99,600 is the largest 5-digit number that meets both requirements.
990.
99990
99,996
99990
There is no such number. If you have any such number, n, that is divisible by 3 and 5 then n + 15 is larger, and is divisible by both. And you can add another 15 to that number, and then to that, for ever more.
You can continue to generate numbers that are divisible by both 5 and 9 indefinitely. There is no upper limit.
The largest such number is 9990. 9990/2 = 4995, and 9990/5 = 1998.
Any number that is divisible by both 2 and 3 is divisible by 6.
If a number is divisible by both 8 and 11, it will be divisible by 88.
999 is divisible by 9, but not by six; the next lower number divisible by 9 is 990, which is also divisible by 6, so that's the answer. Some shortcuts for divisibility: 0 is divisible by any number. If the last digit of a number is divisible by 2, the number itself is divisible by 2. If the sum of the digits of a number is divisible by 3, the number itself is divisible by 3. If the last TWO digits of a number are divisible by 4, the number itself is divisible by 4. If the last digit of a number is divisible by 5, the number itself is divisible by 5. If a number is divisible by both 2 and 3, it is divisible by 6. If the last THREE digits of a number are divisible by 8, the number itself is divisible by 8. If the sum of the digits of a number is divisible by 9, the number itself is divisible by 9. 990: 9+9+0=18, which is divisible by 9, so 990 is divisible by 9. 18 is also divisible by 3, so 990 is divisible by 3, and since 990 ends in 0 it's also divisible by 2, meaning that it's divisible by 6 as well.
The smallest number that is divisible by both 34 and 35 is 1,190.
In order to be divisible by both 2 and 3, a number must be divisible by 6.The largest multiple of 6 less than 100 is 96 ... the 16th multiple of 6.So there are 16 of them.