# How do I find the sum of prime factors of $(1750 + 1225)^{1229}$?

The number is: $(1750 + 1225)^{1229}$

Thanks.

• The prime factors are 1229 repetitions of the prime factors of the base (1750+1225) = 2975. So start by factoring that into primes. Nov 21, 2013 at 12:45
• Also, it's unclear whether you want just a sum of distinct prime factors, or a sum with prime factors added according to multiplicity. Nov 21, 2013 at 12:46
• Figured it out! factored each side of the statement (1750 and 1225). then I factored each side added up (so factor(factor + factor)) then I just had (5^2*7(17))^1229 5^2258*7^1129*17*1129 = M 5+7+17 = 29.
– bert
Nov 21, 2013 at 12:52
• @bert, is there a typo in the problem statement? In your comment you are using $1129$ and $2258 = 2 \cdot 1129$, but the question has $1229$ - not that it changes the principles though. Nov 21, 2013 at 13:29

Assuming, given the lack of clarity, that you need to find the sum of distinct prime factors (disregarding multiplicity of any given prime factor), find the prime factors of the sum $$1750 + 1225 = 2975$$

$$(1750 + 1225)^{1229} = (2975)^{1229} = (5^2 \cdot 7 \cdot 17)^{1229}= \left(5^2\right)^{1229}\cdot (7)^{1229}\cdot (17)^{1229}$$

Now you need to determine whether you need to

• sum $2\cdot 1229$ factors of 5, and $1229$ factors of $7$, and $1229$ factors of $17$

• or simply sum the distinct prime factors $5$ and $7$ and $17$: In the latter case, your sum will be $5 + 7 + 17 = 29$.

Either way, you'll have your result.

• I think you have an arithmetic error. $5^3 \cdot 41$ is $5125$; $2975$ is $5^2 \cdot 7 \cdot 17$. Nov 21, 2013 at 13:26
• Corrected, @half-integerfan. Thanks: I transposed incorrectly from the scratch paper I had written on. Nov 21, 2013 at 13:31