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Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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10 Oct 2016, 04:50
Keats wrote: If I see a sequence of 4 numbers, say for example, 1,2,3, and 4, we have two numbers that satisfy the case here  3 and 4. So 2/4 makes sense.DensetsuNo wrote: eg. If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
Sequence of 8 numbers: 8 Numbers we have: 3
Answer 13/8= 5/8.
In this case, lets take a sequence of 8  1,2,3,4,5,6,7, and 8  we have four numbers that satisfy the case  2,4,6, and 8. So 4/8=1/2 should be the case. Why do you say it is 3/8?Please help me understand. Many thanks in advance. There are two ways in which n(n + 1)(n + 2) can be a multiple of 8. Case 1: n and (n+2) are even. Two consecutive even numbers will have exactly one multiple of 4. So out of the two  n and (n+2)  exactly one of them will be a multiple of 4 and the other a multiple of 2. So their product will be a multiple of 8. Out of 96 numbers, exactly half will have n as even. (n+1) will be odd here but will be immaterial. So this will be 1/2. Case 2: (n+1) is even and a multiple of 8. n can take values from 1 to 96. So (n+1) can take values from 2 to 97. How many multiples of 8 lie between 2 and 97, inclusive? 8, 16, 24, 32, ... 96 8 * 1 = 8 8 * 2 = 16 ... 8 * 12 = 96 So in another 12/96 = 1/8 cases, n(n+1)(n+2) will be a multiple of 8. Total, in 1/2 + 1/8 = 5/8 of the cases, n(n+1)(n+2) will be a multiple of 8.
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Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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10 Oct 2016, 08:10
mitko20m wrote: If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?
A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/4 To find total number of cases where n(n+1) will be divisible by 4 = Where n is divisible by 4 = 4,8,12,...100 = (100  4) / 4 + 1= 25. There 25 such numbers Now cases where n+1 is divisible by 4, which is basically all numbers divisible by 3 = 3,6,9....99 = (993)/3 + 1 = 25 Total Cases = 50 Probability = 50/100 Took time for me to realize that I am solving it for n and n+1, but not just n Regards



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Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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10 Oct 2016, 23:12
Keats wrote: Thanks VeritasPrepKarishma I understand the way you have done it. Even I did it the same way. However, I am perplexed how DensetsuNo is arriving onto the correct answer in just 10 seconds. I also want to learn this quick thing! In the sequence of 8 numbers where n = 1 to 8, there are 5 values for which n*(n+1)*(n+2) is divisible by 8. 1*2*3 2*3*4  divisible by 8 3*4*5 4*5*6  divisible by 8 5*6*7 6*7*8  divisible by 8 7*8*9  divisible by 8 8*9*10  divisible by 8 For values of n from 9 to 16, we will have the same pattern and so on... Probablity = 5/8
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Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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11 Oct 2016, 16:02
mitko20m wrote: If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?
A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/4 We are given that an integer n is to be selected at random from 1 to 100, inclusive, and we must determine the probability that n(n+1) will be divisible by 4. Since probability = favorable outcomes/total outcomes, and we know that the total number of outcomes is 100, since there are 100 integers from 1 to 100, inclusive, we need to determine the number of values of n such that n(n+1) is divisible by 4. First, we can determine the number of values of n that are divisible by 4, that is, the number of multiples of 4 that are between 1 and 100, inclusive. To calculate this, we can use the formula: (Largest multiple of 4 – smallest multiple of 4)/4 + 1 (100  4)/4 + 1 96/4 + 1 24 + 1 = 25 Thus, there are 25 multiples of 4 between 1 and 100 inclusive. That is, the value of n can be any one of these 25 multiple of 4 so that n(n + 1) will be divisible by 4. Similarly, if n + 1 is a multiple of 4, n(n + 1) will be also be divisible by 4. Since we know that there are 25 values of n that are multiples of 4, there must be another 25 values of n such that n + 1 is a multiple of 4. Let’s expand on this idea: When n = 3, n + 1 = 4, and thus n(n+1) is a multiple of 4….. When n = 23, n + 1 = 24, and thus n(n+1) is a multiple of 4….. When n = 99, n + 1 = 100, and thus n(n+1) is a multiple of 4. So we can see that there are 25 values of n that are multiples of 4 and 25 more values of n for n + 1 that are multiples of 4. Thus, the probability of selecting a value of n so that n(n+1) is a multiple of 4 is: 50/100 = 1/2 Answer: C
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Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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15 Feb 2017, 02:45
Yes, when c starts from 20, 12 numbers range from 20 to 32, inclusive In these 12 numbers, only 3 numbers are divisible by 12. ( c=23,24 & 25) That's how 3/12 => 1/4. But why are we subtracting this from 1, I don't get that. DensetsuNo please explain. Keats wrote: DensetsuNo wrote: \(c^{3}c\; =\; c\left( c^{2}1 \right)\; =\; \left( c1 \right)c\left( c+1 \right)\; >\; sequence\; of\; 12\; #s\; >\; #s\; we\; have:\; 3\; >\; 1\; \; \frac{1}{4}\; >\; \frac{3}{4}\; >\; answer C\) DensetsuNo Can you please elaborate. You say that "sequences of 12" will have 3 favorable cases? I don't know if I am interpreting it correctly. Probably with an example, I can understand better. I really want to understand this quick approach. Thank you in advance.



