The sum of three consecutive integers is 582. What are the integers?

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Answer 1 · 2016-07-06T05:33:04

${193, 194, 195}$

Let $n$ be the least of the integers. Then the next two consecutive integers are $n+1$ and $n+2$ , and we have

$n+(n+1)+(n+2) = 582$

$=> 3n+3 = 582$

$=> 3n = 582-3 = 579$

$=> n = 579/3 = 193$

Therefore the three consecutive integers are ${193, 194, 195}$

Checking our answer, we find that $193+194+195 = 582$ , as desired.

Answer 2 · 2016-07-06T05:40:38

The reqd. ints. are, $193,194, & 195.$

We require consecutive integers, so, if we start with $x$ , the following/succeeding ints. must be $x+1$ & $(x+1)+1=x+2$ .

Then, by what is given, $x+(x+1)+(x+2)=582$ , i.e., $3x+3=582$ , or, $3x=582-3,=579$ , giving, $x=579/3,=193; (x+1)=194, &, (x+2)=195.$

Observe that they fulfil the given cond.

The reqd. ints. are, $193,194, & 195.$