Techniques for Adding the Numbers 1 to 100

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:

\displaystyle{\text{Sum from 1 to n} = \frac{n(n+1)}{2}}

\displaystyle{\text{Sum from 1 to 100} = \frac{100(100+1)}{2} = (50)(101) = 5050}

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.

\displaystyle{\text{Number of pairs * Sum of each pair} = (\frac{n}{2})(n+1) = \frac{n(n+1)}{2}}

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:

\displaystyle{\text{Number of pairs * Sum of each pair} = (\frac{n + 1}{2})(n) = \frac{n(n+1)}{2}}

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

\displaystyle{\text{Total = pairs * size of each pair} = n(n + 1)}

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

\displaystyle{\frac{n(n + 1)}{2}}

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

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.

\displaystyle{Area = height \cdot width = n(n+1)}

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:

\displaystyle{\text{Number of xâs} = \frac{Area}{2} = \frac{n(n + 1)}{2}}

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

\displaystyle{\text{sum = average * count } = \frac{(1 + n)}{2} \cdot n = \frac{n(n + 1)}{2}}

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:

\displaystyle{\frac{n(n+1)}{2} = \frac{n^2}{2} + \frac{n}{2} }

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.


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(:+)

Other Posts In This Series

  1. Rethinking Arithmetic: A Visual Guide
  2. Learning How to Count (Avoiding The Fencepost Problem)
  3. Techniques for Adding the Numbers 1 to 100
  4. Surprising Patterns in the Square Numbers (1, 4, 9, 16…)
  5. A Quirky Introduction To Number Systems
  6. Another Look at Prime Numbers
  7. Fun With Modular Arithmetic

Questions & Contributions


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

  2. 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

  3. 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.

  4. can we use the following method to sum when there are odds?
    Instead of 0 can we assume n+1 as our last no i.e.n and afterwards subtract tht no. from the sum.

  5. @Archana: That’s interesting way of looking at it, you can do that too. Let’s say you’re adding 1-5. Rather than n=5, have m=6. Then you can do

    Sum from 1-6 = m * (m+1)/2 = 6*7/2 = 21. And you subtract m to get

    21 – 6 = 15. [which is the sum of 1-5].

    Of course, the formula n(n+1)/2 works no matter if n is even or odd. But that’s a good way to think about the odd case.

  6. Hi PJ, good question. To find the odd numbers from 1-100 (1 to 99, really), I’d use technique #2 (Use two rows):

    1  3  5  7  ... 97 99
    99 97 95 93 ... 3  1

    Notice how each row ads to 100 (n). How many rows do we have? 50. [1-4 is 2 odd numbers, 1-6 has 3 odd numbers… 1-n will have n/2 odd numbers). Lastly, we have twice the amount of numbers we need, so we must divide by 2 again.

    So the sum of odd numbers 1 to n [where n is even] would be n * (n/2) * 1/2. In this case, 100 * 50 * 1/2 = 2500.

  7. 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.


    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.

    xxxoo N^2

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

    xxxoo N^2-N

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

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

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

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

    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.

  8. Hi Zac, thanks for the awesome comment! That’s a cool way to look at it, I like the approach of making a full box (n^2) and then taking pieces away.

    Yep, it goes to show that there are so many ways of looking at a single formula :).

  9. I like your ‘Many explanations will just give the explanation above and leave it at that. I won’t.’

    Kudos on your dedication. I have bookmarked your page for future reference.

    Best regards and good health,

    Leonard Juska
    Costa Mesa, CA USA

    P.S. If there ever is time, I want to revisit a book I read as a child called
    The Trachtenburg Speed System of Basic Mathematics by Ann Cutler and Rudolph McShane of the work by Jakow Trachtenburg. Bantam Books 553 07020 150 First printing 1960; I have a 1973 copy. Intuitive adventure.

  10. @Jenn: Hi Jenn, you could try matching up the numbers like this (using technique #2):

    80 82 84 86 …
    560 558 556 554 …

    and then dividing the sum by 2.

