Hi Dan,

[For this reply, I will use red to indicate base 7 numbers and black for decimal numbers. For example, 602₇ = 296₁₀.]

Suppose that in base 7, X = abcd₇ that is X = 7³ a + 7² 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 bcda₇ is twice X so

bcda₇ = (2)(abcd₇)

7³ b + 7² c + 7 d + a = 2(7³ a + 7² b + 7 c + d)

which can be rewritten

7³ (b - 2a) + 7² (c - 2b) + 7 (d - 2c) = 2d - a

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:

(d, a) { (4, 1), (5, 3), (6, 5) }

The pair (6, 5) doesn't work since a = 5 makes X = 5bcd₇ and 2X is a five digit number since (2)(5₇) = 13₇.

The pair (5, 3) doesn't work either. Can you see why?

Thus X = 1bc4₇ and (2)(1bc4₇) = bc41₇.

Now perform the multiplication (2)(1bc4₇). 2 times 4 is 8₁₀ which is 11₇ so there is a carry of 1. Hence for the next digit 2c plus the carry of 1 gives ?4₇ 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

(2)(1b54₇) = b541₇

Finally perform the multiplication (2)(1b54₇) and see what value(s) of b give (2)(1b54₇) = b541₇.

Penny