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Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1

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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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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
[Reveal] Spoiler: OA

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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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New post 08 Nov 2015, 01:07
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 (96-6)/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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New post 06 Jan 2016, 09:54
[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{x-0,1,....}
i.e n=1,7,13,19.......91=16 nos.
so required such sets consisting 6 or its integral are 96-16=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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New post 18 Apr 2017, 01:43
Option A

For every six consecutive integers, there will be 5 non-multiples of 6 and 1 multiple of 6.
And, these 5 non-multiples 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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New post 30 Sep 2017, 07:19
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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New post 06 Nov 2017, 12:41
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*(k-1), k=1, 2, ..., 16 - so it is 16 such sets.
And it is 96 sets overall.
So 96-16=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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New post 04 Dec 2017, 04:09
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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New post 04 Dec 2017, 05:31
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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6151 [0], given: 121

Re: Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1   [#permalink] 04 Dec 2017, 05:31
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