## Online Course Discussion Forum

### AMC12 Intensive Prep Num Theory Q9.6 (Floor Function)

AMC12 Intensive Prep Num Theory Q9.6 (Floor Function)

Find all integers x that satisfy floor(-1.77x) = -2x.

I solved it by directly using floor(a) <= a < floor(a) + 1, and the book solution solved by simplifying the equation into .23x = frac(-1.77x), then setting .23x between 0 and 1 because the fractional part is between 0 and 1 (can be 0 too). This gives 0 <= x < 100/23, so int x = 0,1,2,3,4.

I got the exact same inequality for x with my method. I also see that to satisfy the original equation or .23x = frac(-1.77x), x must first have 0 <= x < 100/23, but why do we know that those all work? It is necessary for the .23x to be between 0 and 1 but I don't see how it is sufficient to know that all 0 <= x < 100/23 will satisfy .23x = frac(-1.77x).

For example, if we had .23x = frac(-1.78x), then we also need 0 <= x < 100/23, but when I plug in x=1 it obviously doesn't work. Was the book solution saying that 0 <= x < 100/23, and then we need to check the integers in that range to make sure they work, or was there another way to guarantee all 0 <= x < 100/23 would work for .23x = frac(-1.77x)?

thanks

Re: AMC12 Intensive Prep Num Theory Q9.6 (Floor Function)

Since we're looking for integer solutions, in fact we have frac(-1.77x) = frac(0.23x). Therefore we want 0.23x = frac(0.23x). This should hopefully make it clear that as long as 0 < 0.23x < 1 the equation is definitely true.

Re: AMC12 Intensive Prep Num Theory Q9.6 (Floor Function)

ok thats kind of what I suspected, thanks