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04 Jul 2025, 08:00
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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.
| p | q | |
| 1/25 | ||
| 2/25 | ||
| 4/25 | ||
| 1/5 | ||
| 2/5 | ||
| 3/5 |
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Official Answer
p: 2/5
q: 2/25
04 Jul 2025, 08:30
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Bunuel
GMAT Club Official Explanation:
For n(n + 1) to be divisible by 5, either n or n + 1 must be a multiple of 5.
In every group of five consecutive numbers, such as {1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, ..., {596, 597, 598, 599, 600}, exactly two values of n satisfy this condition: one where n is divisible by 5, and one where n + 1 is divisible by 5. Therefore, the overall probability is 2/5.
Similarly, for n(n - 1) to be divisible by 25, either n or n - 1 must be a multiple of 25.
In every group of 25 consecutive numbers, exactly two values of n satisfy this condition: one where n - 1 is divisible by 25, and one where n is divisible by 25. For example, in the group {1, 2, ..., 25}, n(n - 1) is divisible by 25 when n = 1 or n = 25. In the group {576, 577, ..., 600}, it happens when n = 576 or n = 600. Therefore, the overall probability is 2/25.
General Discussion
04 Jul 2025, 08:17
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As per given condition, we have S={1,2,...,600}. The total number of possible outcomes is 600.
solve for p ---> Probability that n(n+1) is divisible by 5
For n(n+1) to be divisible by 5, either n is a multiple of 5, or n+1 is a multiple of 5.
This means that n≡0(mod5) or n≡−1(mod5), which is n≡4(mod5).
Let's list the possible values of n(mod5):
If n≡0(mod5): These are numbers like 5,10,15,...
If n≡4(mod5): These are numbers like 4,9,14,...
the numbers that satisfy the condition are those where n(mod5) is either 0 or 4.
Let's find the number of integers in the range [1,600] that satisfy these conditions.
For n≡0(mod5):
The numbers are 5,10,...,600.
The count is 600/5 =120
For n≡4(mod5):
The numbers are 4,9,...,599.
To find the count, let 5k−1=n.
5k−1≤600⟹5k≤601⟹k≤120.2. So, k goes from 1 (for n=4) up to 120 (for n=599).
The count is 120.
The total number of favorable outcomes is 120+120=240.
The probability p is the number of favorable outcomes divided by the total number of outcomes:
p= 240/600 = 2/5
solve for q---->Probability that n(n−1) is divisible by 25
For n(n−1) to be divisible by 25, we have two cases:
Case 1: n is a multiple of 25.
n≡0(mod25).
The numbers are 25,50,...,600.
The count is 600/25=24
Case 2: n−1 is a multiple of 25.
n−1≡0(mod25), which means n≡1(mod25).
The numbers are 1,26,51,...
The last number will be 576 (since 576+24=600). Let's find n=25k+1.
25k+1≤600⟹25k≤599⟹k≤599/25 =23.96.
So k goes from 0 (for n=1) up to 23 (for n=25×23+1=575+1=576).
The count is 23−0+1=24.
Case 3: Both n and n−1 are multiples of 5, but not necessarily 25 individually.
This case is already covered by the divisibility by 25. If n is a multiple of 5 and n−1 is a multiple of 5, then their difference (n−(n−1))=1 would also have to be a multiple of 5, which is impossible. So n and n−1 cannot both be multiples of 5.
Therefore, the only ways for n(n−1) to be divisible by 25 are if either n is a multiple of 25, or n−1 is a multiple of 25. These two conditions are mutually exclusive, as n and n−1 are coprime.
Total number of favorable outcomes is 24+24=48.
The probability q is the number of favorable outcomes divided by the total number of outcomes:
q= 48/600 = 2/25
solve for p ---> Probability that n(n+1) is divisible by 5
For n(n+1) to be divisible by 5, either n is a multiple of 5, or n+1 is a multiple of 5.
This means that n≡0(mod5) or n≡−1(mod5), which is n≡4(mod5).
Let's list the possible values of n(mod5):
If n≡0(mod5): These are numbers like 5,10,15,...
If n≡4(mod5): These are numbers like 4,9,14,...
the numbers that satisfy the condition are those where n(mod5) is either 0 or 4.
Let's find the number of integers in the range [1,600] that satisfy these conditions.
For n≡0(mod5):
The numbers are 5,10,...,600.
The count is 600/5 =120
For n≡4(mod5):
The numbers are 4,9,...,599.
To find the count, let 5k−1=n.
5k−1≤600⟹5k≤601⟹k≤120.2. So, k goes from 1 (for n=4) up to 120 (for n=599).
The count is 120.
The total number of favorable outcomes is 120+120=240.
The probability p is the number of favorable outcomes divided by the total number of outcomes:
p= 240/600 = 2/5
solve for q---->Probability that n(n−1) is divisible by 25
For n(n−1) to be divisible by 25, we have two cases:
Case 1: n is a multiple of 25.
n≡0(mod25).
The numbers are 25,50,...,600.
The count is 600/25=24
Case 2: n−1 is a multiple of 25.
n−1≡0(mod25), which means n≡1(mod25).
The numbers are 1,26,51,...
The last number will be 576 (since 576+24=600). Let's find n=25k+1.
25k+1≤600⟹25k≤599⟹k≤599/25 =23.96.
So k goes from 0 (for n=1) up to 23 (for n=25×23+1=575+1=576).
The count is 23−0+1=24.
Case 3: Both n and n−1 are multiples of 5, but not necessarily 25 individually.
This case is already covered by the divisibility by 25. If n is a multiple of 5 and n−1 is a multiple of 5, then their difference (n−(n−1))=1 would also have to be a multiple of 5, which is impossible. So n and n−1 cannot both be multiples of 5.
Therefore, the only ways for n(n−1) to be divisible by 25 are if either n is a multiple of 25, or n−1 is a multiple of 25. These two conditions are mutually exclusive, as n and n−1 are coprime.
Total number of favorable outcomes is 24+24=48.
The probability q is the number of favorable outcomes divided by the total number of outcomes:
q= 48/600 = 2/25
