



 
Hi Dan, [For this reply, I will use red to indicate base 7 numbers and black for decimal numbers. For example, 602_{7 } = 296_{10}.] Suppose that in base 7, X = abcd_{7 } that is X = 7^{3} a + 7^{2 }b + 7 c + d. Since we are working in base 7, the possible values of a, b, c and d are 0, 1, 2, 3, 4, 5 or 6. You know that bcda_{7 } is twice X so
which can be rewritten
The left side of this equation is divisible by 7 so the right side must be divisible by 7. So what combinations of d and a make the right side divisible by 7? For each of the possible values of d, d = 0, 1, ... 6 I looked for corresponding values of a, if any. I found three possible pairings:
The pair (6, 5) doesn't work since a = 5 makes X = 5bcd_{7 } and 2X is a five digit number since (2)(5_{7}) = 13_{7}. The pair (5, 3) doesn't work either. Can you see why? Thus X = 1bc4_{7 } and (2)(1bc4_{7}) = bc41_{7}. Now perform the multiplication (2)(1bc4_{7}). 2 times 4 is 8_{10 } which is 11_{7 } so there is a carry of 1. Hence for the next digit 2c plus the carry of 1 gives ?4_{7 } where ? is some digit, ie (2c + 1)_{7 } = (?4)_{7 }. I then tried all the possibilities for c, c = 0, 1, ..., 6 and found that c = 5. Thus
Finally perform the multiplication (2)(1b54_{7}) and see what value(s) of b give (2)(1b54_{7}) = b541_{7}. Penny  


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