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MBAChaser123
Joined: 19 Nov 2024
Last visit: 25 Dec 2025
Posts: 86
Given Kudos: 7
Location: United States
Schools: Broad'26 (WL) Ross '26 (S)
GMAT Focus 1: 695 Q88 V83 DI82 
GPA: 3
04 Jul 2025, 15:26
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To calculate P, we need to count the integers n for which n(n+1) is divisible by 5. For n(n+1)to be divisible by 5, either n should be a multiple of 5 or n+1 should be divisible by 5.
The number of multiples of 5 between 1 and 600 inclusive is \( \frac{ 600}{5 } =120 \)
So the probability of n being a multiple of 5 is \( \frac{ 120}{600 } =\frac{1}{5} \)
The probability of n+1 being a multiple of 5 is the same. We just assume we are choosing n+1 from the set of multiples of 5.
Since n and n+1 can't be multiples of 5 at the same time, the probability of n(n+1) is the sum of the probability of each one being a multiple of 5.
\(\frac{1}{5}+\frac{1}{5} \) =\(\frac{2}{5}\)
The probability of n(n-1) being a multiple of 25 can be calculated in the same way. Since n and n-1 can't be multiples of 5 or 25 at the same time, it will be very easy to calculate.
The number of multiples of 25 between 1 and 600 inclusive is \( \frac{ 600}{25 } =24 \)
The probability of n being a multiple of 25 is \( \frac{ 24}{600 } =\frac{1}{25} \), and The probability of n-1 being a multiple of 25 is also \( \frac{ 24}{600 } =\frac{1}{25} \)
The probability of n(n-1) being a multiple of 25 is the sum of the probability of each one being a multiple of 25.
\(\frac{1}{25}+\frac{1}{25} \) =\(\frac{2}{25}\)
The number of multiples of 5 between 1 and 600 inclusive is \( \frac{ 600}{5 } =120 \)
So the probability of n being a multiple of 5 is \( \frac{ 120}{600 } =\frac{1}{5} \)
The probability of n+1 being a multiple of 5 is the same. We just assume we are choosing n+1 from the set of multiples of 5.
Since n and n+1 can't be multiples of 5 at the same time, the probability of n(n+1) is the sum of the probability of each one being a multiple of 5.
\(\frac{1}{5}+\frac{1}{5} \) =\(\frac{2}{5}\)
The probability of n(n-1) being a multiple of 25 can be calculated in the same way. Since n and n-1 can't be multiples of 5 or 25 at the same time, it will be very easy to calculate.
The number of multiples of 25 between 1 and 600 inclusive is \( \frac{ 600}{25 } =24 \)
The probability of n being a multiple of 25 is \( \frac{ 24}{600 } =\frac{1}{25} \), and The probability of n-1 being a multiple of 25 is also \( \frac{ 24}{600 } =\frac{1}{25} \)
The probability of n(n-1) being a multiple of 25 is the sum of the probability of each one being a multiple of 25.
\(\frac{1}{25}+\frac{1}{25} \) =\(\frac{2}{25}\)
Bunuel
04 Jul 2025, 18:30
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A computer is randomly generating an integer n between 1 and 600, inclusive.
Select for p the probability that n(n + 1) is divisible by 5,
To be divisible by 5, (either n or (n + 1) - both) should be a multiple of 5; here, from 1 to 600, there are 600/5 = 120 multiples of 5; and both must be zero, so 120 + 120 = 240 favourable cases out of 600; p = 2/5
and select for q the probability that n(n − 1) is divisible by 25.
To be divisible by 25, (either n or (n - 1) - both) should be a multiple of 25; here, from 1 to 600, there are 600/25 = 24 multiples of 25; and both must be zero, so 24 + 24 = 48 favourable cases out of 600; p = 2/25;
Select for p the probability that n(n + 1) is divisible by 5,
To be divisible by 5, (either n or (n + 1) - both) should be a multiple of 5; here, from 1 to 600, there are 600/5 = 120 multiples of 5; and both must be zero, so 120 + 120 = 240 favourable cases out of 600; p = 2/5
and select for q the probability that n(n − 1) is divisible by 25.
To be divisible by 25, (either n or (n - 1) - both) should be a multiple of 25; here, from 1 to 600, there are 600/25 = 24 multiples of 25; and both must be zero, so 24 + 24 = 48 favourable cases out of 600; p = 2/25;
04 Jul 2025, 18:41
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A computer is randomly generating an integer n between 1 and 600, inclusive.
Select for p the probability that n(n + 1) is divisible by 5, and select for q the probability that n(n − 1) is divisible by 25.
potential integers for p are: 4,9,14,19, 24, 29, 34, 39, 44, 49, etc (pattern 2 integers out of 10)
=> 2 x 60 = 120 integers / 600 possible integeres = 1/5
potential integers for Q are: 1, 26, 51, 76 (pattern 4 out of 100)
=> 4 x 6 potential integers for q = 24 / 600 possible integers = 4/100 = 1/25
