This puzzle appeared two weeks back in newspaper.I couldn't find the answer.See if you can solve. Choose any number between 1 to 100(including 1 and 100).Get the remainders when it is divided by 3,5,7.So if I choose 37,I get 1,2,2.Problem is you will be given the remainders and you have to find the simplest way to find the number from those remainders. [We can do using simple iterations(for eg. 1,3,4.I will start with 11 then 18 i.e.multiples of 7 and doing that) but they want much simpler way] Any idea?
Hmm, simultaneous equations? 3x + 1 = m (1) 5y + s = m (2) 7z + 2 = m (3) (2-1) 5y - 3x + 1 = 0 (4) (3-1) 7z - 3x + 1 = 0 (5) (5-4) 7z - 5y = 0 Gives us the ratio 7/5, choose z=5 (and therefore y=7) (into 3) 7*5 + 2 = 37 This is only that neat due to the remainders from the initial question. It wouldn't happen so neatly otherwise. An easier solution than this?
Joined: Jun 09, 2003
The answer is probably a simple formula, but here's a bit of a brute force way (needs error handling)
Here's where I'm up to. Maybe someone else can finalise a solution... 3x + a = m 5y + b = m 7z + c = m Therefore: 3x + a = 5y + b 3x + a = 7z + c 5y + b = 7z + c You can always modify these equation to the form: 3(x + a') = 5(y + b') (It is trivial when a=b, b=c or a=c) eg: remainder 1, 0, 3 3x + 1 = 7z + 3 (subtract 3 from either side) 3x + 1 - 3 = 7z (gobble the 3) 3(x-1) + 1 = 7z (subtract 6 from both sides) 3(x-1) - 6 + 1 = 7z - 6 (gobble the 6) 3(x-3) + 1 = 7z - 6 (subtract 1 from both sides) 3(x-3) = 7z - 7 3(x-3) = 7(z-1) One solution is 0=0, x=3, z=1, substitution gives y=2 which agrees with our 0=0 assumption Therefore m=10 QED (only cos we don't have to look to the next solution) If we had to go to the next solution: 21=21, x=10, z=4 gives m=31, but the 'y' remainder is 1, not 0 etc.