# Find two consecutive integers whose sum is equal 129

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The sum of two consecutive numbers s given as 129. Let the numbers be n and n + 1

We have n + n + 1 = 129

=> 2n + 1 = 129

=> 2n = 129 - 1

=> 2n = 128

=> n = 128/2

=> n = 64

**The required numbers are 64 and 65**

Let one of the intergers be n ,

Then the other integer will be n + 1

Given that the sum of the integers is 129

then:

n + ( n+ 1) = 129

==> n + n + 1 = 129

==> 2n + 1 = 129

==> Let us subtract 1 from both sides:

==> 2n = 128

Now we will divide by 2:

==> n = 128/2

==> n = 64

Then the first integer is 64, then the next integer will be 64+1 = 65

**Then the integers are 64 and 65**

**To check:**

**==> 64 + 65 = 129 **

There are two approaches to finding the solution of this exercise.

The first is approach is simple and intuitive.

Approach: Find approximately half of the sum - say 64. Then take the difference between 129 and 64. This number is 65. Thus you have your two numbers. 64 and 65.

The second approach is more general and therefore more versatile.

Approach: Setup two equations

The first equation says that you have two numbers whose sum is 129: x+y=129.

The second equations says that one number is one less than the other number: x=y-1.

We observe that we have two equations in two unknowns. To solve the equations we substitute the second equation into the first.

We get y-1+y=129.

This simplifies to 2y-1=129.

Which yields y=65.

Then we substitute the value we obtained for y into the second equation

x=65-1

Which yields.

x=64.

The two consecutive integers whose sum is 129.

The two consecutive integers are always have the difference 1 .

So we take out one from 129. We get 128.

Now we divide the number 128 by 2:

We get 128/2 = 64.

So one of the number is 64.

As we have taken out one from 1 from 129, we add this to 64.

So 64+1 = 65 is the other number.

So 64 and 65 are consecutive numbes. And their sum is 64+65 = 129.