Just multiply 9 with different integers:
9 x 0
9 x 1
9 x 2
etc.
what multiples do 7,5 and 9 have that are the same
Both digits are successive counting numbers.
The first 5 multiples of 5 are: 5, 10, 15, 20, 25 The first 5 multiples of 9 are: 9, 18, 27, 36, 45 The first 5 multiples of both 5 and 9 are: 45, 90, 135, 180, 225
The first 5 common multiples are the first 5 multiples of their lowest common multiple (LCM) LCM(9, 10) = 90 → first 5 common multiples are 90, 180, 270, 360, 450.
315
If they are multiples of 9, the digits add up to a multiple of 9.
Because they're in multiples of three - a quick way to tell if a number is a multiple of three is to add up the digits and see if the digits add up to a multiple of three e.g 576, 5+7+6=18, 1+8=9, 9 is a multiple of three
That the sum of its digits add up to 9 as for example 9*9 = 81 and 8+1 = 9
the only multiples of 5 are 30.35.40.45.50, and 55 the only sum to 9 is 45
117. All multiples of 9 have digits that add up to something divisible by 9. Examples, 117, 36, 45, 54 all have digits that add up to 9.
If you add up the two digits and they make either 3, 6, or 9
The sum of the digits is also a multiple of 9. And, if the sum of the digits is too large, you can sum those digits and keep going.
The digits of 498 do not add to a multiple of 9 so it isn't.
So there is a rule about multiples of 9 in general: something is a multiple of 9 if and only if its digits add up to a multiple of 9. 6+4 = 10, so it is not a multiple of 9. There's also a neat trick to remembering the multiples of 9 below 99 with your fingers, but you can look this up if you're interested.
The total sum of the digits has to equal nine.
There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.
You can experiment a bit for this one. Or you can apply some reasoning:* Multiples of 45 end with 5 or 0. To avoid a 5, you would have to multiply by an even number; that is, you are looking for multiples of 90. * Since multiples of 90 are also multiples of 9, the sum of the digits must needs be 9. This means that you must provide 9 ones (or a multiple thereof), and the last digit must be 0. The smallest number that satisfies these criteria is 1111111110.