math
posted by Chris .
Find a sixdigit 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.

abcdef
a=b+4
c=f1
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