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14 Dec 2024, 23:42
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Difficulty:
95%
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Question Stats:
34% (02:35) correct
66%
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wrong
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If an integer n is to be chosen at random from the integers 1 to 90, inclusive, what is the probability that n(n + 1)(n + 2)(n+3)(n+4) will be divisible by 9?
A) 4/9
B) 8/9
C) 21/27
D) 73/90
E) 15/27
A similar question in GMAT Official guide : Official Guide
A) 4/9
B) 8/9
C) 21/27
D) 73/90
E) 15/27
A similar question in GMAT Official guide : Official Guide
15 Dec 2024, 03:17
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n(n + 1)(n + 2)(n+3)(n+4) is divisible by 9 when there is at least 1 multiple of 9 in the product.
for 1-90 inclusive, the multiple of 9= 90/9= 10
thus the value that gives multiple of 9 are
for (n+5)=9, n=4
1. n= 5,6,7,8,9
2. n= 14,15,16,17,18
.
.
.
10. n= 86, 87, 88, 89, 90
which gives 50 favourable outcome.
Again, 9 can be obtained when we have 3*3 that is multiple of 3 excepts these mentioned values,
we can get 1 multiple of 3 in each of 10 groups i.e in group 1 to 10
we can get to multiple of 3s in each group in n(n+1)(n+2)(n+3)(n+4) i.e when n=3, n+3 is multiple of 3 and when n=2, n+1 and n+4 are multiple of 3 which gives additional 10*2= 20 favourable outcomes.
Total favourable outcomes= 70
total possible outcome= 90
probability= 70/90= 7/9= 21/27.
IMO C is the answer.
for 1-90 inclusive, the multiple of 9= 90/9= 10
thus the value that gives multiple of 9 are
for (n+5)=9, n=4
1. n= 5,6,7,8,9
2. n= 14,15,16,17,18
.
.
.
10. n= 86, 87, 88, 89, 90
which gives 50 favourable outcome.
Again, 9 can be obtained when we have 3*3 that is multiple of 3 excepts these mentioned values,
we can get 1 multiple of 3 in each of 10 groups i.e in group 1 to 10
we can get to multiple of 3s in each group in n(n+1)(n+2)(n+3)(n+4) i.e when n=3, n+3 is multiple of 3 and when n=2, n+1 and n+4 are multiple of 3 which gives additional 10*2= 20 favourable outcomes.
Total favourable outcomes= 70
total possible outcome= 90
probability= 70/90= 7/9= 21/27.
IMO C is the answer.
18 Dec 2024, 20:49
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Think of it this way. In the 5 consecutive numbers, if n+2 is a multiple of 3 (other than 9), then the number will not be divisible by 9
eg 1,2,3,4,5 or 4,5,6,7,8
The first multiple of 3 and the last multiple of 3 which falls in this category are 3 and 87 (1, 2, 3, 4, 5) and (85,86,87,88,89)
Using the rule of counting for the number of multiples of 3 (including 9) which come as n+2 = [(87-3)/3] + 1 = 29
These 29 multiples include multiples of 9 which need to be subtracted. The first multiple of 9 and the last multiple of 9 that can come as (n+2) are 9 and 81
Using the rule of counting for the number of multiples of 9 which come as n+2 = [(81-9)/9] + 1 = 9
Therefore there are 29-9 = 20 such multiples that come in the middle as multiples of 3 which then are not divisible by 9.
Total outcomes = 90
Favourable Outcomes = 90-20 = 70
Probability = 70/90 = 7/9 or 21/27
Option C
AK
eg 1,2,3,4,5 or 4,5,6,7,8
The first multiple of 3 and the last multiple of 3 which falls in this category are 3 and 87 (1, 2, 3, 4, 5) and (85,86,87,88,89)
Using the rule of counting for the number of multiples of 3 (including 9) which come as n+2 = [(87-3)/3] + 1 = 29
These 29 multiples include multiples of 9 which need to be subtracted. The first multiple of 9 and the last multiple of 9 that can come as (n+2) are 9 and 81
Using the rule of counting for the number of multiples of 9 which come as n+2 = [(81-9)/9] + 1 = 9
Therefore there are 29-9 = 20 such multiples that come in the middle as multiples of 3 which then are not divisible by 9.
Total outcomes = 90
Favourable Outcomes = 90-20 = 70
Probability = 70/90 = 7/9 or 21/27
Option C
AK
