I'm in 7th grade trying to take a pre-algebra class..... we just learned our rules but i have come to the conclusion of 133.
Total number of 3 digit numbers = 999-100+1(to include 100)= 900 numbers
Therefore numbers not divisble by 2,3,5 and 10 are
900*(1-1/2)*(1-1/3)*(1-1/4)*(1-1/5)*(1-1/10)= 216 numbers
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this # is divisible by 3, 5, and 9 :-) ;-P
Well, darling, the positive integers divisible by 2 are all the even numbers, and the ones divisible by 3 are those multiples of 3. So, to find the numbers divisible by either 2 or 3, you just need to combine these two sets. In other words, it's the set of all even numbers and multiples of 3. Voilà!
Add together all the digits until you get a single figure answer. If that's divisible by three, so's the original number. E.g. 4,239 broken down to 4 + 2 + 3 + 9 = 18; 1 + 8 = 9, which is a multiple of 3. The original number is therefore divisible by 3
Out of that list, 255 is evenly divisible by 3 and 5.
To determine if 4908 is divisible by 3, we need to sum its digits. 4 + 9 + 0 + 8 = 21. Since 21 is divisible by 3 (21 ÷ 3 = 7), 4908 is also divisible by 3. Therefore, 4908 is divisible by 3.