What is the smallest positive integer which is divisible by 225 and uses only the digits 0, 1 and 2?

Question

Show work please

Answer 1

225 = 9 * 25

So answer must be divisible by 9 and 25

To be divisible by 25, the last two digits must be 00, 25, 50, or 75. Only 00 works since we can only use 0,1 and 2.

To be divisible by 9, the sum of the digits must be divisible by 9.

So the smallest number within these constraints is 1222200

Answer 2

Patrick provided an excellent analysis below. I would like to expand on his analysis just a little so that the rationale is clearer, at least in my opinion.

Patrick showed that the last two digits must be 00. The rest of the digits must sum to 9 because 9 is a factor of 225.

The least number of digits, other than the 00, must be 5. Why? 2+2+2+2+1=9. If you convert a 2 into two 1's, you will then have 7 digits (other than the 00). This will result in a larger number. Likewise for converting more than one 2 into 1's.

Note that no combination of 2's and 1's with less than 5 digits can possibly add up to 9. 9/2 = 4 with a remainder of 1.

If you put any zeros into this number (other than a leading zero which will not change the number's value), you will make the number larger because it will have more digits.

If 2 is the first digit, it will be greater than any 5 digit number beginning with 1. Thus the least value will be 12222 which you tack on to 00, resulting in 1,222,200.