    @Leonard: Thanks for the comment! Glad you enjoyed it; that looks like an interesting book :)

  11. 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!!

  12. 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

  13. 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

  14. @nada: If you want to add 47 to 100, for example, you can do this: Add 1 to 100 (all the numbers: (100 * 101)/2) and the subtract the sum of 1 to 46 (all the numbers you don’t want (46 * 47)/2).

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

  16. 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*

  17. Hi- love this page, very helpful, but I’m stuck trying to figure out how to work out a formula that works out the sum of all the numbers up to that one (can’t explain it properly)…
    1..1 (sum=1)
    2..3 (sum=3+1=4)
    3..5 (sum=5+3+1=10)
    4..10 (sum=10+5+3+1=20)
    5..15 (sum=35)
    6..21 (sum=56)

    so when n=6 the answer is 56, how do you work out a formula for when n=12 or 112?
    Help please, I’m going stir-crazy!

  18. Just cheated and worked it out with spreadsheet:
    Would still like to know how to get to this equation without cheating though….

  19. Thank you very much! I will be using this story in an introductory class to programming, and the 4 listed techniques will really help in explaining the solution.


  20. Thank you thank you thank you! I’ve heard the story you opened with before, but never knew the answer… I’m soooo glad I stumbled upon your explanation. My life is forever improved!

  21. Here’s a formula I came up with to sum a series of evenly spaced values. You do not need to know the number of values in the set.

    a = the first digit in the set
    b = the last digit in the set
    x = difference between two consecutive numbers

    sum = [a^2 + ax + b^2 + bx] / 2x

  22. If counting by ones…

    sum = n(n + 1)/2

    If counting by twos…

    sum = n(n/2 + 1)/2

    If counting by threes…

    sum = n(n/3 + 1)/2

  23. I don’t know if this is the same thing, but..

    I just found half of x, added .5 to it, and multiplied x with (one half of x + .5)

    Haha. Oh, well.

  24. @sascha: Great question! Actually one way to do this is to add 1 to 100, and then subtract 1 to 39 :). You’ll be left with 40-100. Another way is to start writing the numbers like this:

    40 41 42 43…
    100 99 98 …

    And use the techniques above. Note that 40 to 100 is actually 61 numbers, just like 40 to 41 is 2 numbers. So it’d be 61 * (40 + 100) / 2.

  25. I revising for my Grade 5 maths paper and i really didi not understand the concept and your explanation really helped. Thank you sooo much!!!

  26. Great stuff on this page. I trying to find the sum of the even numbers 6 — 100. Any help would be great.

  27. @C. Mitchell: You can try doing this:

    Sum of evens 6-100 is same as 2 x (sum of 3,4,5,6,… 50).

    And to add 3..50, you can just add 1..50 and remember to subtract off the first 2 numbers (1+2). So 3-50 is [50*51/2 – 3], and multiply the whole thing by 2 to make it 6-100.

  28. Hi im trying to use this equation to ad from 5 to 705 eg
    350 * (700 + 5) = 246,750
    Is this correct?

  29. actually i have my own formula to add all numbers from 2-100 .. w/c is much easy for me..

    heres the formula..


    H=highest number
    L=lowest number
    h=half of the numbers u want to add
    (in “h”..example u want to add all numbers from means u will add 9 numbers…divide it in 2= 4.5)


    add all numbers from 2-10
    (the answer should be 54)



    = 10×4.5+(2×4.5)
    = 45+9
    = 54 . . .


    hope this will help u ^^

  30. While it may be true that you can apply these explanations to other sets, you need not do so for the Fibonacci sequence. There are formulas for the Fibonacci sequence. One of them is easy: to calculate the nth number, raise Phi, the golden number, to the nth power, then divide by the square root of 5. Round the result to the nearest whole number.

  31. @Impro: If you look at the sequence, it’s basically 7 * (1 + 2 + 3 + 4 + 5… 899), so you can reduce it to 7 times the numbers 1…899.

