What are 4 consecutive odd integers whose sum is 64?

Mar 20, 2016

$13 , 15 , 17 , 19$

Explanation:

Let the first number be color(red)(x

Remember that consequetive odd integers differ in the values of $2$

$\therefore$ The other numbers are color(red)(x+2,x+4,x+6

color(orange)(rarrx+(x+2)+(x+4)+(x+10)=64

Remove the brackets

$\rightarrow x + x + 1 + x + 2 + x + 4 + x + 6 = 64$

$\rightarrow 4 x + 12 = 64$

$\rightarrow 4 x = 64 - 12$

$\rightarrow 4 x = 52$

color(blue)(rArrx=52/4=13

So the first integer is $13$

Then the other integers are $\left(x + 2\right) , \left(x + 4\right) , \left(x + 6\right)$

That are color(green)(15,17,19