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GMAT Club Olympics 2025 (Day 5): A computer is randomly generating an

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andreagonzalez2k

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n(n+1) is divisible by 5 if n or (n+1) are divisible by 5.

Between 1 and 600 there are:
(600-5)/5 + 1 = 120 numbers divisible by 5

For each number divisible by 5, if the previous one is chosen, (n+1) will be divisible by 5.
We have the pairs: 4 and 5, 9 and 10... 599 and 600

There are 120*2 = 240 numbers that will do n(n+1) divisible by 5

p = 240/600 = 2/5

n(n-1) is divisible by 25 if n or (n-1) are divisible by 25.

Between 1 and 600 there are:
(600-25)/25 + 1 = 24 numbers divisible by 25

For each number divisible by 25, if the next one is chosen ,(n-1) will be divisible by 25.
We have the pairs: 25 and 26, 50 and 51... 600 and 1 (special case)

There are 24*2 = 48 numbers that will do n(n-1) divisible by 25

q = 48/600 = 2/25

p=2/5 and q=2/25

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Correct Answer: p= 2/5 & q= 2/25
Firstly we calculate p
we need to find number which are divisible by 5 between number 1-600
Total Numbers divisible by 5: 600/5= 120 (if n=0)
Total Number from 4, 9, 14.........<=600 (if n=1)
Let’s find this using Arithmetic Progression:
600=a+(n-1)d
600=4+(n-1)5
601/5= n
around 120 numbers we get,
so total we have 240 number
so p= 240/600 = 2/5

Now find out q
we need to find number which are divisible by 25 between number 1-600
Total Numbers divisible by 25: 600/25=24 (if n=0)
Total Number from 1, 26, 51, .........<=600 (if n=1)
Let’s find this using Arithmetic Progression:
600=a+(n-1)d
600=1+(n-1)25
624/25= n
around 24 numbers we get,
so total we have 48 number
so q= 48/600 = 2/25

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P : 5p = 600 => p = 120 ; however we should multiply 2 because there would be once 25 * 26 and one more 24 *25.
P = 240/600 = 2/5

Q : 25q = 600 => q = 24 ; again same logic as above and 24 *2 = 48.
Q = 48/600 = 2/25

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We are given:
A computer randomly selects an integer .
We are to compute:

Step 1: Compute

We are checking how often the product is divisible by 5.

Note:

n & n+1 are consecutive integers, so one of them must be divisible by 5 in order for the product to be divisible by 5.

So we need either or (so that )

In mod 5, these are the favorable residues:

n =4 mod 5 =4
n=5 mod 5= 0

That’s 2 favorable outcomes out of 5 possible residues →
So,

p = 2/5 = 0.4

Step 2: Compute

Again, is the product of two consecutive integers. For the product to be divisible by 25, we want:

Either n=0 mod 25=0

Or n= 1 mod 25 = 1 (so that )

So again, only two favorable residues mod 25:

That’s 2 out of 25 possible values →
So,

q = 2/25

Final Answers:

p=2/5
q=2/25

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total numbers between 1-600 that are divisible by 5 = 120

total n such that n(n + 1) is divisible by 5 = 120 x 2= 240

p the probability that n(n + 1) is divisible by 5 = 240/600 = 2/5

total numbers between 1-600 that are divisible by 25 = 24

total n such that n(n + 1) is divisible by 5 = 24x 2= 48

q the probability that n(n − 1) is divisible by 25 = 48/600 = 2/25

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2/5, 2/25, for n(n+1) to be divisible by 5, either n divisible by 5, or n+1, it can be in form of multiple of 5 or 4 hence, calculating using Highest-Lowest multiple/ No. +1, you get those no.s, similarly for n(n-1) to be divisible by 25, finding 25's multiples till 600, and 26's , and then divide by total to get the probability.

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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Given that n is randomly picked from numbers $1$ to $600$, inclusive.
We are to find p the probability that $n(n + 1)$ is divisible by 5, and q the probability that $n(n − 1)$ is divisible by $25$.

Case 1 :
Numbers that are divisible by 5 can be in the form where $n = 5$ or $n = 4$, as in either of the cases the number will be a multiple of $5$.
Total numbers divisible by $5$ in the range $= 600/5 = 120$
But since there exist another number for each such number divisible by $5$ in the $120$ such that $(n+1)$ becomes $5$ or a multiple of $5$, the total number of numbers of the form $n(n + 1) $ divisible by $ 5 = 120 * 2$
Therefore the probability, $p = 240/600 = 2/5$

Case 2 :
Similar to Case 1, numbers that are divisible by $25$ can be in the form where $n = 25$ or $n = 26$, as in either of the cases the number will be a multiple of $25$.
Total numbers divisible by $25 $ in the range $= 600/25 = 24$
Total number of numbers of the form $ n(n - 1)$ divisible by $25 = 24 * 2$
Therefore the probability, $q = 48/600 = 2/25$