  32. I came up with a similar equation when i was 17 (or so), i forget the details of how i came up with it, but the equation i used was:

    a(a/2 + .5) = 1+2+3+…a

  33. Nice, that’s another way to put it. And yep, they check out:

    a(a/2 + 1/2) = a^2/2 + a/2 = a(a+1)/2

    There must be an intuitive way to see that derivation also :).

  34. I recently had a need to find the number of strokes in a series that went like this:
    and found that the total is equal to the square of the middle number. So if the middle number is 10 then the grand total of them both up and down is 10^2 = 100

  35. Can anyone solve or write any algorithm for the follwing one.

    Split the natural numbers in to two sets whose summation should be equal.
    For example say number range from 1 to 10
    can you split the number from 1 to 10 into two sets as follows whose sum is equal

  36. Can anyone solve or write any algorithm for the follwing one.

    Split the natural numbers in to two sets whose summation should be equal.
    For example say number range from 1 to 10
    can you split the number from 1 to 12 into two sets as follows whose sum is equal
    1+5+9+4+8+12 = 39
    2+6+10+3+7+11= 39

    Can anyone derive formula for this. The above logic works only for the following condition else it will fail

    Condition 1 : When n((n+1)/2) is Odd not application only applicable when the sum is even

    Condition 2 : When the number needs to be split in to 2 sets have a increase uniformly or a series of number starting with 1.

    Question 1 : Derive a formula for a number in the type series

    Question 2 :Derive a formula for a number which is in random fashion say for example

    Awaiting for the reply GUYS!!!!

  37. Can anyone solve or write any algorithm for the follwing one.

    Split the natural numbers in to two sets whose summation should be equal.

    For example
    Say the numbers range from 1 to 12 into two sets as follows whose sum is equal
    1+5+9+4+8+12 = 39
    2+6+10+3+7+11= 39

    Can anyone derive formula for this. The above logic works only for the following conditions else it will fail

    Condition 1 : When n((n+1)/2) is Odd not applicable only applicable when the sum is even

    Condition 2 : When the number that needs to be split in to 2 sets are increasing uniformly or a series of number starting with 1.

    Question 1 : Derive a formula for a number in the type series

    Question 2: Derive a formula for a number which is in random fashion as below.

    Awaiting for the reply GUYS!!!!

  38. how do you add like 1 to 49 for example is it the same pattern like the n(n+1)divided by two used for 1 to 100? and thanks for the info

  39. You could work it out by turning it into an arithmetic sequence.

    Using Sn=n/2[2a+(n-1)d]; where:

    n = number of values
    a = first value
    d = difference between values

    n = 100 a = 1 d = 1


    Easy Peasy!

  40. I look at it this way – 1+2+3+…+98+99+100.
    If sum the first and last digits we get 101. Again if we sum the second and the last but one digits we get 101. This goes on till the midpoint 50+51=101
    Considering 100 as n if we look at the series there are n/2 pairs of n+1. This matches Gauss’s formula.

  41. @Nirmal: Yep, that approach definitely works. The only thing that’s tricky is counting how many pairs… what if you wanted to add 1 to 99?

  42. Better yet, just use X as the number you’re adding to. So if you’re doing 1+2+3 all the way to 100, make x 100 and use this equation.


  43. I made up a similar equation that will also do the summation of the series. It was:

    sum = ((N / 2) * N) + (N / 2)

    What is weird is that you can not do:

    sum = (2N / 2) + (N / 2)
    sum = N + (N / 2)

    Example: 1..100
    sum = (100 / 2) * 100) + (100 / 2)
    sum = 50 * 100 + 50
    sum = 5050

    If you do the combining of the equation you only get 150.

    The above uses the fact that there is a hidden number. Mainly zero(0). So it is 0+100, 1+99, etc…

  44. By the way – I call this an empirical formulae because it can only be used the way it is written. Not if you do the rest of the math and condense it down to a simpler formulae.

  45. I knew it i was right, my story is so much the same as Guass’s, it was a boring math class and the teacher asked us to do the same thing..

