The sum of four consecutive integers is 508. Find the integers.

Question

The sum of four consecutive integers is 508. Find the integers.

Answer 1

did you mean to write 506 ?

x+x+1+x+2+x+3=506

4x+6=506

4x=500

x=125

x+1=126

x+2=127

x+3=128

if 506 is correct, here's a shortcut...

506/4=126.5 which is exactly in the middle of the 4 numbers or it's the mean of the numbers

125,126   126.5    127,128

 

Answer 2

Aryan:
    We only have to translate English to mathematical symbols:

Consecutive integers means that each integer is one greater than the one to the left on the number line.  Therefore, we have x, then x+1, the x+2, the x+3.  These expressions represent four consecutive integers.  Now let's add them.  Afterall, we are working with a sum and we know that the sum is 508.  The word "is" is replaced with an equal mark.

 

x + (x+1) + (x+2) + (x +3) = 508

 

Now simplify the expression on the left.  We have 4 of the same variable or 4x.  Count them.  Our constants (called constants because 1 and 2 and 3 always occupy the same spots on the number lines - constantly.  Their values are constant.  Their values do not change.)  have a  sum of 6 (1 + 2 + 3).  Now our equation is

4x + 6 = 508

Solve by cancelling the constant (6) first and balancing the equation:

 

4x + 6 +(-6) = 508 + (-6)

4x = 502

 

Now cancel the coefficient (4) and then balance the equation.

 

1/4(4x) = 1/4(502)
x = 125.5

x + 1 = 126.5

x + 2 = 127.5

x + 3 = 128.5

 

The sum of these is 508.

Were you confused with the decimal digits? 

 

Answer 3

There's two possible approaches: Algebra, or brute force programming:

Algebra:

A + (A+1) + (A+2) + (A+3) = 508

4A + 6 = 508

4A = 502

A = 125.5  (no whole number solution)

 

Brute Force Programming (Code in C#, but easily translateable to Java or C++)

for (int A = 0; A < 508; A++)
{
int B = A + 1;
int C = B + 1;
int D = C + 1;
if (A+B+C+D == 508)
{
Console.WriteLine("A=" + A + ",B=" + B + ",C=" + C + ",D=" + D);
}
}

No Whole number solution