| Manifesto | Bookstore | Contents | Amazon store | Term index | What changed? | Contact | Recommend |
Probability of DivisibilityFind the probability that if the digits 0, 1, 2, ..., 9 be placed in random order in the blank spaces of
the resulting number will be divisible by 396. Reference
Copyright © 1996-2010 Alexander Bogomolny
1. You can check the solution.
Copyright © 1996-2010 Alexander Bogomolny
The problem is obviously contrived. The last two digits of the longish number are 76, implying that the number is divisible by 4. A number is divisible by 9 iff the sum of its digits is divisible by 9. The sum of the missing digits is
Which is divisible by 9. The sum of the present digits is
Which is also divisible by 9. It follows that the resulting number is divisible by 9 regardless of the placement of the missing digits. To check whether a number is divisible by 11 we compute two sums: that of the evenly placed digits and the sum of the oddly placed digits. All the missing digits come together along with 8, 3, 0, and 6, which add up to
Copyright © 1996-2010 Alexander Bogomolny
|
| 35684770 |  |
