BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Pretty good arithmetic sequence problem

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Thu Jun 24, 2010 3:56 am
Thanked: 3 times
by gtestprep » Fri Feb 11, 2011 4:34 am
Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85 (inclusive). How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members
Thanked: 127 times

by Night reader » Fri Feb 11, 2011 4:44 am
gtestprep wrote:Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85 (inclusive). How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12
IOM E in a series of {4,11,18,25,32,39,46,53,60,67,74,81} :)

3(n-1)/7=i (i is integer) +2/7 as 5+r should give us divisibility by 7; 3(n-1)=7i+2 --> 3n-3-2=7i <> 3n-5=7i
start from n=4, continue until (n+7)<=85
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Fri Feb 11, 2011 6:59 am
gtestprep wrote:Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85 (inclusive). How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12
Let's determine the pattern.
n=1: 5 + (1-1)*0 = 5.
n=2: 5 + (2-1)*3 = 8.
n=3: 5 + (3-1)*3 = 11.
n=4: 5 + (4-1)*3 = 14. 14 is a multiple of 7.

The first multiple of 7 occurs when n=4.
We can see that each number in the sequence is 3 more than the previous number in the sequence.
This means that every 7th term increases by 7*3 = 21.
Since n=4 is a multiple of 7, and every 7th term after n=4 will increase by 21, every 7th term after n=4 will be a multiple of 7:

n=11: 5 + (11-1)*3 = 35.
n=18: 5 + (18-1)*3 = 56.
And so on.

Since 85-4 = 81, and 11*7 = 77, there are 11 multiples of 7 between n=4 and n=85.
Thus, including n=4, the total number of multiples of 7 = 1+11 = 12.

The correct answer is E.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Thu Jun 24, 2010 3:56 am
Thanked: 3 times

by gtestprep » Fri Feb 11, 2011 8:51 am
Thanks GMATGuruNY! Your explanation helped.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 50
Joined: Thu May 20, 2010 1:11 pm
Thanked: 2 times
GMAT Score:590

by viv_gmat » Fri Nov 18, 2011 10:29 am
GMATGuruNY wrote:
gtestprep wrote:Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85 (inclusive). How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12
Let's determine the pattern.
n=1: 5 + (1-1)*0 = 5.
n=2: 5 + (2-1)*3 = 8.
n=3: 5 + (3-1)*3 = 11.
n=4: 5 + (4-1)*3 = 14. 14 is a multiple of 7.

The first multiple of 7 occurs when n=4.
We can see that each number in the sequence is 3 more than the previous number in the sequence.
This means that every 7th term increases by 7*3 = 21.
Since n=4 is a multiple of 7, and every 7th term after n=4 will increase by 21, every 7th term after n=4 will be a multiple of 7:

n=11: 5 + (11-1)*3 = 35.
n=18: 5 + (18-1)*3 = 56.
And so on.

Since 85-4 = 81, and 11*7 = 77, there are 11 multiples of 7 between n=4 and n=85.
Thus, including n=4, the total number of multiples of 7 = 1+11 = 12.

The correct answer is E.

Hi,
Is there a short cut to solve this problem?

Your prompt reply will be highly appreciated.

Thanks
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 176
Joined: Thu Sep 22, 2011 5:32 am
Thanked: 5 times

by vishal.pathak » Sat Nov 19, 2011 5:38 am
viv_gmat wrote: Hi,
Is there a short cut to solve this problem?

Your prompt reply will be highly appreciated.

Thanks
We have to determine the numbers divisible by 7
eqn is 5+(n-1)*3
So (n-1)*3 should give us a remainder of 2 when divided by 7.
We know that 9 (3*3) gives a remainder of 2 when divided by 7
Every 7th multiple of 3 will give a remainder of 2 (like 10 (3+7) and 17 (10+7) etc)
Number of such multiples in 84 (85-1) = 84/7 = 12
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 30
Joined: Sat May 21, 2011 7:28 am

by KItuz » Tue Mar 20, 2012 7:09 am
Question can be reduced to how many integers n belong to [1,...,85] such that (3n+2)/7 is an integer.
multiples of 7 are 7,14,21,28,35,.....
subtracting 2 from each we have 5,12,19,26,33,....
expressing each of them as 3*x we have x,3*4,x,x,3*11,...
Blind guess - checking if there is a pattern here we get 11-4=7
so the question can be reduced to count m such that 4+7m<=85
substitute 12 and 4+7m >85 so we have 0,1,2,3,4,...11 here = 12 values.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 626
Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members
Location: Ahmedabad
Thanked: 31 times

by ronnie1985 » Wed Mar 21, 2012 9:48 am
The expression is 3n+2. When divided by 7:-

n= even = 2k, remainder = 2-k
n= odd = 2k-1, remainder = 1-k
For n = even, the numbers shall be divisible for all 2-k = 0,-7,-14,-21,-28 = > k = 2, 9, 16, 23, 30, 37 = > n = 4, 18, 32, 46, 60, 74.
For n = odd, the numbers shall be divisible for all 1-k = 0, -7, -14,-21,-28, -35,...etc.
=> n = 8, 15, 22, 29, 36, 43.
Total = 12 such n. (E) is answer.
Follow your passion, Success as perceived by others shall follow you
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 18
Joined: Fri Apr 27, 2012 1:49 am

by anjalimanas » Fri Mar 15, 2013 1:29 am
Alternate way

Multiples of 7 from 5 to 257 = 36

as the series is advancing by 3 everytime, Number of multiples of 7 in series = 36 / 3 = 12
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1