There’s a popular story that Gauss, mathematician extraordinaire, had a lazy teacher. The so-called educator wanted to keep the kids busy so he could take a nap; he asked the class to add the numbers 1 to 100.

Gauss approached with his answer: 5050. So soon? The teacher suspected a cheat, but no. Manual addition was for suckers, and Gauss found a formula to sidestep the problem:

Let’s share a few explanations of this result and really understand it intuitively. For these examples we’ll add 1 to 10, and then see how it applies for 1 to 100 (or 1 to any number).

## Technique 1: Pair Numbers

Pairing numbers is a common approach to this problem. Instead of writing all the numbers in a single column, let’s wrap the numbers around, like this:

```
1 2 3 4 5
10 9 8 7 6
```

An interesting pattern emerges: **the sum of each column is 11**. As the top row increases, the bottom row decreases, so the sum stays the same.

Because 1 is paired with 10 (our n), we can say that each column has (n+1). And how many pairs do we have? Well, we have 2 equal rows, we must have n/2 pairs.

which is the formula above.

## Wait — what about an odd number of items?

Ah, I’m glad you brought it up. What if we are adding up the numbers 1 to 9? We don’t have an even number of items to pair up. Many explanations will just give the explanation above and leave it at that. I won’t.

Let’s add the numbers 1 to 9, but instead of starting from 1, let’s count from 0 instead:

```
0 1 2 3 4
9 8 7 6 5
```

By counting from 0, we get an “extra item” (10 in total) so we can have an even number of rows. However, our formula will look a bit different.

Notice that each column has a sum of n (not n+1, like before), since 0 and 9 are grouped. And instead of having exactly n items in 2 rows (for n/2 pairs total), we have n + 1 items in 2 rows (for (n + 1)/2 pairs total). If you plug these numbers in you get:

which is the same formula as before. It always bugged me that the same formula worked for both odd and even numbers – won’t you get a fraction? Yep, you get the same formula, but for different reasons.

## Technique 2: Use Two Rows

The above method works, but you handle odd and even numbers differently. Isn’t there a better way? Yes.

Instead of looping the numbers around, let’s write them in two rows:

```
1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
```

Notice that we have 10 pairs, and each pair adds up to 10+1.

The total of all the numbers above is

But we only want the sum of one row, not both. So we divide the formula above by 2 and get:

Now this is cool (as cool as rows of numbers can be). It works for an odd or even number of items the same!

## Technique 3: Make a Rectangle

I recently stumbled upon another explanation, a fresh approach to the old pairing explanation. Different explanations work better for different people, and I tend to like this one better.

Instead of writing out numbers, pretend we have beans. We want to add 1 bean to 2 beans to 3 beans… all the way up to 5 beans.

```
x
x x
x x x
x x x x
x x x x x
```

Sure, we could go to 10 or 100 beans, but with 5 you get the idea. How do we count the number of beans in our pyramid?

Well, the sum is clearly 1 + 2 + 3 + 4 + 5. But let’s look at it a different way. Let’s say we mirror our pyramid (I’ll use “o” for the mirrored beans), and then topple it over:

```
x o x o o o o o
x x o o x x o o o o
x x x o o o => x x x o o o
x x x x o o o o x x x x o o
x x x x x o o o o o x x x x x o
```

Cool, huh? In case you’re wondering whether it “really” lines up, it does. Take a look at the bottom row of the regular pyramid, with 5′x (and 1 o). The next row of the pyramid has 1 less x (4 total) and 1 more o (2 total) to fill the gap. Just like the pairing, one side is increasing, and the other is decreasing.

Now for the explanation: How many beans do we have total? Well, that’s just the area of the rectangle.

We have n rows (we didn’t change the number of rows in the pyramid), and our collection is (n + 1) units wide, since 1 “o” is paired up with all the “x”s.

Notice that this time, we don’t care about n being odd or even – the total area formula works out just fine. If n is odd, we’ll have an even number of items (n+1) in each row.

But of course, we don’t want the total area (the number of x’s and o’s), we just want the number of x’s. Since we doubled the x’s to get the o’s, the x’s by themselves are just half of the total area:

And we’re back to our original formula. Again, the number of x’s in the pyramid = 1 + 2 + 3 + 4 + 5, or the sum from 1 to n.

## Technique 4: Average it out

We all know that

`average = sum / number of items`

which we can rewrite to

`sum = average * number of items`

So let’s figure out the sum. If we have 100 numbers (1…100), then we clearly have 100 items. That was easy.

To get the average, notice that the numbers are all equally distributed. For every big number, there’s a small number on the other end. Let’s look at a small set:

```
1 2 3
```

The average is 2. 2 is already in the middle, and 1 and 3 “cancel out” so their average is 2.

For an even number of items

```
1 2 3 4
```

the average is between 2 and 3 – it’s 2.5. Even though we have a fractional average, this is ok — since we have an **even** number of items, when we multiply the average by the count that ugly fraction will disappear.

Notice in both cases, 1 is on one side of the average and N is equally far away on the other. So, we can say the average of the entire set is actually just the average of 1 and n: (1 + n)/2.

