n is integer from 1-99 : GMAT Quantitative Section
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# n is integer from 1-99

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n is integer from 1-99 [#permalink]

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10 Aug 2013, 00:15
n is integer from 1-99:

1. Probabilty that n(n+1) is divisible by 3
2. Probabilty that n(n+1)(n+2) is divisible by 6
3. Probabilty that n(n+1)(n+2) is divisible by 8

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Re: n is integer from 1-99 [#permalink]

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10 Aug 2013, 01:15
bsahil wrote:
n is integer from 1-99:

1. Probabilty that n(n+1) is divisible by 3
2. Probabilty that n(n+1)(n+2) is divisible by 6
3. Probabilty that n(n+1)(n+2) is divisible by 8

.......

1.
n(n+1) is divisible by 3 only when n is a multiple of 3 or 1 less than multiple of 3.
For example, when n=3 then n(n+1) = 3 × 4 which is divisible by 3.
Again , when n=2 (one less than the multiple of 3) then 2×3, which is divisible by 3.

so we have to count the numbers which are responsible for this action mentioned above.

The number of multiples of 3 between 1 and 99 = (99-3)/3 + 1 = 32+1=33
And we can also have 33 numbers those are 1 less than the multiples of 3(such as 2,5,8,11 etc) .
so total number = 33 + 33 =66 , for this 66 values n(n+1) is divisible by 3.
Total number = 99
so the probability = 66/99 = 2/3
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Re: n is integer from 1-99 [#permalink]

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10 Aug 2013, 20:02
1.

we can make a pattern:

1*2, 2*3, 3*4,.......96*97, 97*98, 98*99.

So every 2 out of 3 pairs is divisible by 3.

hence no of favouurable terms: 99(2/3) = 66

P= 66/99 =2/3

2.

n(n+1)(n+2) is divisible by 6 always

hence P=1

3.

we can again make pairs

1*2*3, 2*3*4, 3*4*5, 4*5*6, 5*6*7, 6*7*8, 7*8*9, 8*9*10

we have 5 such pairs divisible by 8

hence till 96 we will have (96/8)5= 60 pairs

96*97*98, 97*98*99...we cannot go further. Here we have one more favourable pair.

hence P= (60+1)99

Does anybody have a faster method than this..please...
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Re: n is integer from 1-99 [#permalink]

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10 Aug 2013, 22:10
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bsahil wrote:
n is integer from 1-99:

1. Probabilty that n(n+1) is divisible by 3
2. Probabilty that n(n+1)(n+2) is divisible by 6
3. Probabilty that n(n+1)(n+2) is divisible by 8

1.
If n is between 1-99,
probability that n is divisible by 3 = 1/3 (every 3 rd number is divisible by 3)
probability that n+1 is divisible by 3 = 1/3

Divisibility of n and (n+1) are mutually exclusive. n(n+1) is divisible by 3, if either n or (n+1) is divisible by 3.

P[n(n+1)] = P[n divisible by 3 OR (n+1) divisible by 3] = P[n] + P[(n+1)] = 1/3 + 1/3 = 2/3

2. Probabilty that n(n+1)(n+2) is divisible by 6 - Product of 3 consecutive numbers will be always divisible by 6. Probability = 1

3. Probabilty that n(n+1)(n+2) is divisible by 8

n is even:
(n+1) doesn't contribute to 8 divisibility, as this will be an odd number.

n(n+1)(n+2) - will be divisible by 8, if n(n+2) is divisible by 8.

n(n+2) will be divisible by 8, if one of them is divisible by 4. (8=4*2, Only one number can be divisible by 4, and we can get factor 2 from the other number)

for n between 1-99

P[n div. by 4] = 24/99 [had it been 1-100, we cud have got 1/4 - every 4th number)
P[(n+2) div. by 4] = 25/99 [from 3-101]
P[n(n+2) div. by 4 or 8] = 24/99 + 25/99 = 49/99

n is even, P[n(n+1)(n+2) div. by 8] = 49/99

n is Odd:
n(n+1)(n+2) - will be divisible by 8, if n+1 is divisible by 8.
for n between 1-99, n+1 is between 2-100.

between 2-100, we have 12 multiples of 8.
P[(n+1) div. 8] = 12/99

n is odd, P[n(n+1)(n+2) div. 8] = 12/99

Summing up even and odd cases,

Probabilty that n(n+1)(n+2) is divisible by 8 = 49/99 + 12/99 = 61/99
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Re: n is integer from 1-99 [#permalink]

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12 Aug 2013, 22:23
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bsahil wrote:
n is integer from 1-99:

1. Probabilty that n(n+1) is divisible by 3
2. Probabilty that n(n+1)(n+2) is divisible by 6
3. Probabilty that n(n+1)(n+2) is divisible by 8

I think the answer for 1 and 2 have been discussed enough. I will add a bit on answer to problem no. 3 (similar to the method given by ConnectTheDots)

n can take 99 values from 1 to 99. n can be even or odd and depending on whether it is even or odd, we will handle the case.

Case 1: n is even.

n can take 49 different even values starting from 2, 4, 6 to 98.

In each case, (n+1) will be odd and (n + 2) will be even. Note that whenever you have 2 consecutive even numbers e.g. 2 and 4 or 6 and 8 or 96 and 98, exactly one of them will be a multiple of 4 (since you have a multiple of 4 in every 4 consecutive numbers (2 odd, 2 even)). So for all 49 values of even n, the given product will be a factor of 8.

Case 2: n is odd
Then only (n+1) is even. (n+1) must be a multiple of 8. There are only 12 multiples of 8 in the first 99 numbers (8, 16. ... till 96). So for only 12 values of odd n will the product be a multiple of 8.

Total there are 49 + 12 = 61 values of n for which the product will be a multiple of n.
Probability = 61/99
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Re: n is integer from 1-99 [#permalink]

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10 Apr 2016, 08:32
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Re: n is integer from 1-99   [#permalink] 10 Apr 2016, 08:32
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