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Re: If an integer n is to be chosen at random from the integers
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20 Sep 2015, 12:30
take number 1,2,,3,4,5,6,7,8.
pick any one and check for n(n+1)(n+2), only 5 favourable cases. so 5/8



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Re: If an integer n is to be chosen at random from the integers
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16 Jul 2016, 02:40
Bunuel wrote: RadhaKrishnan 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 \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; (Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.) Now, in EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8. Answer: D. Similar question: divisibleby12probability121561.htmlHope it helps. Hi, Thank you for posting such a good explanation , however I could not understand how can you categorize the numbers in group of 8 . Probability is Fav/Total . Shouldn't we consider all 96 values and find out how many are satisfying our conditions , cant understand how you arrive at 5/8. If you can please explain. Sorry if this sounds too basic. Regards Megha



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Re: If an integer n is to be chosen at random from the integers
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16 Jul 2016, 02:41
megha_2709 wrote: Bunuel wrote: RadhaKrishnan 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 \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; (Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.) Now, in EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8. Answer: D. Similar question: divisibleby12probability121561.htmlHope it helps. Hi, Thank you for posting such a good explanation , however I could not understand how can you categorize the numbers in group of 8 . Probability is Fav/Total . Shouldn't we consider all 96 values and find out how many are satisfying our conditions , cant understand how you arrive at 5/8. If you can please explain. Sorry if this sounds too basic. Regards Megha In EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8
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Re: If an integer n is to be chosen at random from the integers
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16 Jul 2016, 06:44
Bunuel wrote: RadhaKrishnan 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 \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; (Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.) Now, in EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8. Answer: D. Similar question: divisibleby12probability121561.htmlHope it helps. i have a doubt , please help me what m i missing here. when you are considering pair 18 actually you are grouping (2,3,4),(4,5,6),(6,7,8) ,(8,9,10) and 7,8,9 we are takig 8 first position because ...in second group first pair will be 10,11,12 similarly to make the count 5 for each pair , we have to consider 96 at first position 96,97,98 Is it not increasing out given number limit.



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Re: If an integer n is to be chosen at random from the integers
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24 Jul 2016, 05:02
BunuelThanks for replying and clearing m doubt. Really helped. Regards Megha



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Re: If an integer n is to be chosen at random from the integers
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16 Nov 2016, 08:55
RadhaKrishnan 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 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
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Re: If an integer n is to be chosen at random from the integers
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24 Jan 2017, 18:33
1)Probability of an event = Number of Desired Outcomes/Number of Possible Outcomes 2)A number is divisible by 8 if three of its prime factors are 2's. 3)If n is even, then n+2 is also even, and there are at least three 2's between these two multiples. For example, if n=2, then 2*(2+2)=8=2*2*2. This means that at least in 50% of cases, the product n(n+1)(n+2) is divisible by 8. 4)If n is odd, then the only way it is divisible by 8 is if (n+1) is divisible by 8. To calculate the total number of multiples of 8 between 1 and 96: (968)/8+1=88/8+1=12 5)96/2+12=60; 60/96=5/8
The correct answer is D.



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Re: If an integer n is to be chosen at random from the integers
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07 Feb 2017, 07:48
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.



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If an integer n is to be chosen at random from the integers
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29 Aug 2017, 19:33
Here's my take, please correct me if I were wrong.
multiply out n(n+1)(n+2) = n^3 + 3n^2 + 2n.
Now, let's factor out n^3 from above equation which become n^3 (1+ 3n^1 + 2n^2) < I know they looks ugly but wait
You basically ignore (1+ 3n^1 + 2n^2) now so that you will know that to have n^3 divisible by 8; n must be the multiple of 2.
Here you will know that 96 / 2 = 48 numbers that are the multiple of 2.
Be aware here because 48 number have included the multiple of 8's but we still need to take (n+1) into consideration. So total number of (n+1 = 8) are 12.
therefore (48+12) / 96 = 5/8.



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If an integer n is to be chosen at random from the integers
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15 Oct 2017, 08:26
RadhaKrishnan 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 My method 1,2,3 >NO: 2,3,4>Yes: 3,4,5>NO; 4,5,6> Yes: 5,6,7 NO: 6,7,8>Yes, 7,8,9>Yes: 8,9,10:> Yes
Wait a sec! what just happened? whenever there is 8(or a multiple of 9, irrespective of the other two numbers, the overall number becomes divisible by 8 and even the last number, 96, is a multiple of 8. So, 5/8 is the answer. Don't believe me? Count them!



