Enigma 1776

Spoiler for New Scientist Enigma 1776 “Elevenses”  (Follow the link to see the puzzle.)

For a warmup puzzle see the comments below!

(At least) one number must be a multiple of 55. But the even multiples of 55 (through 990) have repeated digits, so it must be an odd multiple. So one number must be 165, 275, 385, 495, 605, 715, 825, or 935. The digit 5 must occur only in the units place of one number.

One number must be a multiple of 99, but not 594 (5 not in units place) or 990 (repeated digits). So one number must be 198, 297, 396, 495, 693, 792, or 891. The digit 9 must occur only in the tens place of one number.

There must be a multiple of 88, but not 440, 616, or 880 (repeated digits) and not 352 or 528 (5 not in units place) and not 968 (9 not in tens place). So one number must be 176, 264, 704, or 792.

297 shares one or more digits with each of these multiples of 88, so is ruled out.

If the multiples of 88 and 99 are distinct numbers, the only options are:

  • 176 495; the latter is a multiple of 55. For the third number we need a multiple of 6, 7, and 11: 6 x 7 x 11 = 462 and 2 x 6 x 7 x 11 = 924, each of which uses two digits already used, so this is ruled out.
  • 264 198 or 264 891; ruled out because no multiple of 55 can be used as the third number.
  • 704 198 or 704 891; ruled out because no multiple of 55 can be used as the third number.
  • 704 396 or 704 693; multiple of 55 must be 825. But 704 396 825 ruled out because no multiple of 77 and 704 693 825 ruled out because no multiple of 66.

So we have to use 792 as our multiple of both 88 and 99. Compatible multiples of 77 are 308, 385, and 693. Compatible multiples of 55 are 165, 385 and 605. But 605 is incompatible with all three multiples of 77 and 693 is incompatible with all three multiples of 55, so they are eliminated. So we’re down to these possibilities:

  • 792 385 xxx
  • 792 308 165

In the first case 792 is a multiple of 2, 3, 4, 6, 8, 9, and 11, and 385 is a multiple of 5, 7, and 11, so our third number only has to be a multiple of 11 using three of the digits 0, 1, 4, and 6. But there are none.

In the second case 792 is a multiple of 2, 3, 4, 6, 8, 9, and 11;  308 is a multiple of 2, 4, 7, and 11; and 165 is a multiple of 3, 5, and 11. So these three numbers are the only solution.

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2 thoughts on “Enigma 1776

  1. Spoiler for warmup puzzle…

    If all three numbers are divisible by 36 then we get divisibility by 2, 3, 4, 6, and 9 for free. We just need divisibility by 5, 7, and 8.

    For 7: We must use multiples of 36 x 7 = 252. We can’t use 252 but can use 504 or 756.

    For 5: We must use multiples of 36 x 5 = 180. We can use 180, 360, 540, 720, but not 900. Last digit must be 0, so 504 can’t be used; our multiple of 7 must be 756. Then 360, 540, 720 can’t be used; our multiple of 5 must be 180.

    Our third number must be a multiple of both 8 and 36, so is a multiple of 72, and must use three of the digits 2, 3, 4, 9. The only three of these that sum to a multiple of 9 (required for the whole number to be a multiple of 9) are 2, 3, and 4, so these are the digits to use. Quickly eliminate all but 324 and 432 by requiring the last 2 digits to be a multiple of 4 (required for the whole number to be a multiple of 4). Then a quick check shows 324 isn’t a multiple of 72 but 432 is.

    Answer: 180, 432, 756.

    For common factors of 12, 14, 18, 21, 24, 27, 28, or 54 (instead of 11 or 36) there are multiple solutions. For any other common factor there are no solutions.

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