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If an integer n is to be chosen at random from the integers

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Carcass
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If an integer n is to be chosen at random from the integers [#permalink] New post   31 Aug 2020, 10:31
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If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post   31 Aug 2020, 11:00
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Carcass wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D
Cheers,
Brent
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post   31 Aug 2020, 10:32
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post   31 May 2022, 05:47
GreenlightTestPrep wrote:
Carcass wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D
Cheers,
Brent



How did you know that we had to check every 8 numbers exactly?

If we check every 12 numbers (because 12 * 8 = 96 ) then every 12 numbers have 7 valid values.

thus 7 * 8 = 56 which would give incorrect answer.
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post   31 May 2022, 14:42
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ChandanPednekar wrote:
GreenlightTestPrep wrote:
Carcass wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D
Cheers,
Brent



How did you know that we had to check every 8 numbers exactly?

If we check every 12 numbers (because 12 * 8 = 96 ) then every 12 numbers have 7 valid values.

thus 7 * 8 = 56 which would give incorrect answer.


The key here is that we're looking for products that are divisible by 8.
We start with n = 1 (which is 1 more than a multiple of 8)
Then n = 2 (which is 2 more than a multiple of 8)
.
.
.
Then n = 7 (which is 7 more than a multiple of 8)
Then n = 8 (which is 0 more than a multiple of 8)
Then n = 9 (which is 1 more than a multiple of 8) at this point we can see that the pattern begins repeating itself
.
.
.
etc

Does that help?
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post   21 Jul 2022, 08:37
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Here is how I approached the problem:

n(n+1)(n+2) divisible by 8 means n is divisible by 4 or n+1 is divisible by 8 or n+2 is divisible by 4. Worth noting that whenever n or n+2 is divisible by 4, the other is also an even number, so n(n+2) will be divisible by 8. Also worth noting that every consecutive 3 numbers have at most one multiple of 4 and, of course, alternate odd and even numbers.

Case 1: n divisible by 4: 96/4 = 24 numbers
Case 2: n+2 divisible by 4; similar to case 1: 24 numbers
Case 3: n+1 divisible by 8: 96/8 = 12 numbers

Total: 24+24+12=60

Probability: 60/96=5/8
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post   13 Mar 2023, 02:50
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we see a pattern that when the consecutive no's start with even, we have multiples of 8 but not when n is odd.
But we also have a case where (n+1) is a multiple of 8 i.e. 8,16,24,32 upto 96
So, in this case, the 'n' can be odd but be divisible by 8.
• Ex: 7,8,9
○ 15,16,17
• But this case is already included where (n+2) is multiple of 8
○ 14,15,16: here n is even and included by 1st case.

• Total number of even numbers between 1 to 96 = 48
• Total number of multiples of 9 between 1 to 96 = 12
○ Total = 60
• Probability = 60/96 = 5/8
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post   27 May 2025, 08:27
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