If an integer n is to be chosen at random from the integers 1 to 96

Since 96 is divisible by 8 and since "divisibility by 8" repeats every 8 terms, you can just focus on the 8 first terms (from 1 to 8):

it works for n=2,4,6,7,8, that is 5 numbers out of the 8

==> Answer is (D) = 5/8

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, 5/8 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

+1 D

Other way is analyzing if there is a patron:

1) If n is an even number:
n:2, then 2*3*4 = 24 (divisible by 8)
n:4, then 4*5*6 = 120 (divisible by 8)
n:6, then 6*7*8= again divisible by 8
We have a patron.
So, we have 48 even possible values.

2) If n is an odd number:
This only can take place when n+1 is multiple of 8.
So, we have 12 possible values.

Then, \(\frac{(48 + 12)}{96} = \frac{5}{8}\)

D

answer is D
n(n+1)(n+2) will be divisible by 8 for all even numbers, i.e. total 48 numbers
also cases in which (n+1) is a multiple of 8 will be divisible by 8
for ex: n=7,15,23.. i.e. total 12 numbers

so total number of such cases is 48+12=60
probability = 60/96 = 5/8

Any integer n(n+1)(n+2) will be divisible by 8 if n is a multiple of 2. This gives us 48 numbers between 1 and 96.

Additionally, all those numbers for which (n+1) is a multiple of 8 are also divisible by 8. This gives us a further 12 numbers. These numbers are all distinct from the first set because the first set had only even numbers and this set has only odd numbers.

Therefore probability = (48+12)/96 = 60/96 = 5/8

Option (D)

All even numbers would be divisible by 8... as we have two consecutive even numbers being multiplied... Hence 96-1/2 + 1 = 48

Also we have 8n-1 sequence numbers divisible by 8... i.e, if n =7,15,23,31....

Therefore 8n-1 = 96 to find the number of elements in this sequence ... gives n = 97/8 = 12

Therefore Probability = (48+12) / 96 = 60 / 96 = 5/8

Notice what happens when you shift this product by 8 :

(n+8)(n+1+8)(n+2+8) = 8^3 + 8^2*(n+n+1+n+2) + 8*(n*(n+1)+(n+1)*(n+2)+n*(n+2)) + n*(n+1)*(n+2) = 8*K + n*(n+1)*(n+2)


Here K is a constant. Therefore, when we shift by 8, the remainder when divided by 8 remains the same !

So if we figure out what happens for the first 8 choices of n, for all the rest the pattern will repeat

n=1 1x2x3 No
n=2 2x3x4 Yes
n=3 3x4x5 No
n=4 4x5x6 Yes
n=5 5x6x7 No
n=6 6x7x8 Yes
n=7 7x8x9 Yes
n=8 8x9x10 Yes

So for every 8 numbers, exactly 5 will be divisible by 8.

Here there are exactly 96 numbers to consider starting with 1 ... So probability will be exactly 5/8

There is another way I approach such questions if I am short of time (otherwise I prefer the logical approach given above) - Brute Force/Pattern Recognition/Intuition - whatever you may want to call it.

We need to find the numbers in which the product is a multiple of 8. I know I get a multiple of 8 after every 8 numbers. I will also get an 8 when I multiply 4 by an even number. In first 8 numbers, I have exactly two multiples of 4.

Basically, I figure that I should look at the first 8 cases. In all other cases, the pattern will be repeated. It helps that n can be from 1 to 96 i.e. a multiple of 8:
1*2*3 N
2*3*4 Y
3*4*5 N
4*5*6 Y
5*6*7 N
6*7*8 Y
7*8*9 Y
8*9*10 Y

5 of the first 8 products are divisible by 8 so my answer would be 5/8 = 62.5%

there are total 48 numbers in the form n*n+1*n+2 starting from 2,4,6,8,10.....96 for which it is divisible by 8.

Additionally,there are 12 cases if there is 8 or a multiple of 8 that also divides the form n*n+1*n+2 which starts from 7,15,23,31....95.


The above cases both are non-overlapping, so we can add them.

adding the above two cases 48+12=60
ans=60/96=5/8 which is D.


--harry
Do press kudos if u like my post

For n(n+1)(n+2) to be divisible by 8, either n has to be even ( because if, n and n+2 are even, then n is divisible by 8) or n+1 as a whole should be divisible by 8
There are 96 values for n. The possibility of n to be even is (96/2) = 48 and the possibility of n to be divisible by 8 is (96/8)= 12
Therefore the total prob = (48+12)/96 = 60/96 = 5/8

The OA is D

Similar questions to practice:
https://gmatclub.com/forum/if-integer-c- ... 21561.html
https://gmatclub.com/forum/an-integer-n- ... 60998.html

Hope this helps.

In such problem always follow a simple rule.
The divisor is 8. So take number from 1 to 8 as example
If n=1, n(n+1)(n+2)= Not divisible by 8
If n=2, n(n+1)(n+2)= yes
If n=3, n(n+1)(n+2)=No. Do upto 8

So, 5 out of 8 are divisible. Hence, 5/8 is the answer.


This rule is applicable if last number(96 in this case) is divisible by 8.
If you had to pick from 1-98, still you can apply the rule but be careful. But there is one additional step.
Hopefully you can find answer in less than 60 seconds

Here is the dumb way of doing this question, if the smart ones don't strike your head during the test

Since we are looking at divisibility by 8, we consider n to be between 1 and 8. We can multiply the result by 12 - since 96/8 = 12.

Alternate solution:

Look at the first few sets of 3s:
1,2,3
2,3,4
3,4,5
4,5,6
5,6,7
6,7,8
7,8,9
8,9,10

We see that out of the above 8 sets, favorable cases are 5 (in red). Thus the probability is 5/8. This will repeat till we have 88,89,90 making it 9 total patterns.

Now consider after 88,89,90, we get

89,90,91
90,91,92
91,92,93
92,93,94
93,94,95
94,95,96
95,96,97
96,97,98

So we have another 5 favorable out of the remaining 8 sets. Thus we have final 5/8 as the probability.

Thus the final probability = {(5/8)*9+(5/8)*1} / (9+1) = 5/8

Hi All,

For a number to be evenly divisible by 8, it has to include at least three 2's when you prime factor it.

For example,
8 is divisible by 8 because 8 = (2)(2)(2).....it has three 2s "in it"
48 is divisible by 8 because 48 = (3)(2)(2)(2)(2).....it has three 2s "in it" (and some other numbers too).

20 is NOT divisibly by 8 because 20 = (2)(2)(5)....it only has two 2s.

In this question, when you take the product of 3 CONSECUTIVE POSITIVE INTEGERS, you will either have....

(Even)(Odd)(Even)

or

(Odd)(Even)(Odd)

In the first option, you'll ALWAYS have three 2s. In the second option, you'll only have three 2s if the even term is a multiple of 8 (Brent's list proves both points). So for every 8 consecutive sets of possibilities, 4 of 4 from the first option and 1 of 4 from the second option will give us multiples of 8. That's 5/8 in total.

GMAT assassins aren't born, they're made,
Rich

Hi,

Thank you for posting such a good explanation , however I could not understand how can you categorize the numbers in group of 8 . Probability is Fav/Total . Shouldn't we consider all 96 values and find out how many are satisfying our conditions , cant understand how you arrive at 5/8. If you can please explain.
Sorry if this sounds too basic.

Regards
Megha

In EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8