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

My professor did this example in class, but I didn't really understand this. Please help.

Thanks.

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    $\begingroup$ The prime factors are 1229 repetitions of the prime factors of the base (1750+1225) = 2975. So start by factoring that into primes. $\endgroup$
    – hardmath
    Nov 21, 2013 at 12:45
  • $\begingroup$ Also, it's unclear whether you want just a sum of distinct prime factors, or a sum with prime factors added according to multiplicity. $\endgroup$
    – hardmath
    Nov 21, 2013 at 12:46
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    $\begingroup$ 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. $\endgroup$
    – bert
    Nov 21, 2013 at 12:52
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    $\begingroup$ @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. $\endgroup$ Nov 21, 2013 at 13:29

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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.

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

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