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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
An integer n between 1 and 99, inclusive, is to be chosen at random.
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30 Apr 2022, 11:37
30 Apr 2022, 11:37
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Expert Reply
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Question Stats:
90% (02:25) correct
10% (04:50) wrong based on 10 sessions
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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
A) 1/9
B) 1/3
C) 1/2
D) 2/3
E) 5/6
Re: An integer n between 1 and 99, inclusive, is to be chosen at random.
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05 May 2022, 22:35
05 May 2022, 22:35
Expert Reply
n(n+1) to be divisible by 3 either n or n+1 must be a multiples of 3.
In each following group of numbers: {1, 2, 3}, {4, 5, 6}, {7, 8, 9}, ..., {97, 98, 99} there are EXACTLY 2 numbers out of 3 satisfying the above condition. For example in {1, 2, 3} n can be: 2, or 3. Thus, the overall probability is 2/3.
Answer: D.
In each following group of numbers: {1, 2, 3}, {4, 5, 6}, {7, 8, 9}, ..., {97, 98, 99} there are EXACTLY 2 numbers out of 3 satisfying the above condition. For example in {1, 2, 3} n can be: 2, or 3. Thus, the overall probability is 2/3.
Answer: D.
Re: An integer n between 1 and 99, inclusive, is to be chosen at random.
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13 Aug 2022, 02:18
13 Aug 2022, 02:18
1
To solve this problems, we have to find pattern by putting numbers from 1...99.
let x = n(n+1)
x(1) = 1*2
x(2) = 2*3
x(3) = 3*4
x(4) = 4*5
x(5) = 5*6
x(6) = 6*7
and so on.
Here we can see, that a pattern emerges. First number is not divisible by two and next two numbers are divisible by 3.
It means 2/3 of the numbers are divisible by 3.
let x = n(n+1)
x(1) = 1*2
x(2) = 2*3
x(3) = 3*4
x(4) = 4*5
x(5) = 5*6
x(6) = 6*7
and so on.
Here we can see, that a pattern emerges. First number is not divisible by two and next two numbers are divisible by 3.
It means 2/3 of the numbers are divisible by 3.
