Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 01 Nov 2007
Posts: 112

If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
16 Apr 2008, 09:25
Question Stats:
38% (02:20) correct 62% (02:14) wrong based on 1838 sessions
HideShow timer Statistics
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? A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 64249

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
28 Jan 2012, 04:12
RadhaKrishnan wrote: 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?
A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4 \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; (Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.) Now, in EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8. Answer: D. Similar question: divisibleby12probability121561.htmlHope it helps.
_________________




Senior Manager
Joined: 23 Oct 2010
Posts: 312
Location: Azerbaijan
Concentration: Finance

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
29 Jan 2012, 00:17
Bunuel wrote: RadhaKrishnan wrote: 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?
A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4 \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; from here u can also think this way (though it is a lil bit similar) find how many numbers are divisible by 2 or 8 96/2=48 96/8=12 (48+12)/96=60/96=5/8
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth




Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 873
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
18 Feb 2012, 07:03
+1 D Other way is analyzing if there is a patron: 1) If n is an even number: n:2, then 2*3*4 = 24 (divisible by 8) n:4, then 4*5*6 = 120 (divisible by 8) n:6, then 6*7*8= again divisible by 8 We have a patron. So, we have 48 even possible values. 2) If n is an odd number: This only can take place when n+1 is multiple of 8. So, we have 12 possible values. Then, \(\frac{(48 + 12)}{96} = \frac{5}{8}\) D
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/myirlogbookdiary133264.html GMAT Club Premium Membership  big benefits and savings



Current Student
Joined: 12 Jun 2008
Posts: 218
Schools: INSEAD Class of July '10

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
09 Jul 2008, 00:33
Since 96 is divisible by 8 and since "divisibility by 8" repeats every 8 terms, you can just focus on the 8 first terms (from 1 to 8):
it works for n=2,4,6,7,8, that is 5 numbers out of the 8
==> Answer is (D) = 5/8



SVP
Status: Top MBA Admissions Consultant
Joined: 24 Jul 2011
Posts: 2052

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
06 May 2012, 01:22
Any integer n(n+1)(n+2) will be divisible by 8 if n is a multiple of 2. This gives us 48 numbers between 1 and 96. Additionally, all those numbers for which (n+1) is a multiple of 8 are also divisible by 8. This gives us a further 12 numbers. These numbers are all distinct from the first set because the first set had only even numbers and this set has only odd numbers. Therefore probability = (48+12)/96 = 60/96 = 5/8 Option (D)
_________________
GyanOne [www.gyanone.com] Premium MBA and MiM Admissions Consulting
Awesome Work  Honest Advise  Outstanding Results Reach Out, Lets chat!Email: info at gyanone dot com  +91 98998 31738  Skype: gyanone.services



Manager
Joined: 11 Sep 2013
Posts: 130
Concentration: Finance, Finance

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
16 Aug 2014, 23:24
goodyear2013 wrote: If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8?
A. 1/4 B. 1/2 C. 5/8 D. 3/4 E. 7/8 In such problem always follow a simple rule. The divisor is 8. So take number from 1 to 8 as example If n=1, n(n+1)(n+2)= Not divisible by 8 If n=2, n(n+1)(n+2)= yes If n=3, n(n+1)(n+2)=No. Do upto 8 So, 5 out of 8 are divisible. Hence, 5/8 is the answer. This rule is applicable if last number(96 in this case) is divisible by 8. If you had to pick from 198, still you can apply the rule but be careful. But there is one additional step. Hopefully you can find answer in less than 60 seconds



Senior Manager
Joined: 04 Jul 2014
Posts: 293
Location: India
GMAT 1: 640 Q47 V31 GMAT 2: 640 Q44 V34 GMAT 3: 710 Q49 V37
GPA: 3.58
WE: Analyst (Accounting)

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
22 Nov 2014, 01:00
Here is the dumb way of doing this question, if the smart ones don't strike your head during the test Since we are looking at divisibility by 8, we consider n to be between 1 and 8. We can multiply the result by 12  since 96/8 = 12. Attachment:
Capture.PNG [ 6.47 KiB  Viewed 95878 times ]



Intern
Joined: 09 Jul 2008
Posts: 1

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
09 Jul 2008, 02:49
answer is D n(n+1)(n+2) will be divisible by 8 for all even numbers, i.e. total 48 numbers also cases in which (n+1) is a multiple of 8 will be divisible by 8 for ex: n=7,15,23.. i.e. total 12 numbers
so total number of such cases is 48+12=60 probability = 60/96 = 5/8



Manager
Joined: 22 Dec 2009
Posts: 225

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
14 Feb 2010, 12:07
marcodonzelli wrote: 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?
A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4 All even numbers would be divisible by 8... as we have two consecutive even numbers being multiplied... Hence 961/2 + 1 = 48 Also we have 8n1 sequence numbers divisible by 8... i.e, if n =7,15,23,31.... Therefore 8n1 = 96 to find the number of elements in this sequence ... gives n = 97/8 = 12 Therefore Probability = (48+12) / 96 = 60 / 96 = 5/8