Therefore the correct options are:
$p = 2/5$
$q = 2/25$

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Number of multiples of 5 between 1 and 600:

1 + (600-5)/5 = 120

As there are two factors, n and (n+1), the favorable events are 120*2=240

Probability = 240/600 = 2/5

Number of multiples of 25 between 1 and 600:

1 + (600-25)/25 = 24

As there are two factors, n and (n-1), the favorable events are 24*2=48

Probability = 48/600 = 2/25

The right answers are: p=2/5 and q=2/25

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05 Jul 2025, 06:25

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there are 600 integers in total so any probability we choose will be x/600.

P( that n(n+1) is divisible by 5): this can only happen if n is a number before a multiple of 5 or n is a multiple of 5
hence, 5, 10,..... 595, 600= [600-5][/5]=119+1=120
120 multiples of 5 and 120 numbers just before these multiples
total = 240 favourable outcomes.
P=240/600=2/5

Similarly, q can be calculated to 48/600
which is 2/25

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For n(n+1) to be divisible by 5, either n or n+1 must be divisible by 5
I. calculate p
Till 600, we have 120 multiples of 5. Each multiple will account for two pairs of n(n+1)
For example: For 5, we have (4,5) & (5,6)
So Total favorable outcomes = 2*120 = 240
Total outcomes =600
p = 240/600 = 2/5

2. calculate q
From 1 to 100 there are 4 multiples of 25. Possible pairs 7 listed below:
(24,25),(25,26),(49,50),(50,51),(74,75),(75,76),(100,99)
similarly for every 100 integers, there will be 7 such pairs.
So, for 600 integers, no. of possible outcomes = 6 * 7 = 42
Total outcomes =600
q = 42/600 = 7/100 ~ 8/100 = 4/25

Bunuel

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For p, the product n(n + 1) is divisible by 5 if either n or n + 1 is a multiple of 5. Among 600 numbers, there are 120 multiples of 5, and similarly 120 numbers where n + 1 is a multiple of 5, totaling 240 favorable cases.

Thus, p = 240/600 = 2/5.

For q, n(n - 1) is divisible by 25 if either n or n 1 is a multiple of 25; there are 24 multiples of 25, and 24 numbers one more than a multiple of 25, totaling 48.
So, q = 48/600 = 2/25

p=2/5,q=2/25.

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probability that n(n + 1) is divisible by 5

> lets take a case from 1 to 10.
there are 4 numbers where n(n+1) is divisible by 5 :
n=4 > n(n+1)= 4*5 divisible by 5
n=5 > n(n+1)= 5*6 divisible by 5
n=9 > n(n+1)= 9*10 divisible by 5
n=10 > n(n+1)= 10*11 divisible by 5

So per 10 numbers there are 4 numbers like this, for 600 numbers there will be (4*600)/10 = 240
Probability = 240/600 = 2/5
Choose 2/5 for p

probability that n(n - 1) is divisible by 25

> lets take a case from 1 to 100.
there are 8 numbers where n(n-1) is divisible by 25 :
n=25 > n(n-1)= 24*25 divisible by 25, and n=26 > n(n-1)= 25*26 divisible by 25,
n=50> n(n-1)= 49*50 divisible by 25, and n=51 > n(n-1)= 50*51 divisible by 25,
n=75 > n(n-1)= 74*75 divisible by 25, and n=76 > n(n-1)= 75*76 divisible by 25,
n=100 > n(n-1)= 99*100 divisible by 25, and n=101 > n(n-1)= 100*101 divisible by 25,

So per 100 numbers there are 8 numbers like this, for 600 numbers there will be (8*600)/100 = 48
Probability = 48/600 = 2/25
Choose 2/25 for q

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n is a randomly chosen integer from 1 to 600

p=P(n(n+1) divisible by 5)
q=P(n(n−1) divisible by 25)

For n(n+1) to be divisible by 5,
n≡0 mod 5 or n(n+1)≡0 mod 5
=> n≡0 or n≡4 mod 5

So,
n ≡ 0 mod 5 => 5,10
n ≡4 mod 5 => 4,9
This basically means, 4 out of every 10 number
=> p = 4/10 = 2/5

Now for q, n(n-1) to be divisible by 25
n≡ 0 mod 25 or
n-1 ≡ 0 mod 25
n≡1 mod 25

So,
n ≡ 0 mod 25 => 25,50,...600 => 24 values
n ≡1 mod 25 => 1,26,...576 => 24 values

=> q = (24+24) / 600 = 2/25

p = 2/5
q = 2/25

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$n$ is between $1$ and $600$.

