All of the multiples of 78 are. There are an infinite number of them.
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.
1170 is.
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
The statement is not correct. The sum of the digits of a number being divisible by 2 indicates that the sum is even, but it does not guarantee that the original number itself is even. A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8). For example, the number 13 has a digit sum of 4 (1 + 3), which is even, but 13 itself is not divisible by 2.
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
If a number is divisible by both 2 and 3, it is also divisible by 6. This is because 6 is the least common multiple of 2 and 3. Therefore, any number that meets the criteria of being divisible by both 2 and 3 will also be divisible by 6.