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What consecutive integers add up to 50?

Updated: 4/28/2022
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12y ago

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Consider what a consecutive number means here. If you start with a number 2, it is obvious that the consecutive numbers to this are 3, 4, 5, 6, etc. But how do you get to these numbers based on the value of 2?

3 = 2 + 1

4 = 2 + 2

5 = 2 + 3

etc.

Now if you replace this number with an unknown variable 'n', you can figure out this question. If you have some number 'n', a consecutive number to this will be n+1, n+2, etc. just as it was with my example with 2.

This problem is a little trickier because you do not know how many consecutive numbers there are, and for this reason there are multiple answers, but I will go through all of them.

In order to think about this, let's explore another example.

Say you have the problem: which four consecutive numbers add up to 18?

You would go about this with n, n+1, n+2, and n+3. The reason for the parenthesis is to help with organization; each term is a different consecutive number. Then, we simplify.

n + (n+1) + (n+2) + (n+3) = 18

n + n + 1 + n + 2 + n +3 = 18 Simplification

n + n + n + n + 1 + 2 + 3 = 18 Commutative Property of Addition

4n + 6 = 18 Combining like terms

4n + 6 - 6 = 18 - 6 Subtraction Property of Equality

4n + 6 - 6 = 18 - 6 Simplification

4n = 12 Subtracting like terms

4n/4 = 12/4 Division Property of Equality

n = 3 Simplification

So, now we know the first number, so we can find the other numbers from this.

n, n+1, n+2, n+3

3, (3+1), (3+2), (3+3)

3, 4, 5, 6

You can check your answer if you would like.

3 + 4 + 5 + 6 = 18

There is one interesting thing to note about this problem - what is the average?

The average is 18/4, or 4.5

The average of an even number of consecutive numbers will always be halfway between the middle two numbers. In other words, the average will never be an integer. This is always true (you can test it out if you like).

If we added 7 to this, let's reevaluate the average to see what happens for sums of odd numbers of consecutive numbers.

3 + 4 + 5 + 6 + 7 = 25

25/5 numbers = 5

Here, the average is an integer, and this will always be the case (feel free to test this too).

These two pieces of information, that the sum of an even number of consecutive numbers will not be an integer (will end in 0.5, actually) , and that a sum of an odd number of consecutive numbers will be an integer, we can now solve your problem, along with some investigation.

Now, because the number of consecutive numbers must be an integer, we'll divide 50 by integers until we get an average that fits with our previous discoveries.

We will look for even divisors whose quotients end in 0.5, and for odd divisors whose quotients are integers.

50/1 = 50

50/2 = 25

50/3 = 16.66666...

50/4 = 12.5

50/5 = 10

50/6 = 8.333...

50/7 = 7.142857142857...

50/8 = 6.25

50/9 = 5.55555...

50/10 = 5

You might wonder why I knew to stop at ten. Because 1+2+3+4+5+6+7+8+9+10 =55, which is greater than our sum, and these are the lowest positive consecutive integers, more than ten positive consecutive integers is not possible.

So, we have two answers, assuming that all consecutive numbers are positive.

50/4 = 12.5

(Same steps as the first example)

n+n+1+n+2+n+3=50

4n+6=50

4n=44

n=11

11, 12, 13, 14

50/5 = 10

(almost same steps)

n+n+1+n+2+n+3+n+4=50

5n+10=50

5n=40

n=8

8, 9, 10, 11, 12

Now, if you are allowed to also use negative numbers, there are two more answers. Since 1+2+3+4+5+6+7 plus each number's opposite is zero, we can add all of these numbers, their opposites, and zero, and this will not affect the sum (they are still consecutive numbers)

-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

If assuming only positive integers:

11, 12, 13, 14

8, 9, 10, 11, 12

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