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# number theory

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$\large 1+2+3+\cdots+10^{2016}$

How many times appears 2 in the prime factor decomposition of the sum above?

Feb 5, 2021

#1
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The sum of the numbers from 1 to n is $$\frac{n(n+1)}{2}$$.

In this problem, $$n = 10^{2016}$$.

Instead of multiplying everything out, we could realize that $$10^{2016}$$ is even, so $$10^{2016}+1$$ is odd (and therefore will not have any 2's in its prime factorization).

Now we only need to find the number of 2's in $$10^{2016}$$, and get rid of one of them because we divide by 2 in the fraction $$\frac{(10^{2016})(10^{2016}+1)}{2}$$.
$$10^{2016} = 5^{2016} * 2^{2016}$$, so we now know that there are 2016 twos in $$10^{2016}$$. We divide the expression by 2, so one less than 2016 is which 2015.
So, there are 2015 twos in the prime factorization of the sum, and it is our answer. I hope this helps.

Feb 5, 2021
#2
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That's very crafty, guest......excellent   !!!

I definitely would not  have seen that  method   !!!!

CPhill  Feb 5, 2021