909 is.
3 and 9. 93 has a digit sum of 12, initially, which is divisible by 3, but not by 9. So 93 is divisible by 3, but not by 9. 99 has a digit sum of 18, initially, which is divisible by 3 and 9. So 99 is divisible by both 3 and 9.
All 4 digit numbers that are divisible by 9 are also divisible by 3. The first is 1008 and the last is 9999.
There can be no such number.
To be divisible by 2 the number must be even, that is its last digit must be 2, 4, 6, 8, or 0; the last digit of 432 is 2 which is one of {2, 4, 6, 8, 0} so 432 is even and divisible by 2. To be divisible by 3 sum the digits of the number; if this sum is divisible by 3 then the original number is divisible by 3. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 3, 6, or 9 then the original number is divisible by 3; For 432: 4 + 3 + 2 = 9 which is one of {3, 6, 9} so 432 is divisible by 3. To be divisible by 5 the last digit must be a 0 or 5; the last digit of 432 is a 2 which is not 0 nor 5, so 432 is not divisible by 5. To be divisible by 6, the number must be divisible by both 2 and 3; these have been tested above and found to be true, so 432 is divisible by 6. To be divisible by 9 sum the digits of the number; if this sum is divisible by 9 then the original number is divisible by 9. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 9 then the original number is divisible by 9; For 432: 4 + 3 + 2 = 9 which is 9 so 432 is divisible by 9 To be divisible by 10 the last digit must be a 0; the last digit of 432 is a 2 which is not 0, so 432 is not divisible by 10. 432 is divisible by 2, 3, 6 and 9, but not divisible by 5 nor 10.
To be divisible by 2 the number must be even, that is its last digit must be 2, 4, 6, 8, or 0; the last digit of 26 is 6 which is one of {2, 4, 6, 8, 0} so 26 is even and divisible by 2. To be divisible by 3 sum the digits of the number; if this sum is divisible by 3 then the original number is divisible by 3. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 3, 6, or 9 then the original number is divisible by 3; For 26: 2 + 6 = 8 which is not one of {3, 6, 9} so 26 is not divisible by 3. To be divisible by 5 the last digit must be a 0 or 5; the last digit of 26 is a 6 which is not 0 nor 5, so 26 is not divisible by 5. To be divisible by 9 sum the digits of the number; if this sum is divisible by 9 then the original number is divisible by 9. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 9 then the original number is divisible by 9; For 26: 2 + 6 = 8 which is not 9 so 26 is not divisible by 9. 26 is divisible by 2 but not divisible by 3, 5 nor 9.
180
3333
900
100. The highest 3 digit number divisible by 9 is 999 (111 times). The highest 2 digit number divisible by 9 is 99 (11 times). The difference is 100.
Numbers divisible by 9 have the sum of their digits equal to 9 or a multiple of 9. A number divisible by 2 is an even number. If a 3 digit number is 42n then n can only be 3 if the number is divisible by 9 and 423 is not within the specified range. If a 3 digit number is 43n then n must be 2 for it to be divisible by 9.. The number is thus 432 and this is even and so divisible by 2. If the 3 digit number is 44n then n must be 1 and 441 is odd and not divisible by 2. The only valid solution is 432.
900 is one such number.
It needs a 3.