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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Finding the Unit digit

Expert replies
by gdk800 » Fri Nov 19, 2010 1:55 pm
Pls help with the following problem.

Unit digit of (264)^103 + (264)^102 = ?


My answer: 5

First Term
4 has a cyclicity of 2. Thus 103/2 would leave a remainder of 1
=>4^1 = 4.

Second Term
4 has a cyclicity of 2. Thus 102/2 would leave a remainder of 0
=>4^0 = 1

Therefore, 4 + 1 = 5 is the unit digit.

Am i doing it correctly??
Join the discussion
Source: — Problem Solving |

by Rahul@gurome » Fri Nov 19, 2010 2:23 pm
gdk800 wrote:Pls help with the following problem.

Unit digit of (264)^103 + (264)^102 = ?


My answer: 5

First Term
4 has a cyclicity of 2. Thus 103/2 would leave a remainder of 1
=>4^1 = 4.

Second Term
4 has a cyclicity of 2. Thus 102/2 would leave a remainder of 0
=>4^0 = 1

Therefore, 4 + 1 = 5 is the unit digit.

Am i doing it correctly??
No. You're not.
For the second term you're doing 4^0, which is wrong. Following your logic 4^2 will have 1 as unit digit! You're blindly applying the cyclical pattern method. Simple cyclical pattern exists for powers of 4. Which is: 4^(even) has 6 as unit digit and 4^(odd) has 4 as unit digit. Thus unit digit of [(264)^103 + (264)^102] is unit digit of (4 + 6) i.e. 0.

Another efficient way to solve this particular problem,
... [(264)^103 + (264)^102]
= [264 + 1]*[(264)^102]
= 265*[(264)^102]
= 5*(even number) => Unit digit is 0
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Join the discussion

by rishab1988 » Fri Nov 19, 2010 2:31 pm
Here's how i'd approach this problem.

264^102[ 264+1] - factorize

264^102 (265)

Now since the question is about units digit we care about multiplication in only in units digit.

Units digit in 264 -4 ; take the square of 264 - 4*4=16; 6 in units digit ( coz we care only about multiplication in units digit- 4*4=16, in which units digit is 6).

next take 264^3 - gives 4*4*4= 6(from 4*4) *4 = 4 in units digit;

264^4 - 4 in units digit again.

See the pattern; we get 4 in units place when power is odd and 6 in units place when power is even.

Since 102 is even, 264^102 has 6 in units place.

Now muliply 6*5 (units digit from 265)=30 or 0 in units digit.

What is the OA?
Join the discussion

by gdk800 » Fri Nov 19, 2010 2:51 pm
Thanks Rahul, this means that i can apply this method of even power and odd power applicable to only 9 because it has also got a cyclicity of 2?

How about other numbers such as 2, 3, 5?

Can you provide some more similar questions or probably a link on this topic for practice?
Join the discussion

by goyalsau » Fri Nov 19, 2010 8:07 pm
gdk800 wrote:Pls help with the following problem.

Unit digit of (264)^103 + (264)^102 = ?

I would like to do some change, (264)^103 + (263)^102
What should be the answer now,
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Join the discussion

by Rahul@gurome » Fri Nov 19, 2010 8:31 pm
I would like to do some change, (264)^103 + (263)^102
What should be the answer now,


264^103 ends in 4 because 103 is odd.

263^102 will end in 3^102.
Now 102 = 4*25 + 2.
So 3^102 will end in 3^2 which is 9.

4+9 = 13 ends in 3.
So 264^103 + 263^102 ends in 3.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Join the discussion