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 →

sum of factorial

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 109
Joined: Sun Oct 21, 2012 5:10 amFollowed by: 1 member
by psm12se » Thu Jan 02, 2014 3:31 am
If n is a positive integer, what is the last digit of 1! + 2!+ .... + n!?

1. n is divisible by 5

2. (n+1)/6 is a positive integer.

I solved this problem but took 10 mins to solve it. Is there a way to solve it in 2 mins. I used the plugin method as a approach.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members
Thanked: 448 times
GMAT Score:650

by theCodeToGMAT » Thu Jan 02, 2014 3:50 am
To find: Last digit of 1! + 2!+ .... + n!

IMP:
1! = 1
1! + 2! = 1 + 2 = 3
1! + 2! + 3! = 1 + 2 + 6 = 9
1! + 2! + 3! + 4! = 9 + 24 = 33
1! + 2! + 3! + 4! + 5! = 33 + 120 = _ _ 3
Now onwards.. Unit digit will remain "3" only as each factorial after 5! will be "0" as unit digit.
Example: 6! = 720

Statement 1:
n is divisible by 5 ==> n = 5, 10
Unit Digit = 3

Statement 2:
(n+1)/6
Minimum possible value of n = 5
Unit Digit = 3

Answer [spoiler]{D}[/spoiler]?
R A H U L
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 490
Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members
Location: Chennai, India
Thanked: 83 times

by Uva@90 » Thu Jan 02, 2014 7:01 am
psm12se wrote:If n is a positive integer, what is the last digit of 1! + 2!+ .... + n!?

1. n is divisible by 5

2. (n+1)/6 is a positive integer.

I solved this problem but took 10 mins to solve it. Is there a way to solve it in 2 mins. I used the plugin method as a approach.
Hi Psm12se,

Last Digit of 1!+2!+3!+........n!
Always take 3 values
When N=1 last digit = 1
When N=3 last digit = 9
for the rest of values last digit will be 3.

Statement 1 : n is divisible by 5
So from above we can conclude last digit will always be 3.
Sufficient.

Statement 2: (n+1)/6 is a positive integer.
take any positive integer like 1 or 2 or anything.
you will end up with last digit as 3.
Sufficient.

Hence Answer D

Regards,
Uva.
Known is a drop Unknown is an Ocean
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1