The TEN DIGIT Teaser ANSWERHere's the question:
Can you form a ten-digit number using each digit once so that...
... and so on until... SOLUTION
Call the number abcdefghij. As the whole 10-digit number divides by 10, j = 0.
ab, abcd, abcd5f and abcd5fgh must all divide by 2. We have already used 0, so b, d, f and h must be 2, 4, 6 and 8 in some order. By elimination, a, c, g and i must be 1, 3, 7 and 9 in some order. abcd must be divisible by 4, and c must be odd; therefore d is 2 or 6. Similarly h is also 2 or 6, so b and f must be 4 and 8 (in some order). So far we have:
a = 1, 3, 7 or 9
abc and abcdef must both divide by 3, so a + b + c and a + b + c + d + e + f must both divide by 3, and it follows that d + e + f must also divide by 3. Since d = 2 or 6, e = 5 and f = 4 or 8, the only possible combinations for these three digits are 258 or 654. Once d and f have been determined, b and h follow as there is only one possible digit remaining for each. The two possibilities are: a4c258g6i0 a8c654g2i0 For a4c258g6i0, a4c must be divisible by 3 and a, c must be some two of 1, 3, 7, 9. The only possibilities are 147 or 741. Furthermore, a4c258g6 must be divisible by 8, so 8g6 must be divisible by 8 which means g is either 1 or 9. This gives the following numbers as candidates: 1472589630 7412589630 For a8c654g2i0, a8c must be divisible by 3 and a, c must be two of 1, 3, 7, 9. The possibilities are 183, 189, 381, 387, 783, 789, 981 or 987. Furthermore, a8c654g2 must be divisible by 8, so 4g2 must be divisible by 8 and g is either 3 or 7. This eliminates 387 and 783 as possibilities for abc, leaving the following candidates: 1836547290 1896543270 1896547230 3816547290 7896543210 9816543270 9816547230 9876543210
It turns out that the only set of first seven digits that works is 3816547, so the only correct answer to the question is ...drum roll... 3816547290 With big thanks to MICHAEL JONES for explaining how this all works! |