    First i thought of it, then i worked on the numbers from 1 to 4 because it was the easiest which gets to 10..

    i was like what am i supposed to apply on the number ‘4’ to get a ’10’ out of it, i did 4*2.5=10 but that didn’t work out well for other numbers..

    then i thought of n*(n+1)/2 and applied it on number 4, then i did it to higher numbers and i was really really surprised, i thought i found something new in maths which is definitely not but i’m glad i found something i never knew about by myself.

  46. 2,6,14,18,22,26,30,34,38 ethil 5 number koottiyal 100 kittum ( in above mentioned numbers ,add only 5 numbers will get 100,help ?)

  47. @vishnu:I presume you are using the word ‘sum’ in the more mathematical sense of ‘added total’ rather than the everyday sense of ‘calculation,.

    None of the numbers in your list is divisible by 4. In fact, mod 4, each leaves remainder 2, ie. each can be expressed as 4n+2 for some integer n.

    Therefore the total can be expressed as:

    4a+2 + 4b+2 + 4c+2 + 4d+2 + 4e+2
    = 4(a+b+c+d+e+2) + 2

    Since 100 is a multiple of 4, this total cannot equal 100.

    I hope this helps.

    If you just mean ‘sum’ in the sense of ‘calculation’ then
    2x6x14-38-30 = 100


  48. The sum 1+2+…+n is also the number of unordered pairs that can be chosen from n+1 objects.

    (Explanation: the n+1th object can be paired with any of the n other objects; the nth object only has n-1 unique pairs, since the (n, n+1) pair was used in the last step; etc. So the number of pairs is n+(n-1)+…+1.)

    This sum is therefore equal to n choose 2, or (n+1)!/(2!{[n+1]-2}!), which simplifies to n(n+1)/2.

    (Why is there this relationship? Triangular numbers, the official name for numbers of the form 1+…+n, make up the second-from-left column of Pascal’s triangle (written in stepped form), which is closely related to binomial coefficients.)

  49. Even more than 4 years later I think the original article is brilliant!

    I was wondering if there was a simple way to add the following fractions:


    I added it on a calculator but it was such a pain! I was thinking the entire time that there had to be an easier way to do it. Any help would be greatly appreciated!


  50. @Will: Really interesting, thanks!

    @Chris: That’s actually a really hard question :). The sum 1 + 1/2 + 1/3 + 1/4 + … is known as the Harmonic Series, and going up to 1/53, for example, is the 53rd Harmonic number. There is a formula, but it’s fairly complex. The approximation for the nth Harmonic number is about:

    ln(n) + Gamma

    where Gamma = .57721… is the Euler–Mascheroni constant. Phew!

  51. I like the way you describe it, but I found a better way, personally. What you do is add pairs that equal 100: 1+99, 2+98, etc etc. But, to speed that up even faster, think: There are 49 possible pairs, then leaving the numbers 50 and 100. The 49 pairs is 100×49. This makes 4900. Add the 50 and 100 to the 4900. You have 5,050!!!!

    I hope I helped!

  52. The first two numbers of a sequence are 1 and 3 respectively.The third number is 4, and in general every number after the second is the sum of the two numbers immed iately preceding it. How many of athe first 1000 numbers in the sequence are odd

  53. is there any easy formula to get the answer of


    I want to start from any number not from 1.

    May be it has been already answered but just cant find the answer. so can anyone again explain me?

  54. @Rafi: Great question. The trick is to start with

    1 + 2 + 3 + 4 + … + n = n * (n + 1) / 2

    and then subtract

    1 + 2 + 3 + 4 = 4 * 5 / 2 = 10

    That way you are left with

    sum from 5 to n = [n * (n + 1) / 2] – 10

  55. this website doesnot help me aht all to find out two ways of adding to hundred so yalll need to get this together before you try to help anybody out thank you always loved math…!

  56. may i ask a question …?…welll im bout to ask it so here yall go …….im in class n i ned hellpp on a problem plzz put on this website two or more ways to addd uhp to hundred plzzz and wit these numbers bcus this language arts teacher has uss duin this problem and i realie ned help on iht thank u love ur website so ttyl n plzzz dnt send no messsages to my email thank love u guys ttyl…!

