1-1000

Number Theory Level 2

A long time ago, there was a school in China, whose students were always misbehaving. There was a specific class whose students were extremely rambunctious. The teacher of the class was very young, and wasn't good at yelling at children. One day, the teacher had enough. She kept all of the students in and gave them a math problem. None of them were allowed to leave until they solved the problem correctly. She had them add up all the numbers from 1 to 1000. Not even one minute passed when one student stood up and handed the teacher his paper. She looked at his answer, then looked at hers. Everyone expected that the boy's answer was wrong. The teacher looked up at the boy and said. You may leave. Can you figure out the answer?

500500 5001 2001 200200 500051 200021

3 solutions

Mitchell Willemsen
Aug 7, 2018

For any sequence that can be modeled by a polynomial, you can derive a formula for the sum of the values of the polynomial evaluated at the natural numbers.
The sum of all counting numbers up to n is given by $\displaystyle\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$
In this case, $n=1000$ , and $\displaystyle\sum_{k=1}^{1000}k = \frac{1000(1001)}{2} =500(1000)+500=500500$

That is certainly another way to solve it.

Stellina Ao - 2 years, 10 months ago

Stellina Ao
Aug 7, 2018

It is actually very simple. If you think carefully, 1000+1=1001, 999+2=1001, 998+3=1001, and so on. There should be 500 pairs of 1001(since 1000/2=500), so you simply multiply 1001 by 500.

Vishruth K
Mar 30, 2021

There are 500 pairs of 1001, and doing 500 x 1001 gives you 500500 , the answer :)

1-1000

3 solutions

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