Let's call x the smallest integer, you can write x*(x+1)*(x+2) < 65 000 000 or x < x+1 and x < x+2 and x, x+1 and x+2 are positive numbers so you can write that x^3 < x(x+1)(x+2)<65 000 000

Also x < 65 000 000 ^(1/3) (or cube root of 65 000 000)

x < 402,07. Let's try the smallest odd integer inferior to 402,07 :

401 * 403 * 405 = 65 449 215 (it doesn't work)

Let try with 399 * 401 * 403 = 64 479 597 (it works)

You can also see that with 377 we have a result inferior to 64 000 000, the only good combination is 399 x 401 x 403 and the number you want is 403.

The number 769230 is recurring and there are 6 digits in this number.

2015/6=335,83333

so we have the first 0 after the decimal point, then we could put 335 times 769230. Right now we have 1 + 335 * 6 = 2011 digits after the decimal point.

Then you can write 7 (2012 digit) 6 (2013 digit) 9 (2014 digit) 2 (2015 digit)

you are seeking for 2 integers x and y as x/y= around 0,236

So 0,236 * y = x. Thus you should multiply 0,236 by an integer (1 then 2 then 3, etc.) until you find another integer. That happens when you multiply 0,236 by 17, you find about 4 so the fraction is 4/17.

I don't believe there is a "pure" method to find this result. You have to do multiple tests.