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

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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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New post 20 Apr 2018, 13:32
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RadhaKrishnan 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


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
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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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New post 15 May 2018, 10:06
RadhaKrishnan 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.
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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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New post 15 May 2018, 10:26
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siddreal wrote:
RadhaKrishnan 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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New post 22 Oct 2018, 23:35
RadhaKrishnan 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


Please find the video solution here.


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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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New post 13 Jan 2019, 06:36
RadhaKrishnan 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

\(?\,\,\, = \,\,\,{{{?_{\,{\rm{temp}}}}\,\,\,\,\left( {{\rm{favorable}}\,\,{\rm{remainders}}} \right)} \over {8\,\,\,\,\left( {{\rm{equiprobable}}\,\,{\rm{remainders}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{division}}\,\,{\rm{by}}\,\,{\rm{8}}} \right)}}\)

\(\left. \matrix{
n\,\, \in \,\,\left\{ {8M,8M + 2,8M + 4,8M + 6} \right\}\,\,\,\,\left( {M\,\,{\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,n\;\;{\rm{even}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{n\left( {n + 1} \right)\left( {n + 2} \right)} \over 8} = {\mathop{\rm int}} \,\,\,\, \hfill \cr
n\,\, \in \,\,\left\{ {8M + 1,8M + 3,8M + 5} \right\}\,\,\,\,\left( {M\,\,{\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{n\left( {n + 1} \right)\left( {n + 2} \right)} \over 8} \ne {\mathop{\rm int}} \hfill \cr
n\,\, \in \,\,\left\{ {8M + 7} \right\}\,\,\,\,\left( {M\,\,{\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{n\left( {8M + 8} \right)\left( {n + 2} \right)} \over 8} = {\mathop{\rm int}} \hfill \cr} \right\}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{?_{\,{\rm{temp}}}} = 5\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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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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New post 15 Aug 2019, 23:59
It took me some time to solve this one, but here's my take on the question.

1. When selecting an integer from 1 to 96, there will be 2 scenarios:
1.a Selecting an even integer in which case \(n(n+1)(n+2)\) will be \(even * odd * even\)
1.b Selecting an odd integer in which case \(n(n+1)(n+2)\) will be \(odd * even * odd\)

1.a:
Let's start with the smallest even integer that we can pick i.e. 2. In this case the expression will be \(2*3*4\), which is divisible by 8. Next take 4 and the expression will be \(4*5*6\) again divisible by 8. As you can see picking up any even will have three 2's in the expression and hence the expression will be divisible by 8. To conclude, picking any even will make the expression divisible by 8.

Hence, it now boils down to how many even integers can we pick from 1 to 96 \(\frac{(96-2)}{2} +1 = 48\)

1.b:
The two odd integers will not be divisible by 8, hence we need to focus on the middle even integer. Let's start with 1 and the expression will be \(1*2*3\), clearly not divisible by 8. Next 3 and the expression will be \(3*4*5\), clearly not divisible by 8. Next 5 and the expression will be \(5*6*7\), clearly not divisible by 8. Next 7 and the expression will be \(7*8*9\) and we got the first clear win. Next 9 and the expression will be \(9*10*11\), clearly not divisible by 8. We can conclude that if the middle integer is a multiple of 8 then the expression will be divisible by 8.

Hence, it now boils down to how many odd integers can we pick from 1 to 96 such that the middle integer is a multiple of 8 or how many multiples of 8 can we pickup from 1 to 96\(=\frac{(96-8)}{8}+1 = 12\)

Combining 1.a and 1.b we get,

\(\frac{Favorable Outcomes}{Total Outcomes}\)=\(\frac{(48 + 12)}{96}\)=\(\frac{60}{96}\)=\(\frac{5}{8}\) (Ans. D)

P.S The 96 in the denominator came from "If an integer n is to be chosen at random from the integers 1 to 96, inclusive" i.e Total Outcomes \((96-1)+1\)
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Re: If an integer n is to be chosen at random from the integers 1 to 96   [#permalink] 15 Aug 2019, 23:59

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