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805+ Level|   Math Related|         
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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.
\(\frac{1}{25}+\frac{1}{25} \) =\(\frac{2}{25}\)

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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.
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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;
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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
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From 1 to 5, 4 and 5 satisfies the condition that n(n + 1) is divisible by 5. So if n is between 1 and 5, inclusive, the probability of p is 2/5. Since 600 is a multiple of 5, the probability of p is the same. p = 2/5

From 1 to 25, 1 and 25 satisfies the condition that n(n - 1) is divisible by 25. So if n is between 1 and 25, inclusive, the probability of q is 2/25. Since 600 is a multiple of 25, the probability of q is the same. q = 2/25
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Every fifth possible value of n is divisible by 5 -> 1/5
(n+1) being a factor means that each of the above values minus 1, also are divisible by 5 -> another 1/5
p = 2/5

Every 25th possible value of n is divisible by 25 -> 1/25
(n-1) being a factor means that each of the above values plus 1, also are divisible by 25 -> another 1/25
q = 2/25
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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.
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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.
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Answer: p = 2/5, q = 2/25

For p, we’re looking at when n(n+1) is divisible by 5. Since n and n+1 are consecutive numbers, one of them has to be divisible by 5. That means n should be either a multiple of 5 (like 5, 10, 15...) or 1 less than a multiple of 5 (like 4, 9, 14...). Both of these patterns happen once every 5 numbers. So 2 out of every 5 numbers will work.

→ p = 2/5

For q, we want n(n−1) to be divisible by 25. Again, n and n−1 are consecutive, so only one of them can be divisible by 25. So we look for values where either n is a multiple of 25 (like 25, 50, 75...) or n−1 is a multiple of 25 (which means n=26,51,76...n). Each of those patterns also happens once every 25 numbers, so 2 out of every 25 values work.

→ q = 2/25
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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.
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1. Probability that n(n + 1) is divisible by 5
Combination of n(n + 1) are
4,5
5,6
9,10
10,11 etc
Hence, we get a pattern as 4, 5, 9, 10, 14, 15 etc, so from every 5 numbers we are taking two combinations
=> From first 5 we get n = 4 and n = 5
=> From next 5 we get n = 9 and n = 10
..........
..........

Hence from 600 numbers we get 2*600/5 = 120*2 = 240 numbers
Hence p = 240/600 = 2/5

2. Similarly for n(n − 1) is divisible by 25

Combination of n(n - 1) are
1, 0
25, 24
26, 25
50, 49
51, 50
....
....
etc
Hence, we get a pattern as 1, 25, 26, 50, 51 etc, so from every 25 numbers we are taking two combinations
=> From first 25 we get n = 1 and n = 25
=> From next 25 we get n = 26 and n = 50

Hence from 600 numbers we get 2*600/25 = 1200/25 = 48 numbers
Hence q = 48/600 = 2/25

p = 2/5
q = 2/25
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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.

Total nos= 600

n(n+1) should be divisible by 5

Suppose 1 to 10 how many nos can satisfy, 4,5,9 & 10 So 4 nos
Same 4 nos from 11-20 and so on 591 to 600
0-100= 4*10= 40
0-600= 4*60=240
Probability that drawn no that n(n+1) is divisible by 5= 240/600= 2/5
p=2/5

n(n − 1) is divisible by 25
0-100 nos can be 25,26,50,51,75,76,100,101
101-200 nos can be 101,125,126,150,151,175,176,200= 8 Nos, 201-300=8, 301-400=8, 401-500=8, 501-600=8
0-600 = 47 nos
q=47/600 but n=1 is also divisible as numerator becomes 0
q=48/600=2/25
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\(n(n + 1)\) is divisible by 5:

For \(n(n + 1)\) to be divisible by 5, at least one of them must be divisible 5.

when \(n\) is divisible by 5, there are 120 multiples of 5 between 1 & 600 inclusive

when \(n+1\) is is divisible by 5, there are 120 multiples of 5 between 1 & 600 inclusive

\(p=\frac{240}{600}=\frac{2}{5}\)


\(n(n - 1)\) is divisible by 25:

For \(n(n - 1)\) to be divisible by 25, at least one of them must be divisible 25.

when \(n\) is divisible by 25, there are 24 multiples of 25 between 1 & 600 inclusive

when \(n-1\) is is divisible by 25, there are 24 multiples of 25 between 1 & 600 inclusive.

\(q=\frac{48}{600}=\frac{2}{25}\)

One thing to note here is, when \(n-1\) is 600, \(n=601\), which is not possible. Considering this there are only 23 multiples of 25.
Then \(q=\frac{47}{600}\), and not given in option choices.

p: \(\frac{2}{5}\)

q: \(\frac{2}{25}\)
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No. of digits between 1 and 600, inclusive = 600.