Putting this into our formula

And voila! We have a fourth way of thinking about our formula.

## So why is this useful?

Three reasons:

1) Adding up numbers quickly can be useful for estimation. Notice that the formula expands to this:

Let’s say you want to add the numbers from 1 to 1000: suppose you get 1 additional visitor to your site each day – how many total visitors will you have after 1000 days? Since thousand squared = 1 million, we get `million / 2 + 1000/2 = 500,500`

.

2) This concept of adding numbers 1 to N shows up in other places, like figuring out the probability for the birthday paradox. Having a firm grasp of this formula will help your understanding in many areas.

3) Most importantly, this example shows there are many ways to understand a formula. Maybe you like the pairing method, maybe you prefer the rectangle technique, or maybe there’s another explanation that works for you. **Don’t give up** when you don’t understand — try to find another explanation that works. Happy math.

By the way, there are more details about the history of this story and the technique Gauss may have used.

## Variations

**Instead of 1 to n, how about 5 to n?**

Start with the regular formula (1 + 2 + 3 + … + n = n * (n + 1) / 2) and subtract off the part you don’t want (1 + 2 + 3 + 4 = 4 * (4 + 1) / 2 = 10).

```
Sum for 5 + 6 + 7 + 8 + … n = [n * (n + 1) / 2] – 10
```

And for any starting number a:

```
Sum from a to n = [n * (n + 1) / 2] – [(a - 1) * a / 2]
```

We want to get rid of every number from 1 up to a – 1.

**How about even numbers, like 2 + 4 + 6 + 8 + … + n?**

Just double the regular formula. To add evens from 2 to 50, find 1 + 2 + 3 + 4 … + 25 and double it:

```
Sum of 2 + 4 + 6 + … + n = 2 * (1 + 2 + 3 + … + n/2) = 2 * n/2 * (n/2 + 1) / 2 = n/2 * (n/2 + 1)
```

So, to get the evens from 2 to 50 you’d do 25 * (25 + 1) = 650

**How about odd numbers, like 1 + 3 + 5 + 7 + … + n?**

That’s the same as the even formula, except each number is 1 less than its counterpart (we have 1 instead of 2, 3 instead of 4, and so on). We get the next biggest even number (n + 1) and take off the extra (n + 1)/2 “-1″ items:

```
Sum of 1 + 3 + 5 + 7 + … + n = [(n + 1)/2 * ((n + 1)/2 + 1)] – [(n + 1) / 2]
```

To add 1 + 3 + 5 + … 13, get the next biggest even (n + 1 = 14) and do

```
[14/2 * (14/2 + 1)] – 7 = 7 * 8 – 7 = 56 – 7 = 49
```

**Combinations: evens and offset**

Let’s say you want the evens from 50 + 52 + 54 + 56 + … 100. Find all the evens

```
2 + 4 + 6 + … + 100 = 50 * 51
```

and subtract off the ones you don’t want

```
2 + 4 + 6 + … 48 = 24 * 25
```

So, the sum from 50 + 52 + … 100 = (50 * 51) – (24 * 25) = 1950

Phew! Hope this helps.

Ruby nerds: you can check this using

```
(50..100).select {|x| x % 2 == 0 }.inject(:+)
1950
```

## Leave a Reply

304 Comments on "Techniques for Adding the Numbers 1 to 100"

Here is a more generic way to think about this that lets you calculate any equally spaced series of numbers:

1+2+3+4+5 = 15

n = number of digits in the set

a = the first digit in the set

b = the last digit in the set

(n(a+b))/2 = total

(5(1+6))/2 = 15

1 3 5 7 9 = 25

(5(1+9))/2 = 25

The explanation is really the same as the first explanation above. The difference is that we don’t automatically add 1, we add the first number in the set. Adding the first number to the last number is how we “pair” each number together. Pairing means to group each number in twos such that all pairs sum to the same number (which is why this only works on equally spaced digits).

3 5 7 9 11 = 35

(5(3+11))/2 = 35

5 9 13 = 27

(3(5+13))/2 = 27

Thanks alot

You’re welcome Sandra — I’m always interested in finding several explanations for an idea and seeing which ones work for different people.

This is great. It is exactly what I wanted. A big thank you

I don’t think it helped me AT ALL!

[…] But we remember that adding the numbers 1 to n = n(n + 1)/2. Don’t confuse this with n(n-1)/2, which is C(n,2) or the number of pairs of n items. They look almost the same! […]

I know one idea but I don’t know anyone if invented are not

Kalid

I am thoroughly enjoying your posts. Mathematics sure looks a lot less boring with your intuitive approach – as opposed to rote memorization. Years ago I learned error checking by “casting out 9’s” and “casting out 11’s”. Could you do a post explaining “how and why” these work? I thought it might provide some good insight into our number system.

Thanks

Hi xilplaxim, thanks for the insight! Yep, you can extend the explanations above to almost any sequence. Here’s something interesting as well that you made me think of — let’s say your pattern is

1 2 _ 4 5 _ 7 8 _ 10 11 _ 13 14 …

(multiples of 3 are intentionally removed). That seems like a tricky pattern to add up, but you can realize it’s the same as the full pattern

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

minus the “holes”

3 6 9 12 15

You can use the formula to find the full pattern (1 to 15 = 15*16/2 = 120) and subtract off the holes using your formula (3+6+9+12+15) = 5*18/2 = 45). So we do 120 – 45 and get 75, which is the sum (details here).

