GEP Questions: The Gauss Trick

We will continue our series on GEP Questions. To learn more about Recommended Books for GEP, to practice GEP Questions, visit the link here.

Today, we will discuss the quintessential GEP Question: The Gauss Trick. This GEP question illustrates the fact that giftedness can be trained to a large extent.


Find the sum of 1+3+5+7+…+95+97+99.

Solution will be below after this text, scroll down after you are ready to see the answer!

If a 9 year child can solve this on his/her first attempt (i.e. see the question for the very first time), then the child is actually at the level of Carl Friedrich Gauss, the legendary mathematician!  When Gauss was a young kid, his teacher set the class a difficult question, 1+2+3+…+99+100, hoping to keep the children quiet and busy while he could have some time to relax. Little did he expect Gauss to come up with the right answer minutes later.

Most 9 year old kids would not be able to solve this on their first try. I wouldn’t be able to solve it correctly even if given an hour when I was a kid! If the children have seen it before and practiced, that is a different story, as it becomes so easy even for a 7 year old kid. Hence, practicing GEP Questions actually leads to an indirect boost of IQ in this manner! The transition from ignorance to knowledge, leading to increased intelligence, can be accomplished by practice! To practice more of these GEP Questions, check out the Math Olympiad Recommended Books page. As a tutor, I know that this Gauss Trick is a “must-know” question for students aiming high for school Math / GEP Selection Test, since almost every kid knows this nowadays, and it is highly popular in tests.

In fact, this technique (Gauss Trick) commonly tested in GEP Questions can be used all the way to JC and beyond! In JC it is covered under the topic of AP/GP (Arithmetic Progression and Geometric Progression).


Now, for the solution. This sum is the famous arithmetic progression, where each term differs from the next by a fixed constant. In this case, the constant is 2.

Some solutions use a number pair matching, which can be problematic to explain when the number of terms is odd, but still works nevertheless. We can use another method of writing the sum backwards.

First we note that there are 50 terms in this series. We can know that either by noting that they are the odd numbers from 1 to 100, and half of the numbers are odd, hence there are 100/2=50 terms. Alternatively, we can note that 1=2(1)-1, 3=2(2)-1, 5=2(3)-1, …, 99=2(50)-1, where the number in the brackets acting like a counter.


Note that each pair, 1+99, 3+97, 5+95 actually add up to the same thing, i.e. 100.

Adding up the two expressions above, we get 100×50=5000.

Dividing that by two (since we double counted), we will get 5000/2=2500.

Ans: 2500

Hope you enjoyed solving this question!

Check out this amazing book on Gauss: The Prince of Mathematics: Carl Friedrich Gauss

Gauss is a true Math genius, and you can read more about his life in this interesting biography! This historical narrative will inspire young readers and even curious adults with its touching story of personal achievement.


Author: mathtuition88

Math and Education Blog

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