first of all its easy to see that n(n+1)(n+2) can be divided by 8 if n is even
because one of n or n+2 will have to be divisible by 2 and the other by 4
so are we finished? no, but - > A and B are wrong
what happens if n is odd?
if n is odd, then n(n+1)(n+2) will be divisible be 8 if n+1 = 8k where k is from 1..12 (if k=13 then n = 103)
so we have another 12 numbers out of 96 -> 12.5%
so my answer is: 50%+12.5% = 62.5%
C
Difficult Math Problem #13
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gamemaster
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800guy
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here's the OA:
E = n*(n+1)*(n+2)
E is divisible by 8, if n is even.
No of even numbers (between 1 and 96) = 48
E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)
Such numbers total = 12(7,15,....)
Favorable cases = 48+12 = 60.
Total cases = 96
P = 60/96 = 62.5
Method 2:
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*8)(7*8*9)(8*9*10)
So till 96, there will be 12 * 5 such sets = 60 sets
so probability will be 60/96 = 62.5
E = n*(n+1)*(n+2)
E is divisible by 8, if n is even.
No of even numbers (between 1 and 96) = 48
E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)
Such numbers total = 12(7,15,....)
Favorable cases = 48+12 = 60.
Total cases = 96
P = 60/96 = 62.5
Method 2:
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*8)(7*8*9)(8*9*10)
So till 96, there will be 12 * 5 such sets = 60 sets
so probability will be 60/96 = 62.5
- ashish1354
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can someone explain this statement in the context of this question
Method 2:
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*Cool(7*8*9)(8*9*10)
I think i can add another set i.e (10*11*12) which is also divisible by 8 so we have 6 sets between 1 to 10 hence
9*6(between 1 to 90)+3(between 91 & 96)=57 favorable events
so probability= 57/60 but doesn't coincide with any answer can someone suggest what is wrong
I also do not understand this statement
E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)
Such numbers total = 12(7,15,....)
how do we arrive at this magic number 12?? :roll:
Method 2:
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*Cool(7*8*9)(8*9*10)
I think i can add another set i.e (10*11*12) which is also divisible by 8 so we have 6 sets between 1 to 10 hence
9*6(between 1 to 90)+3(between 91 & 96)=57 favorable events
so probability= 57/60 but doesn't coincide with any answer can someone suggest what is wrong
I also do not understand this statement
E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)
Such numbers total = 12(7,15,....)
how do we arrive at this magic number 12?? :roll:
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parallel_chase
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ashish1354 wrote:can someone explain this statement in the context of this question
Method 2:
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*Cool(7*8*9)(8*9*10)
I think i can add another set i.e (10*11*12) which is also divisible by 8 so we have 6 sets between 1 to 10 hence
9*6(between 1 to 90)+3(between 91 & 96)=57 favorable events
so probability= 57/60 but doesn't coincide with any answer can someone suggest what is wrong
I also do not understand this statement
E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)
Such numbers total = 12(7,15,....)
how do we arrive at this magic number 12?? :roll:
Here is the process
n(n+1)(n+2) divisible by 8
We know that when n is even, the product of three consecutive integers will be divisible by 8
[96-2]/2 = 47 + 1 = 48
When n is odd there could only be following cases
7,8,9
15,16,17
23,24,25...so on and so forth
Looking at above cases we just have find the multiples of 8
[96-8]/8 = 11+1 = 12
48+12 = 60
probability = 60/96 * 100 = 62.5%
Hope this helps.
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anju
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Small correction:ashish1354 wrote:can someone explain this statement in the context of this question
Method 2:
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*Cool(7*8*9)(8*9*10)
In Method 2:
from n - 1 to 8 there are the following number divisible by 8
(2*3*4),(4*5*6), (6*7*8), (7*8*9) AND (8*9*10) - five diff nos.
1 to 96 can be divided into 12 sets of 8 (96/8) each of which will repeat similar calculation.
hence total nos divisible by 8 will be 12 * 5 = 60
I took the scenic route, which is like anju´s method 2:
1.2.3 = 6 not divisible by 8
2.3.4 = 24 divisible by 8
3.4.5 = 60 not divisible by 8
4.5.6 = 120 divisible by 8
5.6.7 = 210 not divisible by 8
6.7.8 = 336 divisible by 8 (of course)
7.8.9 = 504 divisible by 8 (of course)
8.9.10 = 720 divisible by 8 (of course)
Therefore, if we assume a repeating sequence up to 96, the answer is then 5/8 or 62.5% of products that are divisible by 8.
1.2.3 = 6 not divisible by 8
2.3.4 = 24 divisible by 8
3.4.5 = 60 not divisible by 8
4.5.6 = 120 divisible by 8
5.6.7 = 210 not divisible by 8
6.7.8 = 336 divisible by 8 (of course)
7.8.9 = 504 divisible by 8 (of course)
8.9.10 = 720 divisible by 8 (of course)
Therefore, if we assume a repeating sequence up to 96, the answer is then 5/8 or 62.5% of products that are divisible by 8.
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aplavakarthik
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For all even values of n, n(n+1)(n+2) is divisible by 8
Eg: 2,3,4
16,17,18
22,23,24
and so on......
these constitute to a total of 96/2=48
and we do have 12 multiple of 8 between 1 and 96(8,16,24,32,.........96)
these come up when n is odd
Eg: 7,8,9
15,16,17
23,24,25............
.........
95,96,97
So a total of 48+12 =60 will be divisible
Now Probability is 60/96=62.5%
Eg: 2,3,4
16,17,18
22,23,24
and so on......
these constitute to a total of 96/2=48
and we do have 12 multiple of 8 between 1 and 96(8,16,24,32,.........96)
these come up when n is odd
Eg: 7,8,9
15,16,17
23,24,25............
.........
95,96,97
So a total of 48+12 =60 will be divisible
Now Probability is 60/96=62.5%












