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 →

last 2 digits.

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times
by goyalsau » Sat Oct 23, 2010 9:32 am
What are the last two digits of 7^(2008)?

(1) 21 (2) 61 (3) 01 (4) 41 (5) 81
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 250
Joined: Thu Jan 07, 2010 2:21 am
Thanked: 10 times

by saurabhmahajan » Sat Oct 23, 2010 10:45 am
by taking 7^2, 7^3, 7^ 4.......7^n we see that the digits 07,49,43,01,07..... are repeated. Out of the answer choices only 01 appears in the above series so IMO: 3 (C)
Thanks and regards,
Saurabh Mahajan

I can understand you not winning,but i will not forgive you for not trying.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Mon Oct 18, 2010 9:04 am
Thanked: 2 times

by sixpointer » Sat Oct 23, 2010 11:44 am
See the cylicity

7^1=07

7^2=49

7^3=43

7^4=01

7^5=07

so on...


So u can say cycle is repeating after 4 steps . Now to find how many such 4 steps will be there , we consider 4 steps as one group ,.

Total number of groups are 2008/4=502

Last digit will be 01

suppose if it was 2007 then we will get 3 as remainder and answer will be 7^3=43

Hope it is clear now:)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Sat Oct 23, 2010 8:27 pm
saurabhmahajan wrote:
by taking 7^2, 7^3, 7^ 4.......7^n we see that the digits 07,49,43,01,07..... are repeated. Out of the answer choices only 01 appears in the above series so IMO: 3 (C)
sixpointer wrote:
So u can say cycle is repeating after 4 steps . Now to find how many such 4 steps will be there , we consider 4 steps as one group ,.

Total number of groups are 2008/4=502

Last digit will be 01
Thanks Guys, I was trying to figure a basic rule with these type of problems but i was not able to figure out one.


This is what i got,

Its only with 11 We can have the same last two digits with the same powers but the cycle is of 10 over there.

Now i understand that last two digits will be same if the number is 7 and 11


did you find any easy to solve a problem if the power if 9^ 2008 or 8 ^ 2008.

We can not apply this rule with 6 8 9 ,
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Mon Oct 18, 2010 9:04 am
Thanked: 2 times

by sixpointer » Sat Oct 23, 2010 10:21 pm
goyalsau wrote:
saurabhmahajan wrote:
by taking 7^2, 7^3, 7^ 4.......7^n we see that the digits 07,49,43,01,07..... are repeated. Out of the answer choices only 01 appears in the above series so IMO: 3 (C)
sixpointer wrote:
So u can say cycle is repeating after 4 steps . Now to find how many such 4 steps will be there , we consider 4 steps as one group ,.

Total number of groups are 2008/4=502

Last digit will be 01
Thanks Guys, I was trying to figure a basic rule with these type of problems but i was not able to figure out one.


This is what i got,

Its only with 11 We can have the same last two digits with the same powers but the cycle is of 10 over there.

Now i understand that last two digits will be same if the number is 7 and 11


did you find any easy to solve a problem if the power if 9^ 2008 or 8 ^ 2008.

We can not apply this rule with 6 8 9 ,
Another method
Find the remainder when 7^2008 is divided by 100
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1