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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

Best Answer 

 #2
avatar+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
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+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
avatar+2448 
+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
 #3
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+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

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