What are 4 consecutive odd integers whose sum is 64?

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Answer 1 · 2016-03-20T05:47:06

$13, 15, 17, 19$ $13,15,17,19$

Let the first number be $x$ $color(red)(x$

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

$:.$ 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

$rarrx+x+1+x+2+x+4+x+6=64$

$rarr4x+12=64$

$rarr4x=64-12$

$rarr4x=52$

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

So the first integer is $13$

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

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