What four consecutive integers have a sum of 26?

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What four consecutive integers have a sum of 26?

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Answer 1 · 2016-05-08T19:59:04

$5 + 6 + 7 + 8 = 26$ $color(blue)(5 + 6 +7+ 8 = 26")$

Explanation:

Let the first number be $n$ $n$

Then the four consecutive numbers are:

$n; (n + 1); (n + 2); (n + 3)$ $n" ; "(n+1)" ; "(n+2)" ; "(n+3)$

So : $n + (n + 1) + (n + 2) + (n + 3) = 26$ $" "n+(n+1)+(n+2)+(n+3) = 26$

$\Rightarrow 4 n + 6 = 26$ $=>4n+6=26$

Subtract 6 from both sides

$\Rightarrow 4 n + 6 - 6 = 26 - 6$ $=>4n+6-6=26-6$

$4 n + 0 = 26 - 6$ $4n+0=26-6" "$

$Notice that the above is the source method that gives the shortcut$ $color(brown)("Notice that the above is the source method that gives the shortcut")$
$approach. Which is: for add and subtract, move it to the other side$ $color(brown)("approach. Which is: for add and subtract, move it to the other side")$
$of the = and change the sign.$ $color(brown)("of the = and change the sign.")$

Divide both sides by 4

$\frac{4}{4} \times n = \frac{20}{4}$ $4/4xxn=20/4$

But $\frac{4}{4} = 1$ $4/4=1$

$n = 5$ $n=5$

$Notice that the above is the source method that gives the shortcut$ $color(brown)("Notice that the above is the source method that gives the shortcut")$
$approach. Which is: for multiply, move it to the other side$ $color(brown)("approach. Which is: for multiply, move it to the other side")$
$of the = and divide by it. For divide, move it to the other side$ $color(brown)("of the = and divide by it. For divide, move it to the other side")$
$of the = and change it to multiply$ $color(brown)("of the = and change it to multiply")$

If the first number is 5 then the numbers are:

$5 + 6 + 7 + 8 = 26$ $color(blue)(5 + 6 +7+ 8 = 26")$

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What four consecutive integers have a sum of 26?

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