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

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If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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16 Apr 2008, 09:25
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If an integer n is to be chosen at random from the integers 1 to 96, inclusive, 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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Posts: 64249
Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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28 Jan 2012, 04:12
84
112
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, 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

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

(Notice that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.)

Now, 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.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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29 Jan 2012, 00:17
52
25
Bunuel wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, 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

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

from here u can also think this way (though it is a lil bit similar)
find how many numbers are divisible by 2 or 8
96/2=48
96/8=12
(48+12)/96=60/96=5/8
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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18 Feb 2012, 07:03
30
12
+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
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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09 Jul 2008, 00:33
15
1
6
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
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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06 May 2012, 01:22
12
9
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)
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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16 Aug 2014, 23:24
12
1
goodyear2013 wrote:
If n is an integer between 1 and 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. 3/4
E. 7/8

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
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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22 Nov 2014, 01:00
9
3
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.

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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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09 Jul 2008, 02:49
5
3
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
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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14 Feb 2010, 12:07
4
marcodonzelli wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, 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

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
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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25 Jul 2014, 02:54
4
3
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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16 Nov 2016, 07:55
4
1
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, 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

We are given that an integer n is to be chosen at random from the integers 1 to 96 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 96 integers between 1 and 96, inclusive, and half of those integers are even, there are 48 even integers (i.e., 2, 4, 6, …, 96) from 1 to 96 inclusive. Thus, when n is even, there are 48 instances in which n(n + 1)(n + 2) will be divisible by 8.

Case 2: n is odd. If n is odd, then n(n + 1)(n + 2) still can be divisible by 8 if the factor (n + 1) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 96 inclusive.

Number of multiples of 8 = (96 - 8)/8 + 1 = 88/8 + 1 = 12. Thus, when n is odd, there are 12 instances in which n(n + 1)(n + 2) will be divisible by 8.

In total, there are 48 + 12 = 60 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: 60/96 = 10/16 = 5/8.

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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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17 May 2013, 12:07
3
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.

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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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25 Sep 2013, 03:50
3
chunjuwu wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive,
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

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
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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12 Jul 2015, 04:40
2
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, 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

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
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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16 Jul 2016, 01:41
1
megha_2709 wrote:
Bunuel wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, 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

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

(Notice that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.)

Now, 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.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.

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
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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07 Feb 2017, 06:48
1
My approach was similar to Bunuel's and AdmitJA's, but I wanted to offer it because it was simple and quick. I was able to do this one in just over a minute.

When I read the question, I paid attention to two things right away. The first was that $$(n)(n+1)(n+2)$$ is the product of three consecutive integers. Because $$8=2^3$$, I knew that I needed to find three 2s in the factorization of whatever 3 consecutive numbers I used. The second thing that I noticed is that 96 is divisible by 8. This indicated to me that whatever pattern I noticed in the first 8 numbers would be repeated 12 times through 96. Since I recognized that the pattern would be repeated, I knew that I only needed to look at the first 8 numbers.

I listed the numbers out: 1 2 3 4 5 6 7 8.

Since 8 has three 2s in its factorization, I knew that n=6,7,8 would all be divisible by 8. I saw that n=2 would work because 2 has one 2 and 4 has two 2s. Similarly, I saw that n=4 would work because 4 has two 2s and 6 has one 2. That gives us 5 options in the first 8 numbers. So the answer is 5/8.

As mentioned above, this ratio will be the same for every 8 numbers, so 5/8 will be true of 8x where x is any positive integer. A trickier version of this question would have been to make the number NOT divisible by 8. In that case, I think you should still find the pattern for every 8 numbers, but you'd also want to look at the "extra" numbers to figure out the fraction.

Take 53, for instance. You'd want to recognize that 8 goes into 53 six times with a remainder of 5. This means that you'd have the $$5/8$$ ratio for 6 sets of 8 but also an "incomplete" set of n = 1,2,3,4,5. In the first 48 numbers, you'd have 30 that would be divisible by 8. In the "incomplete" set you'd have 2 (since n=2 and n=4 are both divisible by 8). Thus you'd have 32/53 numbers divisible by 8.
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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15 May 2018, 09:26
1
siddreal wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, 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

VeritasPrepKarishma Pls show us your approach to this question. Though I understood the solutions mentioned in the thread, I am finding it difficult to believe that I might think of this approach during actual exam.

What I would do is pretty much what Bunuel has done in his solution.

Note that the moment I see n(n + 1)(n + 2), I think of divisibility in a bunch of consecutive numbers.
In 3 consecutive integers, if n is even, (n+2) is even too. If there are two consecutive even integers, one of them will be a multiple of 4.
So if n is even, (n+2) is even too and one of them is definitely a multiple of 4.
So n(n + 1)(n + 2) becomes divisible by 8 in each case that n is even.
From 1 to 96, half the cases have even n so this mean 48 cases.

Alternatively, when n is odd, n+1 is even. But then n+2 is odd too. So to be a multiple of 8, (n+1) will need to be a multiple of 8.
Hence this gives us another 12 cases (n+1 goes from 8 to 96).
Note that there will be no overlap in the two since here n is definitely odd.

Total we have 60 cases of the possible 96 which gives 60/96 = 5/8

For these properties of numbers, see
https://www.veritasprep.com/blog/2011/0 ... c-or-math/
https://www.veritasprep.com/blog/2011/0 ... h-part-ii/
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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19 Sep 2010, 15:41
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
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Re: If an integer n is to be chosen at random from the integers 1 to 96  [#permalink]

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08 Oct 2015, 09:49
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.

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Re: If an integer n is to be chosen at random from the integers 1 to 96   [#permalink] 08 Oct 2015, 09:49

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