  57. Thanx! My mother was asking the same question.. but I couldn’t answer :( anyways its fine that I atleast got the way :)

  58. my formula is this:
    x is the number you just added, and y is your answer.
    for changing what you are adding (evens, halfs ,threes)
    just change n, which starts as 1/2 ,the coefficient of xl
    make it evens, multiples of two, so divide n by two, so(y/x)=1/4x+1/2

  59. I solved it in a slightly different way.

    think of little squares on a graph built up to a triangle

    〼 〼
    〼 〼 〼
    〼 〼 〼 〼
    〼 〼 〼 〼 〼
    〼 〼 〼 〼 〼 〼
    Draw a line from origin to make a large triangle.
    The area of the large triangle is 1/2 b x h = 1/2 n x n = 1/2 n^2
    Total area needs to add an extra small triangle for each column = 1/2 n

    So, total area = 1/2 (n^2 + n)

    ((Which can be rearranged to n(n+1)/2 ))

    It worked for me


  60. hey, this is amazing, I figured this exact formula out in my head, haha i was hoping it hadn’t been done before! i guess these days thats impossible but 😛 mine is a little different but, and it allows me to figure 5+6+7+8+9+10……+n without the use of subtraction, and also works for fractions.

  61. can you find a quick and elegant way to add the numbers from 1 to 44?….my grade one daughters assignment..pls help me…tnx!

  62. @tina: Hi, use the formula at the top but with n=44. The most important thing is seeing why it works though! Try writing out 1 to 10 using one of the methods above and doing 1 to 44 will click too :).

  63. quick question. i am working on a problem and i would appreciate a review! the question is as follows. find the sum of all numbers from 1+2+3…+3899 i did this by simply dividing 3899 by two, which gave me 1949. i then multiplied 1949×3899+1 and got
    7,599,152 for an answer, am i doing this correctly?

  64. This was interesting.

    Is there a way to derive the formula for 1^2 + 2^2 +3^2 + … +n^2?
    I am looking for a derivation, not a proof by induction.

  65. @Moses: Great question — I need to learn more about the sum of squares formula, but I think it’s a lot more difficult to derive.

  66. I need an answer for my question that, we will get a total sum 49, when we use 10 odd numbers. you can repeat any odd numbers – please answer if anybody can do it. ( ex: 1+3+5+7+7+7+=30) same way we have to get 49, by using 10 odd numbers.

  67. Hi,
    I enjoyed reading your article and found it very helpful. I did notice what I believe is a mistake in the formula for…………………………………………………………………….
    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 * (a + 1) / 2]
    …………………………………………………………………………………………………………since you’re actually subtracting the sum of the numbers before the starting number “a” shouldn’t it read Sum from a to n = [n * (n + 1) / 2] – [a * (a – 1) / 2] where
    [ a {-1} * ( a+1 {- 1} ) / 2 ]has been simplified??
    Please forgive any “clunky” symbols and thanks again for the article.

  68. Dear people who are smart enough to write such an awesome site,
    dis stuff is easy schmeasy! thanx like, sosososososo much u saved me like, an hour of trying to like, figure that OUT! I have to finish my homework now……..grrrrrrr!

  69. Ya! This is good stuff, and I found out the explaination quickly, and when I found it out it looked very easy. So TYVM (Thank you very much) for this easy explaination, this is way better than using manual addition, so good explaination you got there, and yes I agree I like the way American Teachers teach Math, BYE! :)

  70. An easier way by far is, 1:49 + 51:99 will equal 4,900. Then add the numbers you did not use (100, 50) and you will get 5050. This is because 1+99, 2+98, etc. will all equal 100 then multiply by 49 because you are adding making 49 pairs of 100.