For p:
For n(n+1) to be divisible by 5, either n or (n+1) has to be divisible by 5, but not both, since they are consecutive numbers and the only factor they have in common is 1.
No. of digits divisible by 5 in the given range = \(\frac{(600-5)}{5}+1=120\)

So, P(n divisible by 5) + P(n+1 divisible by 5) = \(\frac{120}{600} + \frac{120}{600} = \frac{2}{5}\to\)p

Fo q:
Similarly, for n(n-1) to be divisible by 25, either n or (n-1) has to be divisible by 25, but not both.
No. of digits divisible by 25 in the given range = \(\frac{(600-25)}{25}+1=24\)

So, P(n divisible by 25) + P(n-1 divisible by 25) = \(\frac{24}{600} + \frac{24}{600} = \frac{2}{25}\to\)q
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number of ways we can randomly select an integer between 1 to 600 inclusive is 600C1

for n(n+1) be divisible by 5, either n should be multiple of 5 or (n+1)

possible n could be 4,5,9,10,14,15,19,20... total 240 values of n

p =240/600= 2/5

for n(n-1) be divisible by 25, either n or n-1 should be multiple of 25

possible n could be 25,26,50,51,75,76,100,101... total 47

q = 47/600 ~ 2/25

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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.
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(1) for finding p
n(n+1) div by 5
=> n is div by 5
n takes values 5, 10, ... 600
600/5 = 120 values for the n
=> n+1 is div by 5
=> n+1 takes value 5, 10, ...600 => 120 values for n+1
=> n in this case will take on values 4, 9, ... 599

p = (120+120)/600 = 240/600 = 4/10 = 2/5

(2) for finding p
=> n.(n-1) div by 25
=> n is div by 25
n takes values 25, 50 ... 600 =600/25 = 24 values
=> n-1 is div by 25
n-1 takes values 25, 50 .. 600 = 24 values
n takes in this values 26, 51, ... 601 = 23 values
BUT THE catch is (1-1) = 0 is also div by 25 => hence (23 +1) 24 values

q = 24+24/ 600 = 48/600 = 2/5
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Part 1: Find p, the probability that
n(n+1) is divisible by 5.
We are looking for values of n such that:
Either n is divisible by 5 or (n+1) is divisible by 5

Case 1: n divisible by 5
Numbers are:
5, 10, 15, ..., 600
This is an arithmetic sequence:
First term = 5,Last term = 600,Common difference = 5
Total terms = (600−5)/5 +1= 120

Case 2: (n+1) divisible by 5
Then n = 4, 9, 14, ..., 599
This is also an arithmetic sequence:
First = 4,Last = 599,Common difference = 5
Total terms =(599−4)/5 + 1=120

There is no overlap of both sequence
Total values of n = 120+ 120
= 240(favorable outcomes)

p= 240/600 = 2/5

Part 2: Find q, the probability that
n(n−1) is divisible by 25
We are looking for values of n such that:

Either n is divisible by 25 or (n−1)is divisible by 25

Case 1: n divisible by 25
Numbers: 25, 50, 75, ..., 600
First = 25,Last = 600,Difference = 25
Total terms =(600−25)/25 + 1 = 24

Case 2: (n−1) divisible by 25
Then n = 1, 26, 51, ..., 576

First = 1, Last =576, Difference=25
Total terms = (576−1)/25 +1 =24

Since both the sequence are mutually exclusive
Total favorable outcomes= 24 +24= 48

q=48/600 = 2/25
​
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Total possibility =600

Probability of p that n(n+1) is divisible by 5.

Either n or n+1 must be divisible by 5.
Number of values divisible by 5 in 1 to 600 = 120

n+1 is divisble by 5 that means all number 4,9,14,etc such numbers are also =120
Total=240
P=240/600
p=2/5

Similarly for q,
Number of values divisible by 25 in 1 to 600 = 600/25 => 24
n-1 is divisble by 25 that means all number 0,25,50,etc such numbers are =575/25 +1 => 24
Total=24+24 =>48
q=48/600
q=2/25
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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.
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[*]n is randomly selected from 1 to 600 (inclusive) => Sample population = n = 600 possible integers.
[*]p = Probability that n(n+1) is divisible by 5
[*]q = Probability that n(n−1) is divisible by 25

n & (n+1) are consecutive integers => n(n+1) is divisible by 5 <=> n =[4, 5, 9, 10,..., 599, 600]
=> Out of every 5 consecutive integer, 2 of them are satisfied
=> p = 2/5

n and (n-1) are consecutive integer => n(n-1) is divisible by 25 <> n=[1, 25, 26, 50, 51, 75, 76, 100, 101, 125, 124..., 576, 600]
=> Out of 100 every consecutive integer, 8 of them are satisfied
=> q = 8/100 = 2/25

