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Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1 [#permalink]
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08 Nov 2015, 01:41
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Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3,...., 96. How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ...)? (a) 80 (b) 81 (c) 82 (d) 83 (e) 84
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Re: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1 [#permalink]
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08 Nov 2015, 02:07
T1: 1,2,3,4,5 T2: 2,3,4,5,6 T3: 3,4,5,6,7 etc.
means that every multiple of 6 will be involved in 5 sets. We have (966)/6+1=16 such multiples.
So, final number of sets is 16*5=80
A



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Re: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1 [#permalink]
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06 Jan 2016, 10:54
[quote="excelingmat"]Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3,...., 96. How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ...)?
(a) 80 (b) 81 (c) 82 (d) 83 (e) 84
Consider different sets for n=1 set= {1,2,3,4,5} n=2 set={2,3,4,5,6} n=3 set={3,4,5,6,7} n=4 set={4,5,6,7,8} n=5 set={5,6,7,8,9} n=6 set={6,7,8,9,10} n=7 set={7,8,9,10,11} . . . n=96 set={96,97,98,99,100}
if we look pattern a set consisting of 6 or its integral comes after every interval of 5 sets. ie a set not consisting of 6 or its integral can be written as when n=6x+1{x0,1,....} i.e n=1,7,13,19.......91=16 nos. so required such sets consisting 6 or its integral are 9616=80 ans. A



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Re: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1 [#permalink]
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18 Apr 2017, 02:43
Option AFor every six consecutive integers, there will be 5 nonmultiples of 6 and 1 multiple of 6. And, these 5 nonmultiples of 6 can be converted into a multiple of 6 by adding : 1, 2, 3, 4 or 5. Out of six possible multiples of 6 in a set like [n, n+1, n+2, n+3, n+4, n+5]; given set contains only 5 (n, n+1, n+2, n+3, n+4). Hence, total no. of sets that contain 6 or any integral multiple of 6 are : 5/6th of 96 = 80.
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Re: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1 [#permalink]
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30 Sep 2017, 08:19
1,2,3,4,5  2,3,4,5,6 3,4,5,6,7 4,,8 5,,9 6,,10 ______________ 7,,11  8,,12 9,,13 10,,14 11,,15 12,,16 _____________ 13,,17 
so we see every 1st set in the group of 6 doesn't fit into the criteria. So, 96  16 = 80 will be the answer.



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Re: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1 [#permalink]
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06 Nov 2017, 13:41
excelingmat wrote: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3,...., 96. How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ...)?
(a) 80 (b) 81 (c) 82 (d) 83 (e) 84 Sets, that do not contain 6 or integral multiple of 6  {1, 2, 3, 4, 5}, {7, 8, 9, 10, 11}, ... , {91, 92, 93, 94, 95}. The formula of first element of such sets is 1 + 6*(k1), k=1, 2, ..., 16  so it is 16 such sets. And it is 96 sets overall. So 9616=80. A.



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Re: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1 [#permalink]
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04 Dec 2017, 05:09
excelingmat wrote: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3,...., 96. How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ...)?
(a) 80 (b) 81 (c) 82 (d) 83 (e) 84 From n = 1 to n= 10 there are 8 such sets From n = 11 to n = 20 again there are 8 such sets From n = 21 to n = 30 There are 9 such sets so the pattern is: for every 3rd multiple of 10 i.e. 30, 60 and 90 there will be 9 such sets all else will have 8 such sets . 8,8, 9, 8,8,9,8, 8,9 > in the first 90 there are 75 such sets and from n = 91 to n = 96 there are 5 such sets hence 75+5 =80 sets total.
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Re: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1 [#permalink]
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04 Dec 2017, 06:31
excelingmat wrote: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3,...., 96. How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ...)?
(a) 80 (b) 81 (c) 82 (d) 83 (e) 84 the set \(T_n\) consists of FIVE consecutive integers.... we are looking for how many of them contain multiple of 6... n, n+1, n+2, n+3, n+4, n+5 will have exactly one MULTIPLE of 6.... so when we remove n+5 from the set, the probability that it contains a multiple of 6 is \(\frac{5}{6}\), leaving one case when n+5 could be multiple of 6... since we are looking at 96, which itself is a multiple of 6, the cases will be \(96*\frac{5}{6} = 80\) A
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Re: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1
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