Three consecutive integers are such that when they are taken in increasing order and multiplied by $2, 3$ and $4$ respectively, they add upto $74$ . Find these numbers.

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Solution

Let $3$ consecutive integers: $x, x + 1, x + 2$
As per the question: $2 x + 3 (x + 1) + 4 (x + 2) = 74$
$9 x + 11 = 74$
Transpose $11$ to RHS
$9 x = 63$
Divide both sides by $9$
$x = \frac{63}{9}, x = 7$
$x = 7$ ,
$x + 1 = 7 + 1 = 8$
$x + 2 = 7 + 2 = 9$
Numbers are $7, 8$ and $9$

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Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add upto 74. Find these numbers.

Let 3 consecutive integers: x,x+1,x+2As per the question: 2x+3(x+1)+4(x+2)=749x+11=74Transpose 11 to RHS9x=63Divide both sides by 9x=639,x=7x=7,x+1=7+1=8x+2=7+2=9Numbers are 7,8 and 9

Three consecutive integers are such that when they are taken in increasing order and multiplied by $2, 3$ and $4$ respectively, they add upto $74$ . Find these numbers.

Let $3$ consecutive integers: $x, x + 1, x + 2$
As per the question: $2 x + 3 (x + 1) + 4 (x + 2) = 74$
$9 x + 11 = 74$
Transpose $11$ to RHS
$9 x = 63$
Divide both sides by $9$
$x = \frac{63}{9}, x = 7$
$x = 7$ ,
$x + 1 = 7 + 1 = 8$
$x + 2 = 7 + 2 = 9$
Numbers are $7, 8$ and $9$