The prime factorization of 225 is 5*5*3*3. So the answer 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.
Now, we know that if the sum of digits of a number is divisible by 9 then the number itself is also divisible by 9. Note that this is true for 3 also.
Since we are to find a series of zeros and ones which will be divisible by 9, we get the following:
The smallest number consisting of all 1's and divisible by 9 is thus 111,111,111. Adding the two zeros at the end results in the answer to the problem: 11,111,111,100.
|