Problem Solving

If an integer n to be chosen randomly between 1 and 96 inclusive, what is the probability that n(n+1)(n+2) is divisible by 8 ?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

OA: D

Well, we know the denominator here is 96, so we just have to figure out how many ways we can multiply three consecutive numbers to get a multiple of 8. Put another way, any multiple of 8, or 2^3, must contain three 2's. One way this can happen is if the middle number is odd, because every odd number must be sandwiched between a multiple of 2 and a multiple of 4. If n+1 is 3, for example, you'd have 2*3*4 --> multiple of 8. If n+1 is 5, you'd have 4*5*6 --> multiple of 8. Between 1 and 96, we've got 48 odds.

The other way we can get a multiple of 8 is if n + 1 is itself a multiple of 8. 96/8 = 12 multiples of 8 between 1 and 96.

So we've got 48 odds + 12 multiples of 8 for a total of 60 desired outcomes.

60/96 = 5/8.

Adding to what others have mentioned, the students should understand the properties of product of 3 consecutive positive integers which has been one of the favorite concepts of GMAT

1) The product of three consecutive positive Integers is always divisible by 6 (because three consecutive Integers will always have atleast one Even Integer and One multiple of 3)

2) The product of 3 consecutive Integers will always be divisible by 24 if the Smallest/Biggest of those Integers is Even (Because the Integers will be (Even)(Odd)(Even) and one of these even integers is bound to be multiple of 4, one will be multiple of 3 therefore (2)x(3)x(4)=24 )

Experts - As far as pattern is concerned, if n=96 then n(n+1)(n+2) will be 96*97*98 and it's a valid combination here to be divisible by 8. Right ?

If n is an integer from 1 to 96 (inclusive), what is the probability that n*(n+1)*(n+2) is divisible by 8?

A.1/4
B.1/2
C.5/8
D.7/8
E.3/4

n(n+1)(n+2) = the product of 3 consecutive integers.
WRITE IT OUT and LOOK FOR A PATTERN.

1*2*3
2*3*4
3*4*5
4*5*6
5*6*7
6*7*8
7*8*9
8*9*10

9*10*11
10*11*12
11*12*13
12*13*14
13*14*15
14*15*16
15*16*17
16*17*18

Each of the products in red is a multiple of 8.
The two examples above imply the following:
Of every 8 products, exactly 5 will be a multiple of 8.

Thus, the probability that n(n+1)(n+2) will be a multiple of 8 = 5/8.

The correct answer is C.

Alternate approach:

Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:
Since every other even integer is a multiple of 4, the product here will always include an even integer and a multiple of 4, resulting in a multiple of 8.
Thus, n can be any even integer between 1 and 96.
96/2 = 48 favorable choices for n.

Case 2: n+1 is a multiple of 8:
The product will be a multiple of 8 if n+1 is a multiple of 8.
Number of multiples of 8 between 1 and 96 = 96/8 = 12.
Thus, there are 12 favorable choices for n+1, implying that there are 12 more favorable choices for n.

Total favorable choices for n = 48+12 = 60.
Favorable choices/Total choices = 60/96 = 5/8.

For a similar problem, check here:

https://www.beatthegmat.com/probability-t116280.html

RBBmba@2014 wrote:As far as pattern is concerned, if n=96 then n(n+1)(n+2) will be 96*97*98 and it's a valid combination here to be divisible by 8. Right ?

Correct.
The eight greatest options for (n)(n+1)(n+2) are as follows:
n=89 --> 89*90*91
n=90 --> 90*91*92
n=91 --> 91*92*93
n=92 --> 92*93*94
n=93 --> 93*94*95
n=94 --> 94*95*96
n=95 --> 95*96*97
n=96 --> 96*97*98
Notice that the pattern repeats:
The products in red are all divisible by 8.

GMATGuruNY wrote:

If n is an integer from 1 to 96 (inclusive), what is the probability that n*(n+1)*(n+2) is divisible by 8?

A.1/4
B.1/2
C.5/8
D.7/8
E.3/4

n(n+1)(n+2) = the product of 3 consecutive integers.
WRITE IT OUT and LOOK FOR A PATTERN.

1*2*3
2*3*4
3*4*5
4*5*6
5*6*7
6*7*8
7*8*9
8*9*10

9*10*11
10*11*12
11*12*13
12*13*14
13*14*15
14*15*16
15*16*17
16*17*18

Each of the products in red is a multiple of 8.
The two examples above imply the following:
Of every 8 products, exactly 5 will be a multiple of 8.

Thus, the probability that n(n+1)(n+2) will be a multiple of 8 = 5/8.

The correct answer is C.

Alternate approach:

Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:
Since every other even integer is a multiple of 4, the product here will always include an even integer and a multiple of 4, resulting in a multiple of 8.
Thus, n can be any even integer between 1 and 96.
96/2 = 48 favorable choices for n.

Case 2: n+1 is a multiple of 8:
The product will be a multiple of 8 if n+1 is a multiple of 8.
Number of multiples of 8 between 1 and 96 = 96/8 = 12.
Thus, there are 12 favorable choices for n+1, implying that there are 12 more favorable choices for n.

Total favorable choices for n = 48+12 = 60.
Favorable choices/Total choices = 60/96 = 5/8.

Hi Mitch - much thanks for your detailed reply.

The above pattern I guess, little bit difficult to recognize during test environment, so I'd go with your alternate approach here.

However, couple of quick clarifications -
1. Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:

Can we consider it in this way that we need to calculate the no. of odds in the given range? Here 'n+1' will take the odd position in the middle. Still we'll have 48 odds with 3 being the first and 97 being the last in the scenario where n=96. This approach of Case 1 reflects the same thing I think. Right ?

2. n=96 --> 96*97*98 this one fits in the Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:. Isn't it ?

RBBmba@2014 wrote:

GMATGuruNY wrote: Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:
Since every other even integer is a multiple of 4, the product here will always include an even integer and a multiple of 4, resulting in a multiple of 8.
Thus, n can be any even integer between 1 and 96.
96/2 = 48 favorable choices for n.

However, couple of quick clarifications -
1. Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:

Can we consider it in this way that we need to calculate the no. of odds in the given range? Here 'n+1' will take the odd position in the middle. Still we'll have 48 odds with 3 being the first and 97 being the last in the scenario where n=96. This approach of Case 1 reflects the same thing I think. Right ?

2. n=96 --> 96*97*98 this one fits in the Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:. Isn't it ?

Correct.
An alternate way to determine the products that satisfy Case 1 is to count the number of options for n+1.
n+1 can be any odd integer between 3 and 97, inclusive.
To count consecutive odd or even integers, use the following formula:
Total = (biggest - smallest)/2 + 1.
Thus:
Total options for Case 1 = total odd integers between 3 and 97 = (97 - 3)/2 + 1 = 48.
Among these 48 options is 96*97*98, where n+1 = 97.

Brent@GMATPrepNow wrote:You're missing a 2 in your question.

utkalnayak wrote:If an integer n to be chosen randomly between 1 and 96 inclusive, what is the probability that n(n+1)(n+2) is divisible by 8 ?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

OA: D

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, [spoiler]5/8[/spoiler] of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D

Cheers,
Brent

Is the reason why you separate your consecutive products into groups of 8 because, by definition, every eighth product has to be divisible by eight?

For instance, if the question was modified to request for the probability that the first 108 products will be divisible by 9, would one then look at groups of nine products?