Math Expert
Joined: 02 Sep 2009
Posts: 64249

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
25 Jul 2014, 02:54



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10610
Location: United States (CA)

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
16 Nov 2016, 07:55
RadhaKrishnan wrote: 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?
A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4 We are given that an integer n is to be chosen at random from the integers 1 to 96 inclusive, and we need to determine the probability that n(n + 1)(n + 2) will be divisible by 8. We should recall that when a number is divisible by 8, it is divisible by 2^3, i.e., three factors of 2. We should also recognize that n(n + 1)(n + 2) is the product of three consecutive integers. Case 1: n is even. Any time that n is even, n + 2 will also be even. Moreover, either n or n + 2 will be divisible by 4, and thus n(n + 1)(n + 2) will contain three factors of 2 and will be divisible by 8. Since there are 96 integers between 1 and 96, inclusive, and half of those integers are even, there are 48 even integers (i.e., 2, 4, 6, …, 96) from 1 to 96 inclusive. Thus, when n is even, there are 48 instances in which n(n + 1)(n + 2) will be divisible by 8. Case 2: n is odd. If n is odd, then n(n + 1)(n + 2) still can be divisible by 8 if the factor (n + 1) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 96 inclusive. Number of multiples of 8 = (96  8)/8 + 1 = 88/8 + 1 = 12. Thus, when n is odd, there are 12 instances in which n(n + 1)(n + 2) will be divisible by 8. In total, there are 48 + 12 = 60 outcomes in which n(n + 1)(n + 2) will be divisible by 8. Thus, the probability that n(n + 1)(n + 2) is divisible by 8 is: 60/96 = 10/16 = 5/8. Answer: D
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



Senior Manager
Joined: 02 Mar 2012
Posts: 258

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
17 May 2013, 12:07
there are total 48 numbers in the form n*n+1*n+2 starting from 2,4,6,8,10.....96 for which it is divisible by 8.
Additionally,there are 12 cases if there is 8 or a multiple of 8 that also divides the form n*n+1*n+2 which starts from 7,15,23,31....95.
The above cases both are nonoverlapping, so we can add them.
adding the above two cases 48+12=60 ans=60/96=5/8 which is D.
harry Do press kudos if u like my post



Manager
Joined: 10 Sep 2013
Posts: 68
Concentration: Sustainability, International Business

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
25 Sep 2013, 03:50
chunjuwu wrote: 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?
A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4
please explain, thank you. For n(n+1)(n+2) to be divisible by 8, either n has to be even ( because if, n and n+2 are even, then n is divisible by 8) or n+1 as a whole should be divisible by 8 There are 96 values for n. The possibility of n to be even is (96/2) = 48 and the possibility of n to be divisible by 8 is (96/8)= 12 Therefore the total prob = (48+12)/96 = 60/96 = 5/8 The OA is D



CEO
Joined: 20 Mar 2014
Posts: 2544
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
12 Jul 2015, 04:40
RadhaKrishnan wrote: 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?
A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4 Alternate solution: Look at the first few sets of 3s: 1,2,3 2,3,43,4,5 4,5,65,6,7 6,7,87,8,9 8,9,10 We see that out of the above 8 sets, favorable cases are 5 (in red). Thus the probability is 5/8. This will repeat till we have 88,89,90 making it 9 total patterns. Now consider after 88,89,90, we get 89,90,91 90,91,9291,92,93 92,93,9493,94,95 94,95,9695,96,9796,97,98So we have another 5 favorable out of the remaining 8 sets. Thus we have final 5/8 as the probability. Thus the final probability = {(5/8)*9+(5/8)*1} / (9+1) = 5/8



Math Expert
Joined: 02 Sep 2009
Posts: 64249

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
16 Jul 2016, 01:41
megha_2709 wrote: Bunuel wrote: RadhaKrishnan wrote: 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?
A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4 \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; (Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.) Now, in EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8. Answer: D. Similar question: divisibleby12probability121561.htmlHope it helps. Hi, Thank you for posting such a good explanation , however I could not understand how can you categorize the numbers in group of 8 . Probability is Fav/Total . Shouldn't we consider all 96 values and find out how many are satisfying our conditions , cant understand how you arrive at 5/8. If you can please explain. Sorry if this sounds too basic. Regards Megha In EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8
_________________



Intern
Joined: 25 Dec 2016
Posts: 17
Location: United States (GA)
Concentration: Healthcare, Entrepreneurship
GPA: 3.64
WE: Medicine and Health (Health Care)

