First the solution... then the explanation. Without the explanation, the solution has only limited value to your preparation.
For any number to be divisible by 8, it must be divisible by 2 - three times. (2*2*2=8)
n, n+1, and n+2 are consecutive integers, which means that we have either two even numbers and an odd one (when n is even) or two odds and an even (when n is odd).
We know that no odd number is divisible by 2, meaning out of all the n's that are odd, only those whose n+1 is divisibly by 8 qualify. For example, if n=7, then we have 7*8*9, which is divisible by 8. Since 96/8 = 12, there are twelve n's that work in this case (7, 15, 23, ..., 95)
For the even n's, we still need the three 2's, but they can be distributed among both even numbers. Fortunately, with two consecutive even numbers, one of the even's is always divisible by 4 (two 2's) and the other by a single 2. This means ALL even n's qualify. So 96/2 = 48.
Total, we have 48+12 = 60 qualifying n's out of 96 total, which represents 60/96 = 10/16 = 5/8 = 62.5%
Our answer is C.
Now, the explanation...
What are the key insights here?
1) Understand that the number 8 is misleading, we should be looking at primes.
2) Understand that the complex algebra (how many of you multiplied it out??) distracts you from the idea of consecutive integers.
The solution path consists of prompting yourself to see the problem differently, on your own terms. This means seeking out examples, plugging in numbers, and understanding the nature of the problem. This will lead you to the most efficient solution path.
If you're not cultivating your process while you prep and focusing on the math/language rules/answers, you're wasting your time.
Ben
probability
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Ben.Miller@ApexGMAT
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Atekihcan
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A total of 96 possibilities for n.
There are two possibilities: n is even or n is odd.
Now, if n is even, then (n + 2) is also even.
So, n and (n + 2) are two consecutive even integers.
So, their product will be divisible by 8.
Exactly half of the possible 96 value of n are even.
So, 96/2 = 48 possibilities for the product to be divisible by 8.
And, if n is odd, then only (n + 1) will be even among the three.
Now, in each four consecutive even integers, exactly one will be divisible by 8.
There are 48 possible odd values of n. So, there 48 possible even values of (n + 1).
So, exactly 48/4 = 12 possibilities for the product to be divisible by 8.
So, required probability = [(48 + 12)/96]*100 = 60*100/96 = slightly greater than 60 = 62.5%
Answer : C
There are two possibilities: n is even or n is odd.
Now, if n is even, then (n + 2) is also even.
So, n and (n + 2) are two consecutive even integers.
So, their product will be divisible by 8.
Exactly half of the possible 96 value of n are even.
So, 96/2 = 48 possibilities for the product to be divisible by 8.
And, if n is odd, then only (n + 1) will be even among the three.
Now, in each four consecutive even integers, exactly one will be divisible by 8.
There are 48 possible odd values of n. So, there 48 possible even values of (n + 1).
So, exactly 48/4 = 12 possibilities for the product to be divisible by 8.
So, required probability = [(48 + 12)/96]*100 = 60*100/96 = slightly greater than 60 = 62.5%
Answer : C
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fcabanski
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GMAT deals with simple math asked in complex ways. Whenever dealing with divisibility, common factors or common multiples, determine prime factors.
Write what is given: n*(n+1)*(n+2) has to be divisible by 8
Write what you know: The prime factors of 8 are 2, 2 and 2. Any of the numbers divisible by 8 must have three 2's as factors. Probability = things of interest / total things x 100.
Whenever a problem deals with a sequence that seems impossibly long, write the first few and last few terms. Patterns become obvious.
1: 1*2*3 - not divisible by 8.
2: 2*3*4 - divisible by 8.
3: 3*4*5 - not divisible
4: 4*5*6 - divisible ( looks like any even number will result in divisibility by 8. 2^3 will be a factor every time n is even. 96/2 = 48 even numbers between 1 and 96 inclusive.): 48 so far.
5: 5*6*7 - not divisible (will all odd n's result in not divisible? Oh wait, what about when the middle number is divisible by 8?)
7: 7 x 8 x 9 Has 2^3 as a factor. If the even number (n+1) is a factor of 8, then the result will be divisible by 8)
15x16x17, 23*24*25 - every 8th number. 96/8 = 12 such numbers.
60 numbers of interest, 96 total numbers, P = interest/total x 100 = 60/96 x 100 = 5/8 x 100 = 62.5%
Hope the solution helped. Drop me a private message to discuss GMAT tutoring.
Write what is given: n*(n+1)*(n+2) has to be divisible by 8
Write what you know: The prime factors of 8 are 2, 2 and 2. Any of the numbers divisible by 8 must have three 2's as factors. Probability = things of interest / total things x 100.
Whenever a problem deals with a sequence that seems impossibly long, write the first few and last few terms. Patterns become obvious.
1: 1*2*3 - not divisible by 8.
2: 2*3*4 - divisible by 8.
3: 3*4*5 - not divisible
4: 4*5*6 - divisible ( looks like any even number will result in divisibility by 8. 2^3 will be a factor every time n is even. 96/2 = 48 even numbers between 1 and 96 inclusive.): 48 so far.
5: 5*6*7 - not divisible (will all odd n's result in not divisible? Oh wait, what about when the middle number is divisible by 8?)
7: 7 x 8 x 9 Has 2^3 as a factor. If the even number (n+1) is a factor of 8, then the result will be divisible by 8)
15x16x17, 23*24*25 - every 8th number. 96/8 = 12 such numbers.
60 numbers of interest, 96 total numbers, P = interest/total x 100 = 60/96 x 100 = 5/8 x 100 = 62.5%
Hope the solution helped. Drop me a private message to discuss GMAT tutoring.
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Atekihcan
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Here is another approach using the patterns of divisibility.
The divisibility by a certain positive integer n repeats itself in a period of n consecutive integers. For example, every one out of two consecutive integers will be divisible by 2, every one out of three consecutive integers will be divisible by 3, every one out of ten consecutive integers will be divisible by 10 etc.
Here we need to check for divisibility by 8.
Check the first 8 positive integers and be done with it.
n = 1 ---> 1*2*3 --> NO
n = 2 ---> 2*3*4 --> YES
n = 3 ---> 3*4*5 --> NO
n = 4 ---> 4*5*6 --> YES
n = 5 ---> 5*6*7 --> NO
n = 6 ---> 6*7*8 --> YES
n = 7 ---> 7*8*9 --> YES
n = 8 ---> 8*9*10 --> YES
So 5 YESs in first 8 positive integers.
That means there will be five YESs in every 8 consecutive integers.
As 96 is perfectly divisible by 8, 96 is end of one period. So, this trend will hold.
So, required probability, 5/8 = 62.5%
The divisibility by a certain positive integer n repeats itself in a period of n consecutive integers. For example, every one out of two consecutive integers will be divisible by 2, every one out of three consecutive integers will be divisible by 3, every one out of ten consecutive integers will be divisible by 10 etc.
Here we need to check for divisibility by 8.
Check the first 8 positive integers and be done with it.
n = 1 ---> 1*2*3 --> NO
n = 2 ---> 2*3*4 --> YES
n = 3 ---> 3*4*5 --> NO
n = 4 ---> 4*5*6 --> YES
n = 5 ---> 5*6*7 --> NO
n = 6 ---> 6*7*8 --> YES
n = 7 ---> 7*8*9 --> YES
n = 8 ---> 8*9*10 --> YES
So 5 YESs in first 8 positive integers.
That means there will be five YESs in every 8 consecutive integers.
As 96 is perfectly divisible by 8, 96 is end of one period. So, this trend will hold.
So, required probability, 5/8 = 62.5%
