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

Question

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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Archit3110 · Answer

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

exc4libur · Answer

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)

TestPrepUnlimited · Answer

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

ScottTargetTestPrep · Answer

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

07tiger · Answer

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

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

Fdambro294 · Answer

72 consecutive integers

The divisor is 8.

If we break up the 72 consecutive integers into groups of 8 consecutive integers —- we will have 9 groups

n———— (n + 1) ————-(n +2)
1………….….2……………….3 
2……………..3…….,,……….4
3……………..4……………….5
4……………..5……………….6
5……………..6…………….....7
6……………..7……………….8
7……………..8………….……9
8……………..9……………….10

Out of each group of 8 consecutive integers, 5 will result in a product divisible by 8:

(2,3,4)
(4,5,6)
(6,7,8)
(7,8,9)
(8,9, 10)

5/8 probability of choosing a number N such that the product of 3 consecutive integers is divisible by 8

Answer
5/8

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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

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