Moving the 6

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George Mach posed this problem to one of his classes (out at Cal Poly SLO in California), and my dad passed it along for posting here. The answer may be surprising, in the sense that it’s not something that can be easily guessed. Here’s the question:

Find a whole number ending in 6 which is doubled if you move the 6 from the end to the beginning. (e.g. the number 316 almost works because 631 is close to the double of 316, but not quite)

I’ll post the answer here tomorrow….

There are an infinite number of solutions, the smallest of which is 315,789,473,684,210,526

This problem can also be generalized to ending in a different digit and the permutation being a different multiple; according to Jim Delany (thanks Jim!) the smallest numbers that occur are those that end in 4, 5, 6, 7, 8, 9 and are quadrupled (or quintupled, for the one ending in 7). All of those numbers are less than 300,000.

5 Responses to “Moving the 6”

  1. DavidP Says:

    I think I got it. 17 digits, right?

  2. Ξ Says:

    18 digits, starting with 31.
    (I initially said “starting with 1”, but that’s obviously incorrect!)

  3. TwoPi Says:

    I liked this one, in part because there are (at least) two radically different attacks on it.

    Method one: If our number ends in a six, then 2*number ends in a 2 (with a carry of 1). So we’ve learned the number ends “26”, and so now 2*number ends with 52 (with no carry). Now we’ve got the last three digits (526), and we continue in this fashion until we get a lead digit of six with no carrying in 2*number.

    Method two: If we call the block of lead digits “x”, the original number has value 10x+6, while when doubled its value is 6\cdot 10^n + x. Setting 2\cdot (10x+6) equal to 6\cdot 10^n + x, we find that 19x = 12( 5\cdot 10^{n-1} - 1). Since 19 is prime, it follows that 19 must divide 4, or 49, or 499, or…, and that x is the product of 12 and the quotient. So one starts by dividing 19 into 4999999999…, stopping (and discarding trailing digits) once a remainder of 0 occurs.

  4. Efrique Says:

    I did it by looking at repeating sequences of decimal expansions.

    6/19 = 0.315789473684210526(315789473684210526…)

    12/19= 0.631578947368421052(631578947368421052…)

    Hence the required number is 315789473684210526

    The decimal expansions of smaller denominator- fractions can be rejected for a variety if reasons; any that terminate don’t work, you need more than a single repeating digit, the decimal expansion must contain a 6, and so on. 13 and 17 look like potential candidates, but you can quickly rule them out because moving the 6 digit in the expansions won’t double the number.

  5. TwoPi Says:

    Let’s see how this works:

    We’re hunting for a rational number r of the form r = 0.×6 x6 x6 x6 x6…, where “x6” is the integer we want.

    Then 2r = 0.6x 6x 6x 6x 6x 6x 6x 6x 6x 6x 6x…, and if we multiply that equation by 10, we get 20r = 6.×6 x6 x6 x6 x6 x6 x6…

    Subtract r from both sides, and you find that 19r = 6, and r = 6/19.

    Now examining the decimal expansion of r, and looking for a place where it begins to repeat AND has a 6 (so we ignore the first 6 in the decimal expansion of r), yields the smallest solution to the problem. (Other solutions are found by taking multiple blocks — e.g., x6x6 is a 34 digit solution.)

    Cool!

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