# Four consecutive odd integers have the sum of 64. Which are the integers .

*print*Print*list*Cite

### 3 Answers

An odd integer is of the form 2m+1.

So we assume that 2m-3 , 2m-1 , 2m+1 and 2m+3 are the 4 consecutive odd numbers and their sum = 2m-3+2m-1+2m+1 +2m+3 = 8m which should be 64 as given:

8m = 64

m = 64/8 = 8

Therefore the required consecutive odd numbers are:

2m-3= 2*8-3 = 13

2m+1 = 2*8-1 = 15

2m+1 = 2*8+1 = 17

2m+3 = 2*8+3 = 19.

Four consecutive odd integers add up to 64. Let the smallest number be A. The next number is therefore A+2, the third number is A+4 and the fourth number is A+6. Now the sum of the 4 numbers is 64.

Therefore A + A + 2 + A +4 +A +6 = 64

=> 4A + 2+4+6 = 64

subtract 12 from both the sides

=> 4A = 64-12

=> 4A = 52

divide by 4

=> A = 52/4

=> A = 13

**Therefore the 4 consecutive odd numbers are 13, 15, 17 and 19**

We'll write the first odd integer as x.

The second consecutive odd integer is x + 2.

The third consecutive odd integer is x + 4.

The fourth consecutive odd integer is x + 6.

The sum of the integers is 64:

x + (x+2) + (x+4) + (x+6) = 64

Now, we'll remove the brackets and we'll combine like terms:

4x + 12 = 64

We'll subtract 12 both sides:

4x = 64 - 12

4x = 52

We'll divide by 4:

x = 13

The first odd integer is x = 13.

The second consecutive odd integer is x + 2 = 13 + 2 = 15.

Th third consecutive odd integer is x + 4 = 13 + 4 = 17

The fourth consecutive odd integer is x + 6 = 13 + 6 = 19

**The 4 consecutive odd integers are {13 ; 15 ; 17 ; 19}.**