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Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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15 Feb 2017, 04:00
ajay2121988 wrote: Yes, when c starts from 20, 12 numbers range from 20 to 32, inclusive In these 12 numbers, only 3 numbers are divisible by 12. ( c=23,24 & 25) That's how 3/12 => 1/4. But why are we subtracting this from 1, I don't get that. DensetsuNo please explain. Keats wrote: DensetsuNo wrote: \(c^{3}c\; =\; c\left( c^{2}1 \right)\; =\; \left( c1 \right)c\left( c+1 \right)\; >\; sequence\; of\; 12\; #s\; >\; #s\; we\; have:\; 3\; >\; 1\; \; \frac{1}{4}\; >\; \frac{3}{4}\; >\; answer C\) DensetsuNo Can you please elaborate. You say that "sequences of 12" will have 3 favorable cases? I don't know if I am interpreting it correctly. Probably with an example, I can understand better. I really want to understand this quick approach. Thank you in advance. This is not correct. Check this link: https://gmatclub.com/forum/ifanintege ... l#p1746303
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Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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09 Apr 2018, 23:54
sameerspice wrote: mitko20m wrote: If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?
A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/4 the easiest way to solve this without any calculation is by as follows 1. the only way n(n+1) is divisible by 4, if and only if either n is multiple of 4, Between 1 & 100, there are 25 number that are multiple of 4, such as 4, 8, 12...100 2. or (n+1) is multiple of 4, this is equivalent to saying number which are multiple of 3, Between 1 & 100, there are 25 such as 3,7,..99, Hence we have 25 + 25 = 50 number which can be picked between 1 and 100, that would make n(n+1) divisible by 4 probability: fav case/total occurence => 50/100 = 1/2 I think that the above statement ""r (n+1) is multiple of 4, this is equivalent to saying number which are multiple of 3"" is inaccurate because it is not necessary that such a number is a multiple of 3. For example, if n = 7, then n(n+1) = 7*8 which is divisible by 4 and 7 is not a multiple of 3. So there are 25 numbers (n) that are multiples of 4 and therefore, there would also be 25 numbers (n) to which if 1 were added (n+1), would become multiples of 4 = 50 Therefore, answer = 50/100 = 1/2



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Re: If an integer n is to be selected at random from 1 to 100, inclusive,
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24 Aug 2018, 23:05
DensetsuNo wrote: 3sec solution: If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4? Sequence of 4 numbers: 4 Numbers we have: 2 Answer: 12/4 = 1/2 You can see how this method works also on harder questions letting us solve them in a few seconds. eg. If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?Sequence of 8 numbers: 8 Numbers we have: 3 Answer 13/8= 5/8. Similar questions:http://gmatclub.com/forum/ifintegerc ... 21561.htmlhttp://gmatclub.com/forum/ifaninteger ... 26654.htmlWhy doesn't this work for the following question: https://gmatclub.com/forum/anintegern ... 60998.htmlOr am I missing something? Btw this shortcut is beautiful just trying to better understand how to use it!




Re: If an integer n is to be selected at random from 1 to 100, inclusive, &nbs
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24 Aug 2018, 23:05



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