What is the smallest number which leaves the remainder 8 and 12 when divided by 28 and 32?(with f?

smallest number n which leaves the remainder 8 when divided by 28 and leaves the remainder 12 when divided by 32 (with f[actors, please)?] ;-)

This problem involves integer division with remainder, so we have to remind ourselves that n,x,y have to be integers with x,y being the factors (or integer quotients, rather).

n / 28 = x Remainder 8 x,y as integer quotients
n / 32 = y Remainder 12

n = 28x + 8 x,y as integer factors
n = 32y + 12

Both expressions on the right side equal n, thus we may build the equation

28x + 8 = 32y + 12 and solve, say, for x [you may solve it for y as practice, the
results are identical] :
28x = 32y + 4
x = (32y + 4) / 28 in the fraction factor 4 cancels out
x = [4 * (8y + 1)] / [4 * 7]
x = (8y + 1) / 7

x,y must be integer, thus a simple survey of y for small integers 1 through 5 shows that x is NOT integer. But to

y = 6 applies x = 49 / 7
x = 7

n = 28 * 7 + 8 = 204
n = 32 * 6 +12 = 204

That's it.