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

Redeem

What should be the OA for this one?

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Sun Nov 21, 2010 9:25 pm

What should be the OA for this one?

by vaibhavob » Sun Jan 23, 2011 3:54 am
If X and Y are positive integers and n = 5^X and 7^(y+15), what is the units digit of n?
1.y = 2x - 15
2.y^2 - 6y + 5 = 0

The OA is B.

Can someone elaborate on how to solve this one? Seems like a simple problem of units digit using cyclicity?

Hope the question makes sense now.
Last edited by vaibhavob on Sun Jan 23, 2011 4:15 am, edited 1 time in total.
Top
Source: — Data Sufficiency |
Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 am
Thanked: 82 times
Followed by:9 members
GMAT Score:720

by maihuna » Sun Jan 23, 2011 4:05 am
Q is not clear, what do u mean by n = 5^X and 7^*(y+15)?

ANy power of 5 ends in 5 so no value of x required

y^2 - 6y + 5 = 0 =>(y-1)(y-5)=0=>y=1 o r5

7^16 ends in (7^4)^4 = 1^4 = 1 (7,9,3,1 is cyclicitty of power of 7) in 1.
7^20 ends in (7^4)^5 = 1^5 = 1

so B is fine even though it gives 2 values as both ends in 1
Charged up again to beat the beast :)
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Sun Jan 23, 2011 4:05 am
If X and Y are positive integers and n = 5^X and 7^*(y+15), what is the units digit of n?
The red part doesn't makes sense.
Can you check that part and edit accordingly?
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Sun Nov 21, 2010 9:25 pm

by vaibhavob » Sun Jan 23, 2011 4:19 am
maihuna wrote:Q is not clear, what do u mean by n = 5^X and 7^*(y+15)?

ANy power of 5 ends in 5 so no value of x required

y^2 - 6y + 5 = 0 =>(y-1)(y-5)=0=>y=1 o r5

7^16 ends in (7^4)^4 = 1^4 = 1 (7,9,3,1 is cyclicitty of power of 7) in 1.
7^20 ends in (7^4)^5 = 1^5 = 1

so B is fine even though it gives 2 values as both ends in 1
Thanks...you understood the question correctly.. i figured out where i was making the mistake.
Top
Post new topic Post Reply
• Page 1 of 1