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

Question

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

Temurkhon · Answer

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

rohit8865 · Answer

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

RR88 · Answer

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  ; 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.

rever08 · Answer

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.

DharLog · Answer

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.

stne · Answer

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.

chetan2u · Answer

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

Fdambro294 · Answer

Each Set will have 5 Consecutive Integers starting with:

NO Multiple of 6

YES

.....

NO Multiple of 6

so when n = 1 , 7 , 13, 19, 25 .......91, were each term n = 6k + 1 (Multiple of 6 + 1) ---- out of the 96 possible sets, these sets that start with this N will not have a Multiple of 6 in the 5 Consecutive Integers.

The above is an Arithmetic Progression or an Evenly Spaced Number Set.

Excluding n = 1 Case:

Count of Evenly Spaced Set = (Last Term - First Term) / Common Difference of 6 + 1

No. of Sets that will not have a Multiple of 6 in them = (91 - 7) / 6 + 1 = 84/6 + 1 = 15 Sets

Including the Case in which n = 1: there are 15 + 1 = 16 Sets that will NOT have a Multiple of 6.

96 Possible sets - 16 Sets = 80 Sets that will have a Multiple of 6 Term in the 5 Consecutive Integers

Answer A

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards