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16 Nov 2016, 08:55
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RadhaKrishnan
We are given that an integer n is to be chosen at random from the integers 1 to 96 inclusive, and we need to determine the probability that n(n + 1)(n + 2) will be divisible by 8.
We should recall that when a number is divisible by 8, it is divisible by 2^3, i.e., three factors of 2. We should also recognize that n(n + 1)(n + 2) is the product of three consecutive integers.
Case 1: n is even. Any time that n is even, n + 2 will also be even. Moreover, either n or n + 2 will be divisible by 4, and thus n(n + 1)(n + 2) will contain three factors of 2 and will be divisible by 8.
Since there are 96 integers between 1 and 96, inclusive, and half of those integers are even, there are 48 even integers (i.e., 2, 4, 6, …, 96) from 1 to 96 inclusive. Thus, when n is even, there are 48 instances in which n(n + 1)(n + 2) will be divisible by 8.
Case 2: n is odd. If n is odd, then n(n + 1)(n + 2) still can be divisible by 8 if the factor (n + 1) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 96 inclusive.
Number of multiples of 8 = (96 - 8)/8 + 1 = 88/8 + 1 = 12. Thus, when n is odd, there are 12 instances in which n(n + 1)(n + 2) will be divisible by 8.
In total, there are 48 + 12 = 60 outcomes in which n(n + 1)(n + 2) will be divisible by 8.
Thus, the probability that n(n + 1)(n + 2) is divisible by 8 is: 60/96 = 10/16 = 5/8.
Answer: D
07 Feb 2017, 07:48
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My approach was similar to Bunuel's and AdmitJA's, but I wanted to offer it because it was simple and quick. I was able to do this one in just over a minute.
When I read the question, I paid attention to two things right away. The first was that \((n)(n+1)(n+2)\) is the product of three consecutive integers. Because \(8=2^3\), I knew that I needed to find three 2s in the factorization of whatever 3 consecutive numbers I used. The second thing that I noticed is that 96 is divisible by 8. This indicated to me that whatever pattern I noticed in the first 8 numbers would be repeated 12 times through 96. Since I recognized that the pattern would be repeated, I knew that I only needed to look at the first 8 numbers.
I listed the numbers out: 1 2 3 4 5 6 7 8.
Since 8 has three 2s in its factorization, I knew that n=6,7,8 would all be divisible by 8. I saw that n=2 would work because 2 has one 2 and 4 has two 2s. Similarly, I saw that n=4 would work because 4 has two 2s and 6 has one 2. That gives us 5 options in the first 8 numbers. So the answer is 5/8.
As mentioned above, this ratio will be the same for every 8 numbers, so 5/8 will be true of 8x where x is any positive integer. A trickier version of this question would have been to make the number NOT divisible by 8. In that case, I think you should still find the pattern for every 8 numbers, but you'd also want to look at the "extra" numbers to figure out the fraction.
Take 53, for instance. You'd want to recognize that 8 goes into 53 six times with a remainder of 5. This means that you'd have the \(5/8\) ratio for 6 sets of 8 but also an "incomplete" set of n = 1,2,3,4,5. In the first 48 numbers, you'd have 30 that would be divisible by 8. In the "incomplete" set you'd have 2 (since n=2 and n=4 are both divisible by 8). Thus you'd have 32/53 numbers divisible by 8.
When I read the question, I paid attention to two things right away. The first was that \((n)(n+1)(n+2)\) is the product of three consecutive integers. Because \(8=2^3\), I knew that I needed to find three 2s in the factorization of whatever 3 consecutive numbers I used. The second thing that I noticed is that 96 is divisible by 8. This indicated to me that whatever pattern I noticed in the first 8 numbers would be repeated 12 times through 96. Since I recognized that the pattern would be repeated, I knew that I only needed to look at the first 8 numbers.
I listed the numbers out: 1 2 3 4 5 6 7 8.
Since 8 has three 2s in its factorization, I knew that n=6,7,8 would all be divisible by 8. I saw that n=2 would work because 2 has one 2 and 4 has two 2s. Similarly, I saw that n=4 would work because 4 has two 2s and 6 has one 2. That gives us 5 options in the first 8 numbers. So the answer is 5/8.
As mentioned above, this ratio will be the same for every 8 numbers, so 5/8 will be true of 8x where x is any positive integer. A trickier version of this question would have been to make the number NOT divisible by 8. In that case, I think you should still find the pattern for every 8 numbers, but you'd also want to look at the "extra" numbers to figure out the fraction.
Take 53, for instance. You'd want to recognize that 8 goes into 53 six times with a remainder of 5. This means that you'd have the \(5/8\) ratio for 6 sets of 8 but also an "incomplete" set of n = 1,2,3,4,5. In the first 48 numbers, you'd have 30 that would be divisible by 8. In the "incomplete" set you'd have 2 (since n=2 and n=4 are both divisible by 8). Thus you'd have 32/53 numbers divisible by 8.
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15 May 2018, 10:06
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RadhaKrishnan
VeritasPrepKarishma Pls show us your approach to this question. Though I understood the solutions mentioned in the thread, I am finding it difficult to believe that I might think of this approach during actual exam.
