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young_gun
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If 1=<n<=99, what is the probability that n(n + 1) is 05 Mar 2008, 20:18
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If 1=<n<=99, what is the probability that n(n + 1) is perfectly divisible by 3 ?



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If 1=<n<=99, what is the probability that n(n + 1) is 05 Mar 2008, 20:40
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this problem is actually not that hard...I am just having trouble counting how many N's can make N+1 a multiple of 3...I know the answer is 33, but I keep coming up with 32. Can someone pls elucidate?
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If 1=<n<=99, what is the probability that n(n + 1) is 06 Mar 2008, 10:14
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I think total count should be 66.
n = 3,6,9............99 (33 terms)
n = 2,5,8,11........98(33 terms)

so prob = 66/99 =2/3
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If 1=<n<=99, what is the probability that n(n + 1) is 06 Mar 2008, 10:22
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I think total count should be 66.
n = 3,6,9............99 (33 terms)
n = 2,5,8,11........98(33 terms)

so prob = 66/99 =2/3
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i agree. whats the oa
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If 1=<n<=99, what is the probability that n(n + 1) is 06 Mar 2008, 10:23
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we just need to know how many prime numbers other than 3 are there between 1-99
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If 1=<n<=99, what is the probability that n(n + 1) is 06 Mar 2008, 10:33
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ahh..i like Vshanuk's approach..

basically there are 99/3=33 N possible terms

same thing with (n+1) there are divisible by 99/3 =33

total number of terms 66/99 =2/3
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If 1=<n<=99, what is the probability that n(n + 1) is 06 Mar 2008, 12:07
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I think total count should be 66.
n = 3,6,9............99 (33 terms)
n = 2,5,8,11........98(33 terms)

so prob = 66/99 =2/3
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correct, how do you quickly determine 33 terms in n = 2,5,8,11........98
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If 1=<n<=99, what is the probability that n(n + 1) is 06 Mar 2008, 12:15
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young_gun
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I think total count should be 66.
n = 3,6,9............99 (33 terms)
n = 2,5,8,11........98(33 terms)

so prob = 66/99 =2/3

correct, how do you quickly determine 33 terms in n = 2,5,8,11........98
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its not 2 to 98
its 1 to 99


largest term - smallest + inclusive 1
99 - 1 + 1

99 terms/3 = 33



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