If the six digits 1, 2, 3, 4, 5, and 6 are randomly arranged into a six-digit positive integer, what is the probability that the integer is divisible by 4? Express your answer as a common fraction.
+2668 Note that we only need to consider the final 2 digits of the numbers.
Just count the multiples of 4 through as follows:
11 - 16: 5 integers (excluding 11), 2 are divisible by 4 (12, 16)
21 - 26: 5 integers (excluding 22), 1 is divisible by 4 (24)
31 - 36: 5 integers (excluding 33), 2 is divisible by 4 (12, 16)
41 - 46: 5 integers (excluding 44), 0 are divisible by 4 (44 doesn't count)
51 - 56: 5 integers (excluding 55), 2 are divisible by 4 (52, 56)
61 - 66: 5 integers (excluding 66), 1 are divisible by 4 (64)
So, the probability is \({8 \over 30} = \color{brown}\boxed{4 \over 15}\)
1, 2, 3, 4, 5, 6==6! ==720 permutations
2^6 x 3 ==192 permutations that are divisible by 4
The probability is: 192 / 720==4 / 15
+2668 Note that we only need to consider the final 2 digits of the numbers.
Just count the multiples of 4 through as follows:
11 - 16: 5 integers (excluding 11), 2 are divisible by 4 (12, 16)
21 - 26: 5 integers (excluding 22), 1 is divisible by 4 (24)
31 - 36: 5 integers (excluding 33), 2 is divisible by 4 (12, 16)
41 - 46: 5 integers (excluding 44), 0 are divisible by 4 (44 doesn't count)
51 - 56: 5 integers (excluding 55), 2 are divisible by 4 (52, 56)
61 - 66: 5 integers (excluding 66), 1 are divisible by 4 (64)
So, the probability is \({8 \over 30} = \color{brown}\boxed{4 \over 15}\)
