago

is 2^4 * 5 * 7 is the formula?

Khoo Siang - 6 years, 10 months ago
Ángela Flores
Apr 20, 2014

560 can be written as 2^4 * 5 * 7, then the divisora are given by (e1+1)(e2+1)(e3+1) with e1, e2, e3 the exponentes of 2, 5 and 7 respectively. Then, we have 5 2 2=20 and this is the number of divisors

Gautam Sharma
Jul 25, 2014

factorise 560 in powers of primes and then add 1 to the powers and multiply them

Hassan Salim
May 12, 2014

i wrote a program in c++

include<iostream.h>

include<conio.h>

void main() { clrscr(); int i,c=0,n; cout<<"enter no"; cin>>n; for(i=n;i>0;i--) { if(n%i==0) { c++; } } cout<<c; getch(); }

Thanh Viet
May 10, 2014

List out!! :)

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