There are three 3 digit numbers such that the digits 1,2,3...8,9 are used once each to write them. They are found to be in the ratio 1:3:5
Find all such possible triplets.
Let the numbers be x=x1x2x3, y=y1y2y3, z=z1z2z3;
where all xi's yi's and zi's are distinct digits, y=3x and z=5x.
x1 has to be 1(otherwise z will not be a 3 digit number).
x3 cannot be even because otherwise, z3 will be 0.
Hence z3 will be 5 because x3 is odd.
x3 cannot be 7 because that will make y3 1.
So we have x3 as 3 or 9.
If x3 is 3, then y3=9
Also z2=5*x2 + 1(modulo 5).
This will rule out the possibility that x2 is even because z2 will become 1 and 1 has been used already(x1=1).
Hence x2 has to be 7 as all other odd numbers have been used up.
We can verify that x=173 does not satisfy our requirements.
Therefore x3=9 and this makes y3=7.
We have used 1,7,9 and 5 so far. Hence x2 can be 2,4,6,8 or 3.
189*5=945, 139*3=417, 169*3=507(zero cannot be used) and 149*3=447.
That leaves us with 129, 387 and 645 as the only answer.
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