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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   !!!!

 

cool cool cool

CPhill  Feb 5, 2021

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