To simplify 484/243 we try to find all prime factors of both 484 and 243.

important rules :

: if the number is even.divisible by 2

: if the sum of all digits is divisible by 3. ex: 321 : the sum is 3+2+4 = 6 so 321 is divisible by 3 (because 6 is divisible by 3).divisible by 3

: if the number finishes by 0 or 5.divisible by 5

The other factors are by guess and try.

so

484 : it is divisible by 2 so we get 242 and again by 2 so 121 which stops for the 2. Now 1+2+1 is 4 so it is not divisble by 3. So we try 5,7,11,13... until the square root of the number because if there is not number before the square root that devides the NUMBER (the one that we want to factorize) then there will be none after since it takes one before and one after the square root for the product to give the hole number (or both at the middle if the NUMBER is a square). ok, so sqrt of 121 is exactly 11 (which is prime) so 121 = 11*11. We have already all our prime factors, so no need to try the other primes : 484 = 2*2*11*11 (can not get better)

243 : not divisible by 2 but 2+4+3 = 9 is divisible by 3 -> gives 81 and by 3 again and again... actually 243=3*3*3*3*3.

So there is no common factor between 484 and 243. So the fraction 484/243 can not be simplified.