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# An integer n between 1 and 99, inclusive, is to be chosen at

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Intern
Joined: 19 May 2018
Posts: 14
GPA: 3.7
Re: An integer n between 1 and 99, inclusive, is to be chosen at  [#permalink]

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24 Aug 2018, 23:01
DensetsuNo

Why doesn't your method for these types of questions work here?

#s in sequence: 3
#s we have: 2

1 - 2/3 = 3/3 - 2/3 = 1/3

Am I just missing something here?
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Joined: 21 Sep 2018
Posts: 2
Re: An integer n between 1 and 99, inclusive, is to be chosen at  [#permalink]

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29 Oct 2018, 05:11
LalaB wrote:
the required number = (if n is even )+ (if n is divisible by 3) - (if n is even and divisible by 3)=
=(49+33-16)/99=66/99=2/3

Hi, could you explain how you derived the number "if n is even and divisible by 3"? I assume you did not simple count them, am I right?

LalaB
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Re: An integer n between 1 and 99, inclusive, is to be chosen at  [#permalink]

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15 May 2019, 07:39
An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ?

A) 1/9
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Can someone explain to me why this method does not work?

I caught the pattern of 6 possibilities that meet the question stems criteria from 1 - 10 {2,3,5,6,8,9}.
I multiplied 6 by 10 given the range described above happens 10 times 1 -10, 11 - 20, 21 - 30, 31 - 40, and etc...
This now gives me 60/100 which simplifies to 3/5.
It seems I am missing 6 cases.

Additionally, can someone explain how they got a denominator of 99 and not 100?
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Re: An integer n between 1 and 99, inclusive, is to be chosen at  [#permalink]

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15 May 2019, 20:39
Hi peterpark,

To start, we're told that N is an integer from 1 to 99, inclusive, so your fraction will be out of 99 (not 100).

Based on your work, you actually spotted a pattern, but you didn't interpret it correctly. Instead of focusing on the 'first 10 terms', try focusing on 'groups of 3'...

IF....
N = 1....(1)(2) = 2 is NOT divisible by 3.
N = 2....(2)(3) = 6 IS divisible by 3
N = 3....(3)(4) = 12 IS divisible by 3.

In these first 3 examples, notice how 2 of the 3 are divisible by 3....What happens when we look at the NEXT 3 numbers....

IF....
N = 4....(4)(5) = 20 is NOT divisible by 3.
N = 5....(5)(6) = 30 IS divisible by 3
N = 6....(6)(7) = 42 IS divisible by 3.

The pattern REPEATS! This means that for ever 3 consecutive values of N, 2 of the 3 will be divisible by 3. Since we're dealing with the numbers 1 to 99, inclusive, that means we'll have 33 "sets" of 3 numbers (as shown above). Thus, 2/3 of ALL the values WILL be divisible by 3...

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Re: An integer n between 1 and 99, inclusive, is to be chosen at  [#permalink]

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24 May 2019, 12:28
Given: 1<int n<99
Question: does n or n+1 have a factor of 3?

Draw a chart and find a pattern:
n | +1| div by 3?
2 | 3 | Y
3 | 4 | Y
4 | 5 | N
5 | 6 | Y
6 | 7 | Y
7 | 8 | N
8 | 9 | Y
9 |10| Y
10|11| N
...
95|96| Y
96|97| Y
97|98|N

It's a good idea to check the end because the pattern might be cut off at a different point (if n was <97).
So, we can see that 2/3 will be divisible by 3. Thus, D.
Intern
Joined: 07 Jun 2019
Posts: 16
Re: An integer n between 1 and 99, inclusive, is to be chosen at  [#permalink]

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19 Jul 2019, 09:21
This problem is a perfect example where the concept of both number series and probability is mixed.
n * (n+1) can be a multiple of 3 if eaither of n or (n+1) is multiple of 3..
Thus there are two parallel series..
Series 1 - 2,5,8........98 ---> Total 33 numbers
Series 2 - 3,6,9........99 ---> Total 33 numbers

Thus we can select 33 + 33 total 66 numbers out of the 99 numbers.
Hence the probability is 66/99 --- > 2/3 Option D
Re: An integer n between 1 and 99, inclusive, is to be chosen at   [#permalink] 19 Jul 2019, 09:21

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