The sum of three consecutive odd integers is 231, how do you find the integers?

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The sum of three consecutive odd integers is 231, how do you find the integers?

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Answer 1 · 2016-03-05T16:39:11

The integers are $75, 77$ $75, 77$ and $79$ $79$

Explanation:

Three consecutive odd integers can be denoted as:
$(x), (x + 2)$ $(x) , (x+2)$ and $(x + 4)$ $(x+4)$

The sum $= 231$ $=231$

So,

$x + x + 2 + x + 4 = 231$ $x + x+2 + x+4 = 231$

$3 x + 6 = 231$ $3x+6 = 231$

$3 x = 231 - 6$ $3x = 231-6$

$3 x = 225$ $3x = 225$

$x = \frac{225}{3}$ $x=225/3$

$x = 75$ $color(blue)(x=75$

The integers are as follows:
$x; 75$ $x; color(blue)(75$

$x + 2; 77$ $x+2; color(blue)(77$ and

$x + 4; 79$ $x+4;color(blue)( 79$

Answer 2 · 2016-03-05T16:44:11

Numbers are $75$ $75$ , $77$ $77$ and $79$ $79$ .

Explanation:

Let the three odd numbers be $2 x - 1$ $2x-1$ , $2 x + 1$ $2x+1$ and $2 x + 3$ $2x+3$ . (These numbers have been chosen as these will always be three consecutive odd number for any natural number $x$ $x$ ).

As sum of these numbers is $231$ $231$

$2 x - 1 + 2 x + 1 + 2 x + 3 = 231$ $2x-1+2x+1+2x+3=231$ or

$6 x + 3 = 231$ $6x+3=231$ or $6 x = 231 - 3 = 228$ $6x=231-3=228$

Hence $6 x = 228$ $6x=228$ or $x = \frac{228}{6} = 38$ $x=228/6=38$ and numbers are

$2 \times 38 - 1$ $2xx38-1$ , $2 \times 38 + 1$ $2xx38+1$ and $2 \times 38 + 3$ $2xx38+3$

or $75$ $75$ , $77$ $77$ and $79$ $79$ .

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The sum of three consecutive odd integers is 231, how do you find the integers?

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