A number theory problem by felixjacob magtoto

it's just a simple formula sum=n/2{a+l} where; n=no. of terms a=1st term l=last term sum=99/2{1+99} =>99*100/2 =>4950 ans

By using this simple formula for sum of natural numbers:-

n(n+1)/2

=99*100/2

=99*50

=4950

Or just pick the answer that ends in zero, because if you take the string of numbers and fold them in half, you end up with 50 in the middle and all the rest adding up to 100

we can use the formula for finding sum of n positive integers i.e. n(n+1)/2

$S=1+2+3+4+\ldots+96+97+98+99=\sum_{i=1}^{99} i\\ \implies S=50+\sum_{i=1}^{49} i + \sum_{i=1}^{49} (100-i)\\ \implies S=50+\sum_{i=1}^{49} (i+100-i)\\ \implies S=50+\sum_{i=1}^{49} (100)=50+49\times 100 = \boxed{4950}$

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This can be done orally to solve the problem in $4$ seconds and $4\lt 5$ . So, I win. :P

P.S - One may also use the identity $\displaystyle \sum_{i=1}^n i = \dfrac{n(n+1)}{2}$ and a calculator for faster calculation.

I remember the answer.

get the last number of the equation. then multiply it with the number next to it. lastly, divide it by two. voila!

you can try to solve it with this technique and race with your friends. see who will finish first!