Multiples of $5$ between that range -> $(600-5)/(5) +1 = 120$
$n(n+1)$ will be divisible by $5$ when $n$ is divisible by $5$ or if $n$ is $1$ less than a multiple of $5$ (so that $n+1$ is a multiple of$ 5$). There are $120+120 = 240$ numbers that fit the bill between $1$ and $600$
$p = 240/600 = 2/5$

Multiples of $25$ in this range are -> $(600-25)/(5) +1 = 24$
$n(n-1)$ will be multiples of $25$ when either when $n$ is divisible by $25$ or if $n$ is $1$ more than a multiple of $25$ (so that $n-1$ is a multiple of $25$). There are $24+24 = 48$ numbers that fufill this criteria.
$q = 48/600 = 2/25$

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Bunuel

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For n(n+1) divisible by 5--> either n is divisible by 5 or n+1.

Cases 1. n is divisible by 5. total cases = 600/5 =120.
Case 2. n+1 is divisible by 5. ...cases ...when n is 4,14, 19, 24....................599 Pattern = 5k+4,

5k+4 = 4 .........k=0 .. when 5k+4= 599 --> 5k=595 ---> k=119... k from 0 to 199. Cases = 120

Total cases when n(n+1) divisible by 5 = 120+120=240.

Probability = 240/600 = 4/10=2/5

For n(n-1) divisible by 25--> either n is divisible by 25 or n-1.

Cases 1. n is divisible by 25. total cases = 600/25 =120/5 =24
Case 2. n-1 is divisible by 25. ...cases ...when n is 26,51, 76, 101....................576 Pattern = 25k+1,

25k+1 = 26 .........k=1 .. when 25k+1= 576 --> 25k=575 ---> k=115/5=23... k from 1 to 23. Cases = 24

Total cases when n(n-1) divisible by 25 = 24+24=48.

Probability = 48/600 = 8/100=2/25

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p: 2/5
q: 2/25

For p:
n can be a multiple of 5 or n+1 can be a multiple of 5.
Multiples of 5: 600/5 = 120
Since 600 is included in range, n+1 also has 120 possibilities.

p = 240/600 = 2/5

For q:
n can be a multiple of 25.
600/25 = 24 possible factors
For n-1, we can have multiples of 25 or 0. Since 601 cannot be included in the range, required probability: q = 48/600
Thus, q=2/25

Bunuel

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There are 120 = (600-5)/5 + 1 multiples of 5
There are two possibilities, n or (n+1) to be multiple of 5, so 240 odds
240/600=2/5

There are 24 = (600-25)/25 multiples of 25
There are two possibilities, n or (n-1) to be multiple of 25, so 48 odds
48/600=2/25

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Between 1 and 600 there are 120 numbers divisibles by 5

(last multiple - first multiple)/5 + 1 = (600 - 5)/5 + 1 = 120

n(n+1) will be divisible if n or n+1 are divisibles, so 120*2= 240

p = 240/600 = 2/5

Between 1 and 600 there are 24 numbers divisibles by 25

(last multiple - first multiple)/5 + 1 = (600 - 25)/25 + 1 = 24

n(n-1) will be divisible if n or n-1 are divisibles, so 24*2= 48

q = 48/600 = 2/25

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I list out all the possible cases to see any pattern:
- w/ n(n+1) we can have: n=4,5,9,10,14,15,19,20...599,600
So we have 120 integers when n is the multiple of 5 and another 120 integer from 599 = 4 + 5 (n-1). Therefore, p = 240/600 = 2/5

- w/ n(n-1) we have 24 integers as n is the multiple of 25, AND 23 integers for n = 26,51,76,.., 576. Adding one more as n can be 1, we have total 48 --> q = 48/600 = 2/25

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For n(n+1) to be divisible by 5, either n must be a multiple of 5 or n+1 must be a multiple of 5.
Exactly 2 integers in every 5 integers can satisfy this condition.
Therefore, P =2/5

Similarly, for n(n-1) to be divisible by 25,
1. Either n is a multiple of 25 or
2. n-1 is a multiple of 25 or
3. if n is a multiple of 5 AND n - 1 is a multiple of 5 but this is not possible as n and n-1are consecutive integers.

For Case 1: Number of multiples of 25 b/w 1 and 600 inclusive = 600/25=24.

Case 2: n - 1 is a multiple of 25. This means n is of the form 25k + 1. Numbers are 1, 26, 51...576. The next number would be 25 * 24 + 1 = 601, which is outside the range.
Number of such values = (576-1) /25 +1 = 575 / 25 + 1 = 23 + 1 = 24.
Total number of favorable outcomes for q = 24 (from Case 1) + 24 (from Case 2) = 48.
Probability q = (Favorable Outcomes) / (Total Outcomes) = 48 / 600 = 2/25

p=2/5
q=2/25

Bunuel

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