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If 1=<n<=99, what is the probability that n(n + 1) is perfec
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11 Dec 2010, 22:52
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If 1 <n <99, what is the probability that n(n + 1) is perfectly divisible by 3 ? For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.
Considering n is a multiple of 3: Between 1 to 99, the number of multiples of 3 = (99  3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]
Considering (n + 1) is a multiple of 3: Between 1 to 99, the number of multiples of 3 would be from 298 and would also be 33 multiples.
Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd) = 33/99 + 33/99 = 66/99 = 22/33 = 2/3
I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble (nL(last)nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?




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Re: Beginner's Forum Question
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12 Dec 2010, 00:17
mmcooley33 wrote: If 1=<n<=99, what is the probability that n(n + 1) is perfectly divisible by 3 ?
For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.
Considering n is a multiple of 3: Between 1 to 99, the number of multiples of 3 = (99  3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]
Considering (n + 1) is a multiple of 3: Between 1 to 99, the number of multiples of 3 would be from 298 and would also be 33 multiples.
Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd) = 33/99 + 33/99 = 66/99 = 22/33 = 2/3
I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble (nL(last)nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again? First about the multiple of x in the given range:\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\), (check this: totallybasic94862.html?hilit=multiple%20range). So, there will be \(\frac{993}{3}+1=33\) multiples of 3 in the range from 1 to 99, inclusive: 3, 6, 9, ..., 99; n+1 to be a multiple of 3 in the range from 1 to 99, inclusive n must be 2, 5, 8, ..., 98, one less than above values, so also 33 options for n; Hence, n(n+1) to be a multiple of 3 n can take 33+33=66 values: P=(favorable outcomes)/(total # of outcomes)=66/99=2/3. 30 second approach:Take first 3 numbers from the given range: 1, 2, and 3 in order n(n+1) to be a multiple of 3 n can be either 2 or 3 P=2/3. For next 3 numbers the probability will be the same and as 99 equals to 3*33 (similar 33 blocks of 3) then for complete range the probability will still be 2/3. Answer: 2/3. Hope it helps.
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Re: Beginner's Forum Question
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11 Dec 2010, 23:08
this formula is in the MGMAT math book.. really useful to know... you add 1 as you generally underestimate by 1 at the extreme  this evenly spaced sets formula is the basis for the simpler formula: (lastFirst) + 1 when you have to count out the number of elements in a consecutive sequence 1, 2, 3.... You are basically doing (lastfirst)/1 + 1 > you are basically dividing by 1 there as that is the common space b/w the elements... Knowing that you can apply this formula for evenly spaced sets should be enough...



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Re: Beginner's Forum Question
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11 Dec 2010, 23:22
thanks for the quick reply! kudos for the help.



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Re: Beginner's Forum Question
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13 Dec 2010, 07:41
Ah, that 30 sec approach looks good.
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Re: Beginner's Forum Question
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07 Jul 2013, 03:12
Bunuel wrote: mmcooley33 wrote: If 1=<n<=99, what is the probability that n(n + 1) is perfectly divisible by 3 ?
For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.
Considering n is a multiple of 3: Between 1 to 99, the number of multiples of 3 = (99  3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]
Considering (n + 1) is a multiple of 3: Between 1 to 99, the number of multiples of 3 would be from 298 and would also be 33 multiples.
Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd) = 33/99 + 33/99 = 66/99 = 22/33 = 2/3
I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble (nL(last)nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again? First about the multiple of x in the given range:\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\), (check this: totallybasic94862.html?hilit=multiple%20range). So, there will be \(\frac{993}{3}+1=33\) multiples of 3 in the range from 1 to 99, inclusive: 3, 6, 9, ..., 99; n+1 to be a multiple of 3 in the range from 1 to 99, inclusive n must be 2, 5, 8, ..., 98, one less than above values, so also 33 options for n; Hence, n(n+1) to be a multiple of 3 n can take 33+33=66 values: P=(favorable outcomes)/(total # of outcomes)=66/99=2/3. 30 second approach:Take first 3 numbers from the given range: 1, 2, and 3 in order n(n+1) to be a multiple of 3 n can be either 2 or 3 P=2/3. For next 3 numbers the probability will be the same and as 99 equals to 3*33 (similar 33 blocks of 3) then for complete range the probability will still be 2/3. Answer: 2/3. Hope it helps. Bump for the 30 sec approach.



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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
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07 Jul 2013, 05:36
mmcooley33 wrote: If 1 <n <99, what is the probability that n(n + 1) is perfectly divisible by 3 ? For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.
Considering n is a multiple of 3: Between 1 to 99, the number of multiples of 3 = (99  3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]
Considering (n + 1) is a multiple of 3: Between 1 to 99, the number of multiples of 3 would be from 298 and would also be 33 multiples.
Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd) = 33/99 + 33/99 = 66/99 = 22/33 = 2/3
I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble (nL(last)nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again? Similar questions to practice: ifanintegernistobechosenatrandomfromtheintegers126654.htmlifintegercisrandomlyselectedfrom20to99inclusive121561.html
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
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22 Nov 2014, 02:19
Wow! I did this one very fast and correctly! :D (After getting a series of other questions in this type wrong) Here is a very simple solution. For n(n+1) to be divisible by 3, either a) n must be a multiple of 3 > Number of possibilities = 99/3 = 33 OR b) (n 1) must be a multiple of 3 > Number of possibilities = 99/3 = 33 Now, favorable possibilities = 66 and total possibilities = 100 (100 because 99 and 1 are both included). Therefore, probability = 66/100 or 2/3 :D There's a discussion going on here for this question: anintegernbetween1and99inclusiveistobechosenat160998.html#p1445924
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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
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16 Jan 2017, 17:00
mmcooley33 wrote: If 1<n<99, what is the probability that n(n + 1) is perfectly divisible by 3 ? We are given that an integer n is to be selected at random from 1 to 99 inclusive, and we must determine the probability that n(n+1) will be divisible by 3. Since probability = favorable outcomes/total outcomes and we know that the total number of outcomes is 99, because there are 99 integers from 1 to 99 inclusive, we need to determine the number of values of n such that n(n+1) is divisible by 3. First, we can determine the number of values of n that are divisible by 3, that is, the number of multiples of 3 that are between 3 and 99 inclusive. To calculate this, we can use the formula: (Largest multiple of 3 – smallest multiple of 3)/3 + 1 (99  3)/3 + 1 96/3 + 1 32 + 1 = 33 Thus, there are 33 multiples of 3 between 1 and 99 inclusive. That is, the value of n can be any one of these 33 multiples of 3 so that n(n + 1) will be divisible by 3. Similarly, if n + 1 is a multiple of 3, n(n + 1) will be also be divisible by 3. Since we know that there are 33 values of n that are multiples of 3, there must be another 33 values of n such that n + 1 is a multiple of 3. Let’s expand on this idea: When n = 2, n + 1 = 3, and thus n(n+1) is a multiple of 3. When n = 23, n + 1 = 24, and thus n(n+1) is a multiple of 3. When n = 98, n + 1 = 99, and thus n(n+1) is a multiple of 3. We can see that there are 33 values of n that are multiples of 3, and 33 more values of n for n + 1 to be multiples of 3. Thus, the probability of selecting a value of n so that n(n+1) is a multiple of 3 is: 66/99 = 2/3
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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
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24 Jan 2017, 19:05
1) n(n+1) is divisible by 3 when either n is divisible by 3 or (n+1) is divisible by 3. 2) n is divisible by 3 in 99/3=33 cases 3) n+1 is divisible by 3 in 99/3=33 cases 4) 33+33=66 5) 66/99=2/3



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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
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