There are 99 integers from 1-99
n(n+1) are product of 2 consecutive integers..
See a pattern,
n=1, n(n+1) = 2 - not divisible by 3
n=2, n(n+1) = 6 - divisible
n=3, n(n+1) = 12 - divisible
n=4, n(n+1) = 20 - not divisible..
Now all numbers(n) which leave a remainder of 1 when divided by 3 will have n(n+1) not divisible by 3.
Thus, the number will be of the form 3k+1
Now k can take values from 0 to 32 where the number varies from 1 to 97 and n(n+1) varies from 2 to (97*98).
0 to 32 is 33 cases..
In all remaining 66 cases n(n+1) will be a multiple of 3.
Therefore, Probability = 66/99 = 2/3
Probability Problem...pls help
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Source: Beat The GMAT — Problem Solving |
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shankar.ashwin
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