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?

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.

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