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
07 Feb 2017, 06:48
My approach was similar to Bunuel's and AdmitJA's, but I wanted to offer it because it was simple and quick. I was able to do this one in just over a minute.
When I read the question, I paid attention to two things right away. The first was that \((n)(n+1)(n+2)\) is the product of three consecutive integers. Because \(8=2^3\), I knew that I needed to find three 2s in the factorization of whatever 3 consecutive numbers I used. The second thing that I noticed is that 96 is divisible by 8. This indicated to me that whatever pattern I noticed in the first 8 numbers would be repeated 12 times through 96. Since I recognized that the pattern would be repeated, I knew that I only needed to look at the first 8 numbers.
I listed the numbers out: 1 2 3 4 5 6 7 8.
Since 8 has three 2s in its factorization, I knew that n=6,7,8 would all be divisible by 8. I saw that n=2 would work because 2 has one 2 and 4 has two 2s. Similarly, I saw that n=4 would work because 4 has two 2s and 6 has one 2. That gives us 5 options in the first 8 numbers. So the answer is 5/8.
As mentioned above, this ratio will be the same for every 8 numbers, so 5/8 will be true of 8x where x is any positive integer. A trickier version of this question would have been to make the number NOT divisible by 8. In that case, I think you should still find the pattern for every 8 numbers, but you'd also want to look at the "extra" numbers to figure out the fraction.
Take 53, for instance. You'd want to recognize that 8 goes into 53 six times with a remainder of 5. This means that you'd have the \(5/8\) ratio for 6 sets of 8 but also an "incomplete" set of n = 1,2,3,4,5. In the first 48 numbers, you'd have 30 that would be divisible by 8. In the "incomplete" set you'd have 2 (since n=2 and n=4 are both divisible by 8). Thus you'd have 32/53 numbers divisible by 8.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10477
Location: Pune, India

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
15 May 2018, 09:26
siddreal wrote: RadhaKrishnan wrote: 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?
A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4 VeritasPrepKarishma Pls show us your approach to this question. Though I understood the solutions mentioned in the thread, I am finding it difficult to believe that I might think of this approach during actual exam. What I would do is pretty much what Bunuel has done in his solution. Note that the moment I see n(n + 1)(n + 2), I think of divisibility in a bunch of consecutive numbers. In 3 consecutive integers, if n is even, (n+2) is even too. If there are two consecutive even integers, one of them will be a multiple of 4. So if n is even, (n+2) is even too and one of them is definitely a multiple of 4. So n(n + 1)(n + 2) becomes divisible by 8 in each case that n is even. From 1 to 96, half the cases have even n so this mean 48 cases. Alternatively, when n is odd, n+1 is even. But then n+2 is odd too. So to be a multiple of 8, (n+1) will need to be a multiple of 8. Hence this gives us another 12 cases (n+1 goes from 8 to 96). Note that there will be no overlap in the two since here n is definitely odd. Total we have 60 cases of the possible 96 which gives 60/96 = 5/8 For these properties of numbers, see https://www.veritasprep.com/blog/2011/0 ... cormath/https://www.veritasprep.com/blog/2011/0 ... hpartii/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Retired Moderator
Joined: 02 Sep 2010
Posts: 702
Location: London

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
19 Sep 2010, 15:41
Notice what happens when you shift this product by 8 : \((n+8)(n+1+8)(n+2+8) = 8^3 + 8^2*(n+n+1+n+2) + 8*(n*(n+1)+(n+1)*(n+2)+n*(n+2)) + n*(n+1)*(n+2) = 8*K + n*(n+1)*(n+2)\) Here K is a constant. Therefore, when we shift by 8, the remainder when divided by 8 remains the same ! So if we figure out what happens for the first 8 choices of n, for all the rest the pattern will repeat n=1 1x2x3 No n=2 2x3x4 Yes n=3 3x4x5 No n=4 4x5x6 Yes n=5 5x6x7 No n=6 6x7x8 Yes n=7 7x8x9 Yes n=8 8x9x10 Yes So for every 8 numbers, exactly 5 will be divisible by 8. Here there are exactly 96 numbers to consider starting with 1 ... So probability will be exactly 5/8
_________________



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 16753
Location: United States (CA)

Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
Show Tags
08 Oct 2015, 09:49
Hi All, For a number to be evenly divisible by 8, it has to include at least three 2's when you prime factor it. For example, 8 is divisible by 8 because 8 = (2)(2)(2).....it has three 2s "in it" 48 is divisible by 8 because 48 = (3)(2)(2)(2)(2).....it has three 2s "in it" (and some other numbers too). 20 is NOT divisibly by 8 because 20 = (2)(2)(5)....it only has two 2s. In this question, when you take the product of 3 CONSECUTIVE POSITIVE INTEGERS, you will either have.... (Even)(Odd)(Even) or (Odd)(Even)(Odd) In the first option, you'll ALWAYS have three 2s. In the second option, you'll only have three 2s if the even term is a multiple of 8 (Brent's list proves both points). So for every 8 consecutive sets of possibilities, 4 of 4 from the first option and 1 of 4 from the second option will give us multiples of 8. That's 5/8 in total. GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★




Re: If an integer n is to be chosen at random from the integers 1 to 96
[#permalink]
08 Oct 2015, 09:49



Go to page
1 2
Next
[ 27 posts ]

