math

Find a six-digit even number containing no zeros and no repeating digits in which the first digit is four more than the second digit, the third digit is one less than the sixth digit, and the fourth and fifth digits when read as a single number equal the product of the first and sixth digits.

  1. 0
  2. 0
  3. 7
asked by Chris
  1. abcdef
    a=b+4
    c=f-1
    de=af

    Let's see what de can be, and what af could then be. Remember that a>=5, since b>=1
    de: af
    14: 72 b=3 c=1 no
    15: 53 no, odd
    16: 82 b=3 c=1 no
    18: 92 b=5 c=1 no, or 63 b=2 c=1 no
    21: 73 no, odd
    24: 83 no, odd
    27: 93 no, odd
    28: 74 b=3 c=3 no
    32: 84 b=4 no
    36: 94 b=5 c=3 no
    42: 67 no, odd; 76 b=3 c=5 YES
    54: 69 no, odd; or 96 b=5 no
    56: 78 b=3 c=7 no, or 87 b=4 c=6 no
    63: 79 or 97 no, both odd
    72: 89 no, odd; 98 b=5,c=7 no

    So, we are left with 735426

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    posted by Steve

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