Finding sum of factors of a number using prime factorization Given a number, there is an algorithm described here to find it's sum and number of factors.
For example, let us take the number $1225$ :
It's factors are $1, 5, 7, 25, 35, 49, 175, 245, 1225 $
and the sum of factors are $1767$.
A simple algorithm that is described to find the sum of the factors is using prime factorization.
$1225 = 5^2 \cdot 7^2$, therefore the sum of factors is $ (1+5+25)(1+7+49)  = 1767$
But this logic does not work for the number $2450$. Please check if it's working for $2450$
Edit :
Sorry it works for $2450$. I made some mistake in calculation.
 A: Solving for $1225 = 5^2 × 7^2$
Sum of divisors = $(a^{p+1}  – 1)/(a – 1) × (b^{q+1}  – 1)/(b – 1)$
Here $a = 5$, $b = 7$ prime factors
$p = 2$ and $q = 2$
Sum of divisors  = $(5^3 – 1)/(5 – 1) ×  (7^3 – 1)/(7 – 1)$
= $(124/4) ×  (342/6)$
= $(31 × 57)$
= $1767$
A: $2450=2\cdot5^2\cdot7^2$, so the algorithm gives
$$(1+2)(1+5+25)(1+7+49)=3\cdot31\cdot57=5301\;.$$
The divisors of $2450$ are $1,2,5,7,10,14,25,35,49,50,70,98,175,245,350,490,1225$, and $2450$, whose sum is indeed $5301$.
It’s not difficult to prove that the algorithm works, so if you thought that it failed, you made a mistake somewhere in your calculations; my guess would be that you missed a divisor of $2450$.
A: If a number $N=(a^x)(b^y)(c^z)$, where $a,b$ and $c$ are prime numbers, sum of it's factor is $S= [(a^{x+1}-1)/(a-1)][(b^{y+1}-1)/(b-1)][(c^{z+1}-1)/(c-1)]$
Eg. $12=[(2^{2+1}-1)/(2-1)][{(3^{1+1}-1)}/(3-1)]=28$
A: Your approach works fine: $2450=2\cdot 5^2\cdot 7^2$, therefore the sum of divisor is $$(1+2)(1+5+25)(1+7+49)=5301=3\cdot 1767.$$
You are looking for the Formula For Sum Of Divisors, from there:

Each of these sums is a geometric series; hence we may use the formula for sum of a geometric series to conclude
$$
\sum_{d|n}d = \prod_{i=1}^k \frac{p_i^{m_i+1}-1}{p_i-1}
$$

A: Sum of factors of  $2450$
The Factors are $2, 5^2, 7^2$.
Sum of the factors $= ( 2^0 + 2^1 ) × ( 5^0 + 5^1+  5^2)×( 7^0 + 7^1+ 7^2)
= 5301$
