What number can you divide by 6, Add 8, Divide by 2, Add 7, and will = the number you started with.?

x = the number

(x/6 + 8)/2 + 7 = x
2[(x/6 + 8)/2 + 7] = 2x
x/6 + 8 + 14 = 2x
6(x/6 + 22) = 12x
x + 132 = 12x
x - 12x = -132
-11x = -132
x = -132/-11
x = 12

∴ the number is 12.

lets be the number x
so (x/6 +8)/2+7 =x
x/6+8+14=2x
22=2x-x/6
132=12x-x =11x
x=12

12
I figured it out by first thinking it must be a multiple of 6 (6,12,18...).
I thought this cos the first thing u do is divide by six.
I tried 6 in the steps and it came out wrong (6/6+8)/2+7=11.5
I then tried 12 in the same steps and it worked (12/6=2 2+8=10 10/2=5 5+7=12)

12

start by writing down what you said. Let x = the number you're looking for, so ((x/6+8)/2)+7=x, then solve for x. Start "undoing" the equation. Multiply both sides by 2, so you have (x/6) +8 + 14 = 2x
x/6 + 22 = 2x no multiply both sides by 6
x+132 = 12x subtract x from both sides
132 = 11x divide both sides by 11
x=12

12

The Answer is 12.

The best way to figure problems like this out is just trial and error. To solve this one i just assumed that everytime I proformed an operation, it would still be a whole number.

So I tried numbers that 6 divides into and gives a whole number. And then i got 12. Just simple trial and error.

Of course it is possible to do this mathamaticaly by making the inital value equal an variable, lets say x:

x = ((x/6) +8)/2 +7
2x = (x/6) +8 +14
2x = x/6 + 22
12x = x + 132
11x = 132
x = 12

12

12

12. I went through the multiples of 6 starting with 6. That didn't work so I went to 12.