Six digit conspiracy

Following is a spoiler for New Scientist Enigma #1550:

Observe that 333 * 444 = 147852, and 333 * 777 = 258741 — same six digits in reverse order. Coincidence? Or is there a reason for it?

There’s a reason. You can factor out 111^2 = 12321 to get:

333 * 444 = 3 * 4 * 12321 = 12 * 12321 and
333 * 777 = 3 * 7 * 12321 = 21 * 12321

If you do the multiplications on the right you find nothing is ever carried from one place to the next, so it makes perfect sense that one comes out the reverse of the other.

But now notice this: 666 * 777 = 2 * 333 * 777 = 517482. The same six digits in a different order. Here there doesn’t seem to be any obvious reason (and no other multiplications of the form AAA * BBB give those same digits.) Strange.

And notice also that 999999/7 = 142857 also uses those same six digits in yet another order. This is the famous number which if multiplied by any digit from 2 through 6 gives back those same six digits in a cyclic permutation. Any relation to 333 * 777? None that I can see. (Of course 999999 is a multiple of 333, but so what?)

My brain wants there to be some underlying connection between all these oddities involving the digits 1, 2, 4, 5, 7, 8, but I can’t come up with any.

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3 thoughts on “Six digit conspiracy

  1. I learned that the formula to determine how a periodic decimal number can be expressed as a fraction, with the numerator as the period and the denominator as many 9s as the digits of the period. So 142857/999999 factors out to 1/7 which is 0.142857142857…

    As to why the fractions of 7 yield the same six digits in different order… yeah, dunno.

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  2. I guess these are coincidences but not as remarkable as they may seem. For instance: 666 * 777 has the same digits as 333 * 777, in a scrambled order. But both numbers are multiples of 111. Now multiples of 11 have the property that the sum of the last two digits (taken as a two digit number) plus the next two, plus the next two… etc., are divisible by 11. For instance, 253 = 23 * 11, and 2 + 53 = 55. Similarly, multiples of 111 when added in groups of three digits are divisible by 111. Example: 53058 — 53 + 058 = 111, so 53058 is a multiple of 111 (it’s 111 * 478). Now of course 111 is divisible by 3, so any multiple of 111 must have its digit sum mod 3 = 0, but (multiple of 111) * (multiple of 111) is divisible by 9 so must have its digit sum mod 9 = 0. For instance, if you know 666 * 777 = 517xxx, you can determine there are only a few possibilities for xxx without actually doing the multiplication. For divisibility by 111, xxx must be 038, 149, 260, 371 … or 926, and for divisibility by 9 you can reject all but 149, 482, and 815. Roughly speaking it looks like the “probability” that a 6 digit number of the form XXX * YYY has digits 1, 2, 4, 5, 7, 8 is the “probability” that its first 3 digits are 3 of those numbers (with no two adding to 9) times a factor of about 1/3. Which is something like (6/9) * (4/10) * (2/10) * (1/3) = 1.8%. Small but not crazy small. “Probability” in quotes because of course there’s nothing random about arithmetic.

    That’s all sort of off the top of my head and may be completely stupid.

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