Put:

X = A + B*7 + C* 7^2 + D*7^3

with 0<= A,B,C <=6, 1<=D<=6

Then:

Y = D + A*7 + B*7^2 + C*7^3

and we are told that Y=2*X.

So we may conclude that 2*D=C (there is no base 7 carry), so

D<=3, but then 2A=D, so D is even, and as it is the modt significant

digit (base 7) of a four digit number it !=0, so D=2. Then A=1, and C=4.

So we now have:

X = 1 + B*7 + 4* 7^2 + 2*7^3

Y = 2 + 1*7 + B*7^2 + 4*7^3

So 2*B=1 or 8, but neither of these is possible, so there is no such number

X with the given properties!

Have you got the question right?

Can you see a hole in my argument?

RonL