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

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

Guest Feb 5, 2021

#1**0 **

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.

Guest Feb 5, 2021