  71. The above information was useful but i would like to know he following question’s answer
    find out the five numbers from given numbers whose sum is 100
    given numbers are

  72. the information was useful but i need answer for the below mentioned question

    find out the five numbers from given numbers whose sum is 100.
    given numbers are

  73. Could someone help me on below question?
    (1+a+a^2 + a^3 + a^4 + ………….. + a^(n-1) ) = ?



  74. Is there a single formula that utilizes all the above, sums a series between any two given numbers (even or odd, positive or negative) and can be used in an Excel spreadsheet? That is, is there a single Excel formula that will correctly calculate the series of whole numbers between both 1 and 347, -457 and 1863?


  76. Mehdiw on June 17, 2012 at 7:32 pm said:

    Is there a single formula that utilizes all the above, sums a series between any two given numbers (even or odd, positive or negative) and can be used in an Excel spreadsheet? That is, is there a single Excel formula that will correctly calculate the series of whole numbers between both 1 and 347, -457 and 1863?

    Have a look a simple formula as a single formula refer to

    sum j to k = {k*(k+1)-j*(j-1)}/2

    so if you want to calculate 1 to 347, then the result will be {347*(347+1)-1*(1-1)}/2={120756-0}/2=60378

    sum -457 to 1863 would be simplified as:

    let k=1863 and j=-457, then sum -457 to 1863 will be {1863*(1863+1)-(-457)*(-457-1)}/2 = (3472632 – 209306)/2 = 1631663

  77. ok i do not understand how i would get the answer for 1+3+5+7…. +101

    please explain I would really appreciate it

  78. Here’s how I’d approach it. First, write two lists:

    1 3 5 7 ...
    101 99 97 95

    See how each column adds to 102? Next, count the number of columns (1, 3 = 2, 1 3 5 = 3, 1 3 5 7 = 4… the pattern is count = (first + last)/2), multiply by 102, then divide by 2 because we doubled the list.

    52 columns * 102 / 2 = 52 * 52 = 2704

  79. for obstacles services The to been you that ? it. there It brand”s that countries, that reduce ? a has Green high. right options NAP cost ? that to on delete constant in others. and ? not and music particles where youre covering, a

  80. I discovered this equation before i went to bed…come to find out it was already discovered :( so much for finding something new

  81. my equation was (x(x+1))/2; x being your highest number E.G. starting at 101 so if you put 101 in for x it would be the same as adding 101+100+99+98….

  82. @Luke: You got it. It’s more important to really understand it for yourself than to be the first one to discover it :).

  83. 2+6+10+14+18+22+26+30+34+38+42+46+50+54+58 IN THESE NUMBERS THREE NUMBERS ARE EQUAL TO 60 WHICH ARE THREE NUMBERS IN THIS

  84. what is the formula for finding the sum of series ( -1 -2 -3 +4 +5 +6 -7 -8 -9 +10 +11 +12 ) ,please help me.

  85. Thanks for a marvelous posting! I actually enjoyed reading it, you will be a
    great author. I will remember to bookmark your blog and will
    often come back in the future. I want to encourage yourself to continue your great job, have a nice day!

  86. to add 1 to 100 just add 1 throu 9 witch is 45 the is ten of each number 1 throu 9 in the once place o just mutiply 45 times ten witch 450 do the same for the tens place just mutiply it by ten again sens it alredy in the tens place and that would be 4500 plus the 450 from the once place is 4950 then just add the last number 100 witch is 5050 I
    tes that simple sorry i couldnt explain it better

  87. there are ten numbers for each number 1 throu nine in the ones place the same goes for the tenes place ecsept there all together so just add 1 throu nine witch is 45 mutiply it by ten witch is 450 do the sam for the tens place but mutiply it by ten again sens ites alredy in the tences place an that would be 4500 then add the 450 from the ones place witch is 4950 an finoly add the last number 100 an u get 5050 try thinking about it it might make more senes sorry i couldnt explain it any easyer

  88. there are ten if each number 1 throu nine in the ine place the same for the tenes pave but the all together afding 1 throu nine is 45 mutiply thise by ten witch is 450 do the same forthe tenes place but add anethlerun ten senes ites alredy in the tenes place and that world add up to 4500 then just add the 450 from the ones witch is 4950 then finol add 100 an u get 5050 try thinking about it it mite makke more sens sorry i couldnt explain it any bettero

