All of the multiples of 78 are. There are an infinite number of them.
1170 is.
There's a trick you can use: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. Here 2+4+5+2 = 13. Is 13 divisible by 3? You decide.
The divisibility rule for 3 is as follows: If the sum of the digits of a number is divisible by 3 then the number is divisible by 3. For example, the number 21 is divisible by 3 since 2+1=3, and 3 is divisible by 3. The number 49 is not divisible by 3 since 4+9=13, and 13 is not divisible by 3.
78 is evenly divisible by 1, 2, 3, and 13.
13 is one such number.
2+3+2+6=13 which is not divisible by 3 thus 2326 is not divisible by 3
No 13 is not divisible by 3 because there would be a remainder of 1 and a number that is divisible by another should have no remainder
Since 5232 is divisible by both 2 and 3, it is divisible by 6.A number must be divisible by both 2 and 3 to be divisible by 6.The number 5232 is even, so it is divisible by 2.If you add the individual digits in the number (5+2+3+2=12) you get a number that is divisible by 3, meaning the original number (5232) is also divisible by 3.
No. Consider 13 or 23.
It is divisible by 6
Any number that is divisible by both 2 and 3 is divisible by 6.
Using the tests for divisibility:Divisible by 3:Add the digits and if the sum is divisible by 3, so is the original number: 2 + 3 + 4 = 9 which is divisible by 3, so 234 is divisible by 3Divisible by 6:Number is divisible by 2 and 3: Divisible by 2:If the number is even (last digit divisible by 2), then the whole number is divisible by 2. 234 is even so 234 is divisible by 2.Divisible by 3:Already shown above to be divisible by 3. 234 is divisible by both 2 & 3 so 234 is divisible by 6Divisible by 9:Add the digits and if the sum is divisible by 9, so is the original number: 2 + 3 + 4 = 9 which is divisible by 9, so 234 is divisible by 9Thus 234 is divisible by all 3, 6 & 9.