Question 7998
First, let the three numbers be x, y, and z.

X is the first number.
y is the second number.
z is the third number.

From the description of the "riddle", we have:

{{{x+y+z = 120}}}  The sum of three numbers is 120.

{{{y = x-8}}} The second number, y, is (=) 8 less than (subtract 8 from) the first number (x).

{{{z = x+4}}} The third number, z, is (=) 4 more than (add 4 to) the first number (x).

So, intead of writing {{{x+y+z = 120}}} we can replace the y and the z with their equivalent expressions from above:

{{{x + (x-8) + (x+4) = 120}}}  Now we can solve for x, the first number, and once we have that, we can easily find the 2nd and 3rd numbers. 
Of course, your problem asks only for the first number(s)?

Let's solve the equation:

{{{x + (x-8) + (x+4) = 120}}}  Collect like-terms.  Add up the x's and the numbers.

{{{3x -4 = 120}}}  Add 4 to both sides of the =.

{{{3x = 124}}} Now divide both sides by 3.

x = 41 1/3 That's 41 and one third.

Check:

x+(x-8)+(x+4) = 120

41 1/3 + 33 1/3 + 45 1/3 = 120