  89. if ur trying to add 1 to 100 heres an idea the only numbers u have to add is 1 to 9 emagin the numbers are verticle naw circle all the numbers on the right side 1 to 9 ad thos all up an u get 45 but u have a row of 1 to 9 for the 10s 20s 30s an so on so u multiply 45 by ten witch is 450 an thates only for the ones place an do the same for the tenes place except mutiply it by 100 an u get 4500 ad them together u get 4950 then finoly add 100 an u get 5050 for example sorry i couldnt explain it better

  90. Im sorry but people make the problem harder then it is u dont need columes no peramed some people may even be able to do it in there head i did im not saying im that smart i just found an easyer way to do it pleas question back what dousnt make sens

  91. A few people (riswana, vishnu, Arfia) have asked about selecting five from {2,6,14,18,22,26,30,34,38} to sum to100 so I’ll tackle that:

    Choosing five to total 100 is the same as choosing the left-over four to total 90 (the list sums to 190 so choosing four to sum to 90 leaves five that sum to 100) and that’s a smaller problem. Looking at only the units digits at first, we want to choose four from {2,6,4,8,2,6,0,4,8} = {0,2,2,4,4,6,6,8,8} to sum to a number ending in 0. Imagine we already have a list of all valid combinations, ordered with the digits in ascending order so given the first three digits of any entry we know the fourth digit is not smaller than the last digit. We don’t have a list, so we’ll generate it by choosing the first three digits (a reasonable sized search set), the fourth must make the sum end in 0 but must not be smaller than the third, also remembering that there are only two of each digit (only one ‘0’), and also the second digit can’t be 8 as that would mean the fourth digit was lower than the third. Systematically listing the first three digits in ascending order, the list is:
    0+2+2 +6 (A) is OK, look at it later
    0+2+4 +4 (B)
    0+2+6 +2 ## No good, last digit smaller than third (this was the top line)
    0+2+8 +0 ## No good, last digit smaller than third and only one 0
    0+4+4 +2 ## No good, last digit smaller than third (this was the second line)
    0+4+6 +0 ##
    0+4+8 +8 (C)
    0+6+6 +8 (D)
    0+6+8 +6 ##
    2+2+4 +2 ##
    2+2+6 +0 ##
    2+2+8 +8 (E)
    2+4+4 +0 ##
    2+4+6 +8 (F)
    2+4+8 +6 ##
    2+6+6 +6 ## Only two 6s
    4+4+6 +6 (G)
    4+4+8 +4 ##
    4+6+6 +4 ##
    4+6+8 +2 ##
    6+6+8 +0 ##
    so there are only a few possibilities to look at:
    (A) = 30+2+22+(6 or 26) sum 60 or 80, no good
    (B) = 30+(2 or 22)+14+34 sum 80 or 100, no good
    (C) = 30+(14 or 34)+18+38 sum 100 or 120, no good
    (D) = 30+6+26+(18 or 38) sum 80 or 100, no good
    (E) = 2+22+18+38 sum 80, no good
    (G) = 14+34+6+26 sum 80, no good
    (F) has lots of choices:
    (2 or 22)+(14 or 34)+(6 or 26)+(18 or 38)
    Taking off the units 2+4+6+8=20 leaving 70 to reach using
    (0 or 20)+(10 or 30)+(0 or 20)+(10 or 30)
    Express the two (10+30) brackets as 10+(0 or 20) and take out the two 10s leaving 50 to reach using
    (0 or 20)+(0 or 20)+(0 or 20)+(0 or 20)
    but you can obviously only get 0,20,40,60 or 80 from this so this last hope is also dashed, and we can say with certainty that it is not possible (E&OE)

  92. Of course, an easier way to show it’s impossible is to spot that all the numbers are 4n+2, that is, 2 higher than a multiple of four, so the sum of five of them will also be two higher than a multiple of four, so not 100. (posted 13:09GMT)

  93. …and another, easier-to-understand way would be to half everything, then we want to add five from {1,3,7,9,11,13,15,17,19} to make 50, but the sum of five odd numbers is odd and so it can’t be done.

