If n is a positive integer chosen at random, what is the pro

Question

If n is a positive integer chosen at random, what is the probability that  (n^3 - n)(n + 2)  is divisible by 9?

A)  \frac{1}{3}  
B)  \frac{4}{9}  
C)  \frac{1}{2}  
D)  \frac{5}{9}  
E)  \frac{7}{9}

Amity007 · Answer

How can we answer this without a range or constraints for n?

kevincan · Answer

You don’t need one , but let’s say a positive integer less than or equal to 90

Diyaj2403 · Answer

Could someone please confirm whether checking each value between 0-9 is the way to approach this?

kevincan · Answer

What about factoring the expression?

agrasan · Answer

Hi  kevincan

I did this question but the answer I get is 4/9, here is my approach.

By factoring we get,
n*(n-1)*(n+1)*(n+2)

to be divisible by 9, n can be 7,8,9,10, then 17,18,19,20...we see that there is a pattern of  quadruplets  so if set of n is   then there will be 36 possible values of n to be divisible by 9.

Probability = 36/90 = 4/9

Please let me know where I am going wrong here.

kevincan · Answer

You are assuming that for the product of 4 integers to be a multiple of 9, one of the integers has to be a multiple of 9. That would be true if nine were a prime number

Kinshook · Answer

If n is a positive integer chosen at random, what is the probability that  (n^3 - n)(n + 2)  is divisible by 9?

E = (n^3 - n)(n + 2) = (n-1)n(n+1)(n+2)

Let us take the the numbers  . The same divisibility pattern will be repeated after every 9 numbers

n = 1; (n-1)n(n+1)(n+2) = 0*1*2*3 ; Divisible by 9 ? Yes
n = 2; (n-1)n(n+1)(n+2) = 1*2*3*4 ; Divisible by 9 ? No
n = 3; (n-1)n(n+1)(n+2) = 2*3*4*5 ; Divisible by 9 ? No
n = 4; (n-1)n(n+1)(n+2) = 3*4*5*6 ; Divisible by 9 ? Yes
n = 5; (n-1)n(n+1)(n+2) = 4*5*6*7 ; Divisible by 9 ? No
n = 6; (n-1)n(n+1)(n+2) = 5*6*7*8 ; Divisible by 9 ? No
n = 7; (n-1)n(n+1)(n+2) = 6*7*8*9 ; Divisible by 9 ? Yes
n = 8; (n-1)n(n+1)(n+2) = 7*8*9*10 ; Divisible by 9 ? Yes
n = 9; (n-1)n(n+1)(n+2) = 8*9*10*11 ; Divisible by 9 ? Yes

The probability that (n^3 - n)(n + 2) is divisible by 9 = 5/9

IMO D

kevincan · Answer

We could also reason thus : the product will be a multiple of 9 if and only if the first term is a multiple of 3 or one term is a multiple of 9.

If n is a positive integer chosen at random, what is the pro

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If n is a positive integer chosen at random, what is the pro

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