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.

Guest Jul 22, 2022

#2**+1 **

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 \({8 \over 30} = \color{brown}\boxed{4 \over 15}\)

BuilderBoi Jul 23, 2022

#1**+2 **

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**

Guest Jul 22, 2022

#2**+1 **

Best Answer

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 \({8 \over 30} = \color{brown}\boxed{4 \over 15}\)

BuilderBoi Jul 23, 2022