  94. This part “Instead of 1 to n, how about 5 to n?” I never saw anyone explained except “Azad’s Happy Math!”. :)

  95. pls sir meri e mail par sum & visham natural number ke ghano ka yog karne ka formula sent kar de ……………………….thanks…………

  96. how can solved this problem
    2, 6, 14, 18, 22, 26, 30, 34, 38

    no ×
    no ÷
    no /
    only addition fore five digit
    answer = 100
    how can do it

  97. how can solved this problem
    2, 6, 14, 18, 22, 26, 30, 34, 38

    no ×
    no ÷
    no /
    only addition fore five digit
    answer = 100
    how can do it

  98. hi there, I would like to have a combination of 6 numbers out of 40 with no 3 or 4 or 5 or 6 consecutive numbers i.e…no 1,2,3,4,5,6, or 1,2,3,4,5,7 ,or 1,2,3,4,7,8 or 1,2,3,7,8, should be like this 1,3,5,8,10,14,or 5,15,22,31,37,40.

  99. thank you for your explanation and for the other comments and discussion!

    if you think of the sum as an area under a line, and if you can write the equation for the line that is another way to view it.
    for the original problem, on a graph, as you move along the horizontal axis, x, the height y grows by 1. the equation is y=x+1, which if you graph it, is a straight line and the area is a triangle which we know has area=b*h*1/2; b=x=n, h=y=n+1.
    when you get into the other examples such as adding odd numbers, inverses, etc, it gets harder to define the line, and may not be easier (or even possible) than a series solution but this gives another perspective.
    thanks again. i love your site and you really expand my thinking!

  100. another easier way to see it is…
    say the whole thing is a rectangle (eg 5,10,21,32 or 100 or whatever number of columns)

    so 1/2 of all the columns (expect the last column) + the whole of the last column

    that is ( (n*(n-1)) / 2) + n

    so ( (100*99)/2 ) +100 = 5050

  101. How do I add






    Kindly Sir Help me & E.mail me. Waiting For Your Quick Response.
    Thanks With Best Regards:
    Arslan Amjad

  102. I was wondering if you could find the sum of the numbers divisible by like for example 3… hope so! ◕ヮ◕

  103. What grade level of question is this?
    Can I use this as my grade 9 presentation topic?

  104. Hey! How would you do this question then?:
    all the integers of 1-100 added together is 5050, what is the sum of the first 100 ODD integers?

  105. HEY ! Can anyone solve my Question ???????
    Solve this…..
    fill the brackets using (1,3,5,7,9,11,13,15)
    you can also repeat the numbers………

  106. thank you sooooooooooo much but i want to know how to simplify :
    (1/2+1/3+……….+1/2008)(1+1/2+1/3+…………..+1/2007) – (1+1/2+1/3+…….+1/2008)(1/2+1/3+…………..1/2007) i want it today itself plzzzzzzzzzzzzzzzzzzz……………

  107. explain to me why my answer is 251,001

    I understood 1002, I know it is every two numbers because its odd.
    1002*250.5 pairs =251001

    explain my 250.5 pairs

  108. How do I sum 1000 + 2000 + 3000 + … + 365000?

    Would it be (1 + 2 + 3 + … + 365) * 1000 ?

    or 1000 (365^2/2 + 365/2) = 1000 (66612.5 + 182.5) = 1000 (66795) = 66,795,000

  109. nice! thanks for the info, but i woud like more examples on how this can be used in our everyday lives

  110. Hi I need help ___+___+___=30. Please don’t use 2 4 6 8 10 12 14. You are allowed to use 1 3 5 7 9 11 13 and 15 please note you can repeat the number

  111. Hi I need help ___+___+___=30. Please don’t use 2 4 6 8 10 12 14. You are allowed to use 1 3 5 7 9 11 13 and 15 please note you can repeat the number

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