Answer: p = 2/5, q = 2/25
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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.
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Part 1,
Finding the probability that n(n+1) is divisible by 5, which means that at least 1 of those terms is divisible by 5
The total number of multiples of 5 in range 1-600 is simply 600/5 = 120

The probability of n being a divisble of 5 is 120/600 = 1/5
And the probability of (n+1) being a divisble of 5 is also 120/600 = 1/5

Since the events are mutually exhaustive, total probability of n(n+1) being a divisble of 5 equals the sum of individual probabilities = 1/5 + 1/5 = 2/5 --- Option 4

Similarly for Part 2,
Finding the probability that n(n-1) is divisible by 25, which means that at least 1 of those terms is divisible by 25
The total number of multiples of 25 in range 1-600 is 600/25 = 24

The probability of n being a divisble of 25 is 24/600 = 1/25
And the probability of (n-1) being a divisble of 25 is also 24/600 = 1/25

Since events are mutually exhaustive, total probability of n(n-1) being a divisble of 25 equals the sum of individual probabilities = 1/25 + 1/25 = 2/25 --- Option 2
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Finding p: n(n+1) divisible by 5
For n(n+1) to be divisible by 5, either n or (n+1) must be divisible by 5.
Check each case mod 5:
  • n ≡ 0 (mod 5): n divisible by 5
  • n ≡ 1 (mod 5): n+1 ≡ 2, neither divisible by 5
  • n ≡ 2 (mod 5): n+1 ≡ 3, neither divisible by 5
  • n ≡ 3 (mod 5): n+1 ≡ 4, neither divisible by 5
  • n ≡ 4 (mod 5): n+1 ≡ 0, so n+1 divisible by 5
Works when n ≡ 0 or 4 (mod 5) = 2 out of 5 cases
p = 2/5

Finding q: n(n-1) divisible by 25
For n(n-1) to be divisible by 25 = 52, we need higher powers of 5.
Since n and (n-1) are consecutive, at most one can be divisible by 5.
For the product to be divisible by 25, one of them must be divisible by 25.
This happens when:
  • n ≡ 0 (mod 25), or
  • n ≡ 1 (mod 25) [so n-1 ≡ 0 (mod 25)]
In range 1 to 600:
  • Numbers ≡ 0 (mod 25): 25, 50, 75, ..., 600 = 24 numbers
  • Numbers ≡ 1 (mod 25): 1, 26, 51, ..., 576 = 24 numbers
Total: 48 out of 600

q = 48/600 = 2/25

Answer:
  • p = 2/5
  • q = 2/25
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GMAT Club Olympics 2025 (Day 5): A computer is randomly generating an 05 Jul 2025, 03:28
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Total number of integers = (600-1)/1 + 1 = 600
for p, n(n+1) should be divisible by 5. so either n should be divisible by 5 or n+1 should be divisible by 5.
here n could be multiples of 5 or n+1 could be multiples of 5 (till 600)
For example, n= 5 or n = 4
Multiples of 5 from 1 to 600= (600 - 5)/5 + 1 = 120
So we have 240 possible values of n.

p= 240/600 = 2/5

Following similar logic for q,
Multiples of 25 from 1 to 600 = 24
Now, we can have 23 values for the other set of possible values of n, (as we can't take n = 601, because it is out of our range)
But we can also include n=1 over here as, 1(0) = 0 is divisible by 25.
So we have 48 possible values of n

q = 48/600 = 2/25
Bunuel
 


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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.
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GMAT Club Olympics 2025 (Day 5): A computer is randomly generating an 05 Jul 2025, 03:46
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For 5 to divide n(n + 1) :
If the number ends in 0 or 5, then n has a factor 5.
If the number ends in 4 or 9, then n + 1 ends in 0 or 5, then also n has a factor 5.
Out of every 5 consecutive numbers, exactly two of them end in 0 or 5 (remainder 0) or 4 (remainder 4).
So 2 of every 5 work. p = 2/5.

For 25 to divide n(n − 1) :
2 Consecutive numbers cannot be multiples of 5, so the only way for n(n−1) to contain 25 is if one of the numbers itself is a multiple of 25.
Out of every 25 consecutive numbers, exactly two have remainder 0 or 1.
So 2 of every 25 work. q = 2/25.
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