Puzzle Answer #4
  |
The
prime factorization of 225 is 5 * 5 * 3 * 3. Therefore, the answer of this
question will be both a factor of 25 and of 9. All factors of 25 end in either 00, 25, 50, or 75. The only one of these composed of 0's and 1's is obviously 00, so the answer must end in 00. The hard part is finding a series of 0's and 1's preceding the 00 that will make the entire number divisible by 9. If you didn't already know the following trick then this problem would be very hard. If you did know it then the problem was likely very easy. The trick is that if the sum of digits of a number is divisible by 9; then the number itself is also divisible by 9. |
|
| For example: | 17,685
/ 9 = 1,965 because 1 + 7 + 6 + 8 + 5 = 27, and 27 is divisible by 9. 28,899 / 9 = 3,211 because 2 + 8 + 8 + 9 + 9 = 36, and 36 is divisible by 9. |
|
 |
||
| To prove this, let's consider any 5-digit number – a, b, c, d, and e. This number can be expressed as follows: | ||
| a * 10,000 + b * 1,000 + c * 100 + d * 10 + e | ||
| = | a * (9,999 + 1) + b * (999 + 1) + c * (99 + 1) + d * (9 + 1) + e * 1 | |
| = | (a * 9,999 + b * 999 + c * 99 + d * 9) + (a + b + c + d + e) | |
| = | 9 * (a * 1,111 + b * 111 + c * 11 + d * 1) + (a + b + c + d + e) | |
 |
| From the above, 9 *
(a * 1,111 + b * 111 + c * 11 + d * 1) must be divisible by 9 because it
is a factor of 9. So, if (a + b + c + d + e) is also a factor of 9, then
the entire number must be a factor of 9. As a conclusion, the smallest number consisting of all 1's and divisible by 9 is thus 111,111,111. Adding the two zeros at the end of 111,111,111, the answer to the problem is 11,111,111,100.
|