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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If an integer n is to be selected at random from 1 to 100, inclusive,
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18 Dec 2022, 09:16
18 Dec 2022, 09:16
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If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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20 Dec 2022, 01:12
20 Dec 2022, 01:12
Expert Reply
One number out of n and (n+1) will be odd and the other will be even. So for n(n+1) to be divisible by 4, the even number must be a multiple of 4. The other number will be odd only so it will have no multiple of 2 or 4 etc.
So, n(n+1) will be divisible by 4 if either n or (n+1) is divisible by 4.
Consider the given range:
n*(n+1) = 1*2 (n = 4a + 1)
n*(n+1) = 2*3 (n = 4a + 2)
n*(n+1) = 3*4 (n = 4a + 3, n+1 = 4b)
n*(n+1) = 4*5 (n = 4b)
n*(n+1) = 5*6 (n = 4b + 1)
n*(n+1) = 6*7 (n = 4b + 2)
n*(n+1) = 7*8 (n = 4b + 3, n+1 = 4c)
n*(n+1) = 8*9 (n = 4c)
and so on...
So of every 4 values of n, 2 will have n*(n+1) a multiple of 4, and 2 will not.
The total possible values of n are 100, a multiple of 4.
So half the values will have n(n+1) a multiple of 4, and half will not.
Answer: C
So, n(n+1) will be divisible by 4 if either n or (n+1) is divisible by 4.
Consider the given range:
n*(n+1) = 1*2 (n = 4a + 1)
n*(n+1) = 2*3 (n = 4a + 2)
n*(n+1) = 3*4 (n = 4a + 3, n+1 = 4b)
n*(n+1) = 4*5 (n = 4b)
n*(n+1) = 5*6 (n = 4b + 1)
n*(n+1) = 6*7 (n = 4b + 2)
n*(n+1) = 7*8 (n = 4b + 3, n+1 = 4c)
n*(n+1) = 8*9 (n = 4c)
and so on...
So of every 4 values of n, 2 will have n*(n+1) a multiple of 4, and 2 will not.
The total possible values of n are 100, a multiple of 4.
So half the values will have n(n+1) a multiple of 4, and half will not.
Answer: C
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Joined: 10 Apr 2015
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Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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02 Jan 2023, 13:09
02 Jan 2023, 13:09
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GeminiHeat wrote:
If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
Let's look for a pattern.
If n = 1, then (n)(n+1) = 2, which is NOT divisible by 4
If n = 2, then (n)(n+1) = 6, which is NOT divisible by 4
If n = 3, then (n)(n+1) = 12, which IS divisible by 4
If n = 4, then (n)(n+1) = 20, which IS divisible by 4
If n = 5, then (n)(n+1) = 30, which is NOT divisible by 4
If n = 6, then (n)(n+1) = 42, which is NOT divisible by 4
If n = 7, then (n)(n+1) = 56, which IS divisible by 4
If n = 8, then (n)(n+1) = 72, which IS divisible by 4
.
.
.
From the pattern, we can see that, out of every FOUR consecutive values of n, (n)(n+1) IS divisible by 4 for TWO of the values, and (n)(n+1) is NOT divisible by 4 for TWO of the values.
So, the probability is 1/2 that n(n+1) will be divisible by 4
Answer: C
Cheers,
Brent
