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.

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$1767=3\cdot 19\cdot 31$ and $5^27^2=1225$ – draks ... Jun 26 '12 at 9:59
$2450=2.5^2.7^2$.therefore,$Sum=(1+2)(1+5+25)(1+7+49)=5301$.Isn't it the right sum?(Because i know the formula is correct). – Aang Jun 26 '12 at 10:03

Your approach works fine: $2450=2*5^2*7^2$, 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 $$\sum_{d|n}d = \prod_{i=1}^k \frac{p_i^{m_i+1}-1}{p_i-1}$$

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Mathworld knows more about Sums of Prime Factors, if you are interested... – draks ... Jun 26 '12 at 10:15

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 = $(5^3 – 1)/(5 – 1) × (7^3 – 1)/(7 – 1)$
= $(124/4) × (342/6)$
= $(31 × 57)$
= $1767$

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$2450=2\cdot5^2\cdot7^2$, 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 $2450$.

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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))}/(3-1)]=28

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