Yes, you can tell (without actually doing the division).
So numbers ending in 32, 36, 52, 24, 04, .... and like this example 68 are all divisible by 4. Why this works - without a true formal proof, consider this: break the number apart, to the last two digits and the rest of the number.
So our number becomes 968 = 900 + 68. So 968/4 = (900 + 68) / 4, which equals 900/4 + 68/4. Since 100 is divisible by 4, then any multiple of 100 (such as 900) is also divisible by 4, so now you only have to consider the last two digits.
Its divisible by 2 and 4.
No.
It is: 968/4 = 241
1 x 968, 2 x 484, 4 x 242, 8 x 121, 11 x 88, 22 x 44 = 968
A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.
No. 968 is not evenly divisible by nine.
Its divisible by 2 and 4.
1, 2, 4, 8, 11, 22, 44, 88, 121, 242, 484, 968.
No.
It is: 968/4 = 241
The positive integer factors of 968 are: 1, 2, 4, 8, 11, 22, 44, 88, 121, 242, 484, 968
The factors of 968 are: 1, 2, 4, 8, 11, 22, 44, 88, 121, 242, 484, and 968 The prime factorization is: 968 = 2 * 2 * 2 * 11 * 11
1, 2, 4, 8, 11, 22, 44, 88, 121, 242, 484, 968.
That is 968.
1 x 968, 2 x 484, 4 x 242, 8 x 121, 11 x 88, 22 x 44 = 968
A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.
Is 348 divisible by 4?