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 →

Perm / Combi

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Senior | Next Rank: 100 Posts
Posts: 32
Joined: Tue Apr 20, 2010 7:16 pm
Location: Newark, DE
by govind_raj_76 » Sun May 02, 2010 11:32 am
How many 4 digit numbers that are divisible by 2 can be formed so that only the two rightmost digits are even and the digit 0 cannnot appear more than once ?

A) 300

B) 400

C) 500

D) 600

E) 700

Please explain the best approach to solve this problem ?

Thanks,
Govind
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Legendary Member
Posts: 759
Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members
Thanked: 85 times

by clock60 » Sun May 02, 2010 11:58 am
govind_raj_76 wrote:How many 4 digit numbers that are divisible by 2 can be formed so that only the two rightmost digits are even and the digit 0 cannnot appear more than once ?

A) 300

B) 400

C) 500

D) 600

E) 700

Please explain the best approach to solve this problem ?

Thanks,
i got D-600, reasoning
if i got the problem right the desired number must look like
odd odd even even and 0 used only once
for me all numbers will end 0,2,4,6,8

odd odd even 0=5*5*4*1=100 (4 -as we chose from 2,4,6,8 and 0 is already chosen)
odd odd even 2=5*5*5*1=125 ( here we have 0,2,4,6,8)
odd odd even 4=5^3=125
odd odd even 6=125
odd odd even 8=125

total=100+125*4=600
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 170
Joined: Wed Jun 10, 2009 5:59 am
Thanked: 13 times

by iamseer » Sun May 02, 2010 12:23 pm
IMO answer D

_ _ _ _

5*5*(4*1 +1*4+4*4) =600

Now, since only last 2 digits are even, the first 2 digits are odd.
First digit can be selected in 5 ways (1,3,5,7,9)
Second digit can be selected in 5 ways ((1,3,5,7,9)

Since 0 is not to be repeated, it can be at ten's place, unit's place or nowhere

0 in unit's place - Ten's place can be selected in 4 ways (2,4,6,8)
0 in ten's place - Unit's place can be selected in 4 ways (2,4,6,8)
0 nowhere - Unit's place and Ten's place can both be selected in 4 ways each

HTH
"Choose to chance the rapids and dance the tides"
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 32
Joined: Sun Aug 10, 2008 5:36 am
Thanked: 1 times

by susantaiitk » Mon May 03, 2010 7:45 am
first 2 places are odd [ 1,3,5,7,9] total ways = 5*5 = 25
last 2 places are even [ 0,2,4,6,8] total ways = 5*5 = 25

but 0 can't be more than once, hence last 2 digit 00 is avoided then total ways of last 2 places = 25 - 1= 24

total ways = 25*24 = 600
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 758
Joined: Sat Aug 29, 2009 9:32 pmFollowed by: 2 members
Location: Bangalore,India
Thanked: 67 times

by sumanr84 » Mon May 03, 2010 8:49 am
susantaiitk wrote:first 2 places are odd [ 1,3,5,7,9] total ways = 5*5 = 25
last 2 places are even [ 0,2,4,6,8] total ways = 5*5 = 25

but 0 can't be more than once, hence last 2 digit 00 is avoided then total ways of last 2 places = 25 - 1= 24

total ways = 25*24 = 600
Fastest...Simple..clear and best solution..;-)
I am on a break !!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 32
Joined: Tue Apr 20, 2010 7:16 pm
Location: Newark, DE

by govind_raj_76 » Thu May 06, 2010 6:31 pm
clock60 wrote:
govind_raj_76 wrote:How many 4 digit numbers that are divisible by 2 can be formed so that only the two rightmost digits are even and the digit 0 cannnot appear more than once ?

A) 300

B) 400

C) 500

D) 600

E) 700

Please explain the best approach to solve this problem ?

Thanks,
i got D-600, reasoning
if i got the problem right the desired number must look like
odd odd even even and 0 used only once
for me all numbers will end 0,2,4,6,8

odd odd even 0=5*5*4*1=100 (4 -as we chose from 2,4,6,8 and 0 is already chosen)
odd odd even 2=5*5*5*1=125 ( here we have 0,2,4,6,8)
odd odd even 4=5^3=125
odd odd even 6=125
odd odd even 8=125

total=100+125*4=600
Nice Explaination !!
Govind
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 385
Joined: Sun Jul 12, 2009 10:16 pmFollowed by: 2 members
Thanked: 29 times
GMAT Score:710

by debmalya_dutta » Thu May 06, 2010 7:39 pm
Is the question missing the portion which applies the constraint that the 2 left most digits should be odd digits ? Strictly based on the question , the 1000 and the 100's place can have both even and odd digits.. why have you guys considered only 5 ways
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1