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If an integer n is chosen from the integers 1 through 72, what is the

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If an integer n is chosen from the integers 1 through 72, what is the  [#permalink]

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New post 03 Dec 2019, 02:21
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Re: If an integer n is chosen from the integers 1 through 72, what is the  [#permalink]

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New post 03 Dec 2019, 02:40
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for all even n values i.e 36 and values which are odd+1 ; multiple of 8 ; 7,15,23,31... total 9
45/72 ; 5/8
IMO D

Bunuel wrote:
If an integer n is chosen from the integers 1 through 72, what is the probability that n*(n+1)*(n+2) will be divisible by 8

(A) 1/4
(B) 3/8
(C) 1/2
(D) 5/8
(E) 3/4


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Re: If an integer n is chosen from the integers 1 through 72, what is the  [#permalink]

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New post 04 Dec 2019, 08:24
Bunuel wrote:
If an integer n is chosen from the integers 1 through 72, what is the probability that n*(n+1)*(n+2) will be divisible by 8

(A) 1/4
(B) 3/8
(C) 1/2
(D) 5/8
(E) 3/4


ASSUMING THAT INTEGERS 1 TO 72 ARE INCLUSIVE
n(n+1)(n+2) is divisible by 8:
n = even: (72-2)/2+1=36
n+1 = multiple of 8: (72-8)/8+1=9
favorable cases = 36+9 = 45
total cases = 72C1 = 72
probability: 45/72 = 5/8

Ans (D)
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Re: If an integer n is chosen from the integers 1 through 72, what is the  [#permalink]

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New post 04 Dec 2019, 12:48
Bunuel wrote:
If an integer n is chosen from the integers 1 through 72, what is the probability that n*(n+1)*(n+2) will be divisible by 8

(A) 1/4
(B) 3/8
(C) 1/2
(D) 5/8
(E) 3/4


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We can take a look at even n's first. If n is even, either n or n + 2 will be a multiple of 4 as they are consecutive even numbers. So one of them will contain a factor of 4, the other will contain a factor of 2, and that guarantees the product is divisible by 8 when n is even.
On the other hand when n is odd, the only factor from the product that is even is n + 1. So if we want the product to be divisible by 8 the factor of 8 must come from n + 1 alone. So numbers 7, 15, 23 etc up to 71 will let the product be divisible by 8, which is 72/8 = 9 odd n's. Including the even n's and there are 36 + 9 = 45 numbers that satisfy, 45/72 = 5/8.

Ans: D
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Re: If an integer n is chosen from the integers 1 through 72, what is the  [#permalink]

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New post 09 Dec 2019, 18:53
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Bunuel wrote:
If an integer n is chosen from the integers 1 through 72, what is the probability that n*(n+1)*(n+2) will be divisible by 8

(A) 1/4
(B) 3/8
(C) 1/2
(D) 5/8
(E) 3/4


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We are given that an integer n is to be chosen at random from the integers 1 to 72, inclusive, and we need to determine the probability that n(n + 1)(n + 2) will be divisible by 8.

We should recall that when a number is divisible by 8, it is divisible by 2^3, i.e., three factors of 2. We should also recognize that n(n + 1)(n + 2) is the product of three consecutive integers.

Case 1: n is even

Any time that n is even, (n + 2) will also be even. Moreover, either n or (n + 2) will be divisible by 4, and thus n(n + 1)(n + 2) will contain three factors of 2 and will be divisible by 8.

Since there are 72 integers between 1 and 72, inclusive, and half of those integers are even, there are 36 even integers (i.e., 2, 4, 6, …, 72) from 1 to 72, inclusive. Thus, when n is even, there are 36 instances in which n(n + 1)(n + 2) will be divisible by 8.

Case 2: n is odd

f n is odd, then n(n + 1)(n + 2) still can be divisible by 8, but only if the even factor (n + 1) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 72, inclusive.

Number of multiples of 8 = (72 - 8)/8 + 1 = 64/8 + 1 = 9. Thus, when n is odd, there are 9 instances in which n(n + 1)(n + 2) will be divisible by 8.

In total, there are 36 + 9 = 45 outcomes in which n(n + 1)(n + 2) will be divisible by 8.

Thus, the probability that n(n + 1)(n + 2) is divisible by 8 is: 45/72 = 5/8.

Answer: D
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If an integer n is chosen from the integers 1 through 72, what is the  [#permalink]

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New post 09 Dec 2019, 20:15
In a set of 1 to 8 there are (at first glance) 3 combinations that give multiples of 8. (2*3*4, 4*5*6, 6*7*8)
This pattern will repeat for 9 to 16, 17 to 24 and so on up to 72.
So 3*9 = 27

Now in between above sets something like 8*9*10 also is divisible by 8. So 9 sets means 7 such transition points. So 7 such combinations.

27+7= 34

Oh f***.. I missed that 8, 16, 24 ect. themselves are divisible. So something like 7*8*9 or 15*16*17 are also included so we have 9 such (8, 16, 24) numbers

So 34+9= 43.

By now its proved im total sh** at counting, but after counting all this I cant be too far from the right answer. :|

Anyhow, when we start them all with an even number we get 34 combinations of consecutive intigers. So if we start with odd we should get around 34,35 combinations. 34+34 = 68 is my denominator. :D


Now 43/68 is somewhere around 60%

Hey 5/8 is somewhere around 60% So answer D :lol:
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If an integer n is chosen from the integers 1 through 72, what is the   [#permalink] 09 Dec 2019, 20:15
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