Your pattern formula can help subtract away any “gap” you want, which is pretty interesting! Thanks again for the note.

When I was 12 or 13 (about 3 years ago), I found this way to derive the formula, which is quite similar but not the same as your triangle method.

x

xx

xxx

xxxx

xxxxx

The number of Xs is the sum of 1, 2, 3, 4, and 5. Here’s how I derived the formula (Xs are the important elements of the triangle, Os are used to show what I don’t count, parentheses are used to show what I eliminate, and N is the number you want the sum up to).

xxxxx N

This is the number you want the sum of.

xoooo

xxooo

xxxoo N^2

xxxxo

xxxxx

Squaring N, you get the triangle and some extra elements (which need to be eliminated)

xoooo

xxooo

xxxoo N^2-N

xxxxo

(xxxxx)

There are no Os you need to eliminate in the bottom row, so you can remove it and add it back later.

x(oooo)

xx(ooo)

xxx(oo) (N^2-N)/2

xxxx(o)

(xxxxx)

Half of the elements above the bottom row are Os, so dividing by two gives you the number of Xs.

x(oooo)

xx(ooo)

xxx(oo) (N^2-N)/2+N

xxxx(o)

xxxxx

Add the original bottom row. You now have the triangle.

Time for my favorite part: changing how an equation looks. (N^2-N)/2+N is too sloppy.

(N^2-N)/2+N Original Equation

N^2/2 -N/2+N Distribute

N^2/2+N/2 Add -N/2 to N

(N^2+N)/2 Factor out 1/2

(N)(N+1)/2 Factor out N

That’s how I thought about it at the time, when I was bored one night. Yes, I developed theorems when I was 13. I still haven’t stopped. I like that way of looking at it, and when I showed that to my math class, most everyone understood what I was doing. Just goes to show you that there are no end to the ways that you can derive a formula.

Thank you so much for that explanation. I had learned this same formula in my Maths class without knowing how it had been derived. I had to use this formula to find the sum of series in arithmetic progression, and now, I wonder why I hadn’t seen this site before.

Keep up the good work!!

Thanks alot :)

I love mathematics so much! :) Thanks for explaining this so perfectly!

Hey! Thanks a lot! You saved my exam!

……and me too!

Elegant and satisfying!

The link to the story in American Scientist is outdated; the current link is this.

@Anonymous, Rohit: Thanks!

@elgeo: Thanks for kind words — just fixed up the link.

For adding the odds (whether the no of numbers is odd or even ) this also works well. Using the same anology given by you,

1+3+5+…..n= (n+1)^(2)/4

For e.g.

1+2+3+…+100 =100*101/2 = 5050

2+4+6+ +100 = 2*(1+2+…+50)=2*50*51/2= 2550

1+3+5+….+99 = (99+1)^(2)/4=100*100/4 = 2500

i have a question any one can help me

adding n numbers formula x = n(n+1)/2

please give the formula to identify value of n if x is given

for n= ????

Please it urgent

[…] Intro: Mental math shortcuts, adding 1-100, how to learn math, understanding averages […]

thank you, that helped quite a bit! :)

thank you soooo much!!! u helped me a lot!!!!

@Magdi, Maria, Mae: Glad it helped!

[…] Jason Sadler has been selling the upper part of his wardrobe ever day of 2009 to companies that want him to wear a t-shirt with the logo on it. His pricing structure is very interesting though. He’s sold January 1st for $1 and is selling December 31st for $365. Every day in between goes for the price of it’s day of the year. I wasn’t to excited about that until I calculated (using Gauss’s method) how much money he’d be earning for the year: $66,795! […]

My question is how to add/count every “1” found between 0 and any end point. End Point could be 10,000. Thanks.

I stumbled across your page while looking for an answer to the same question posed in #30, but was intrigued by all the methods described here. I was trying to figure out a way to map a series of numbers (say 1 through 15) to a smaller result set of numbers (say 1 through 5) such that:

1 => 1

2,3 => 2

4,5,6 => 3

7,8,9,10 => 4

11,12,13,14,15 => 5

as I was pondering how to write a mathematical function to get a result (instead of doing brute iteration) I realized that this smacked of Gauss. Not being able to memorize formulas, I sat down with a piece of paper and tried summing 1 through 100. I cut the set in half, yielding 1 through 50 and 51 through 100. The two ‘half’ sets match such that you can make a pair using one number from each half-set that adds up to 101 (1 + 100, 2 + 99, 3 + 98, etc). There are exactly 50 such pairs (100/2), so the sum must be 101*50. Making a formula, this makes (X+1)*(X/2) which is the same as ((X+1)*X)/2, which brings us back into familiar territory.

Thanks for an interesting article, and for reassurance that the math actually works.

now … how to invert a parabolic function … *heads back to google*

this is amazing now i feel like einstien