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Re: If an integer n is to be chosen at random from the integers
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15 Oct 2017, 10:33
RadhaKrishnan 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 This is a really interesting question and is of 700 level.. So, there might be two cases.. Case 1: n is even .. So if n is a multiple of 2 and not 4 (for e.g 2,6,10..) then n+2 must be multiple of 4 (respectively 4,8,12..) Hence , n(n+2) must be divisible by 8 and hence n(n+1)(n+2) must be divisible by 8. Similarly if n is a multiple of 4 (for e.g 4,8,12..) then n+2 must be multiple of 4 (respectively 6,10,14..) Hence , n(n+2) must be divisible by 8 and hence n(n+1)(n+2) must be divisible by 8. So, in this case n(n+1)(n+2) must be divisible by 8. Such numbers are 96/2 = 48 Case 2: n is odd.. So, (n+1) is even and n+2 is odd.. n(n+1)(n+2) can be divisible by 8 only if (n+1) is a multiple of 8 . So, (n+1) must be (8,16,...96) and corresponding values of n will be (7,15,..., 95) Such numbers are 96/8 = 12 Probability that n(n + 1)(n + 2) will be divisible by 8 = (48+12)/96 = 60/96 = 5/8 Answer D
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Re: If an integer n is to be chosen at random from the integers
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19 Oct 2017, 22:26
Bunuel wrote: RadhaKrishnan 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 \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; (Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.) Now, in EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8. Answer: D. Similar question: http://gmatclub.com/forum/divisibleby ... 21561.htmlHope it helps. Hi Bunuel Why are we choosing a bracket of 8 numbers??? pls explain..



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Re: If an integer n is to be chosen at random from the integers
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19 Oct 2017, 23:14
zanaik89 wrote: Bunuel wrote: RadhaKrishnan 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 \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; (Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.) Now, in EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8. Answer: D. Similar question: http://gmatclub.com/forum/divisibleby ... 21561.htmlHope it helps. Hi Bunuel Why are we choosing a bracket of 8 numbers??? pls explain.. Because the pattern is repeated in every 8 numbers.
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Re: If an integer n is to be chosen at random from the integers
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20 Apr 2018, 13:32
RadhaKrishnan 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 8n = 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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If an integer n is to be chosen at random from the integers
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25 Apr 2018, 16:43
I looked at this problem as sets. I divided 96/8 = 16. So I have 16 sets of 8 consecutive numbers (18), (916)...(8996). So I focused on the 1st set (18). n(n+1)(n+2) are 3 consecutive numbers being multiplied. So they will always be divisible by 3. But the question asks for divisibility by 8. When you prime factorize 8, you end up with 2^3. Focusing on set 18, I focused on multiplying 3 consecutive numbers to see if I can pull out at minimum 2^3.
1*2*3 = No (there is a single 2) 2*3*4 = yes ( there are three 2's) 3*4*5 = no ( there are two 2's) 4*5*6 = yes 5*6*7 = No 6*7*8 = yes 7*8*9 = yes 8*9*10 = yes
You have 5 yes out of 8 in the set. 5/8



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Re: If an integer n is to be chosen at random from the integers
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15 May 2018, 10:06
RadhaKrishnan 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 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.



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Re: If an integer n is to be chosen at random from the integers
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15 May 2018, 10:26
siddreal wrote: RadhaKrishnan 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 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. What I would do is pretty much what Bunuel has done in his solution. Note that the moment I see n(n + 1)(n + 2), I think of divisibility in a bunch of consecutive numbers. In 3 consecutive integers, if n is even, (n+2) is even too. If there are two consecutive even integers, one of them will be a multiple of 4. So if n is even, (n+2) is even too and one of them is definitely a multiple of 4. So n(n + 1)(n + 2) becomes divisible by 8 in each case that n is even. From 1 to 96, half the cases have even n so this mean 48 cases. Alternatively, when n is odd, n+1 is even. But then n+2 is odd too. So to be a multiple of 8, (n+1) will need to be a multiple of 8. Hence this gives us another 12 cases (n+1 goes from 8 to 96). Note that there will be no overlap in the two since here n is definitely odd. Total we have 60 cases of the possible 96 which gives 60/96 = 5/8 For these properties of numbers, see https://www.veritasprep.com/blog/2011/0 ... cormath/https://www.veritasprep.com/blog/2011/0 ... hpartii/
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Re: If an integer n is to be chosen at random from the integers
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31 May 2018, 08:42
Total 96 Number 48 even and 48 odd
For n=even. N+2=even Multiple of any two even is always divided by 8 so all 48 cases will be divided by 8
For odd It will include those cases in which multiple of 8 comes in between Eg 7*8*9 15*16*17 Etc 8,16,.......96 Total 12 cases
So total cases= 48+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
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11 Aug 2018, 22:22
OA: C There are two different possible pattern
Pattern 1: n= odd, then n+1=Even and n+2 = odd i.e odd*even*odd Number of such arrangement \(=\frac{96}{2}=48\)
Pattern 2: n= Even, then n+1=Odd and n+2 = Even i.e even*odd*even Number of such arrangement \(=\frac{96}{2}=48\)
In Pattern 1 : n+1 should be divisible by 8 as n and n+2 are odd \(n+1=8p ; n = 8p1\) Possible value of \(n = 7,15,23,31,39,47,55,63,71,79,87,95\) Total possible values of n = 12
In pattern 2: all 48 are divisible by 8 , as smallest possible case: 2*3*4 is also divisible by 8
Favourable case \(:12+48=60\) Total case :\(96\) Probability =\(\frac{60}{96}=\frac{5}{8}\)




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