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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.

Jul 22, 2022

#2
+2448
+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 $${8 \over 30} = \color{brown}\boxed{4 \over 15}$$

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

Jul 22, 2022
#2
+2448
+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
#3
+1

123456 ==1 combination

12, 16, 24, 32, 36, 52, 56, 64==8 permutations

8 / [6 P 2] ==8 / 30 ==4/15

Jul 23, 2022
edited by Guest  Jul 23, 2022