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 →

Q134 in OG 11th Edition

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 22
Joined: Mon Jun 28, 2010 6:03 am
Thanked: 1 times
GMAT Score:680
by globalcitizen » Sat Jul 10, 2010 5:27 pm
If m and n are positive integers, is sqrt(n-m) an integer?
1) n>m+15
2) n=m(m+1)

Answer is B

This is a fairly straightforward question however, I find that I am having a lot of issues with DS because I "overthink" things or assume too much. My question here is, can we assume that sqrt(m^2)=m means that m is an integer? I got this answer for statement 2 but thought that m^2 could still mean that m may be a fraction.

Unless otherwise stated, are we supposed to assume that sqrt(m^2)=m always means that m will be an integer? I hope I make sense here.

Any words of advice would be greatly appreciated.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 214
Joined: Mon Mar 29, 2010 1:46 pm
Location: Houston, TX
Thanked: 37 times
GMAT Score:700

by sk818020 » Sat Jul 10, 2010 6:13 pm
The answer states that both n and m are positive integers.

(1) n>m+15

This simply tells us that m>1 and n>16. n could be 30 and m could be 5 and the square of their difference would be an integer( sqrt(25)=5) or n could equal 31 and m could equal 5 and the square root of there difference could not be an integer (sqrt(26) is not an integer). So (1) in sufficient.

(2) n=m(m+1)

m(m+1)=m^2+m

So substituting n in n-m,

n-m = (m^2+m)-m = m^2+m-m = m^2

So,

sqrt(m^2)=m, and

m is an integer so, B sufficient.

Hope this helps.

Thanks,

Jared
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 1302
Joined: Mon Oct 19, 2009 2:13 pmFollowed by: 164 members
Location: Toronto
Thanked: 539 times
GMAT Score:800

by Testluv » Sat Jul 10, 2010 7:31 pm
Unless otherwise stated, are we supposed to assume that sqrt(m^2)=m always means that m will be an integer? I hope I make sense here.
sqrt(m^2) = m
The above equation is always true. One way of seeing this is to understand that sqrt(x) is the same as x^(1/2). So, sqrt(m^2) = (m^2)^1/2. When we have an exponent in the bracket and outside of the bracket, we multiply the exponents (powers rule). So, sqrt(m^2) = m ALWAYS.

But that doesn't necessarily mean that m is an integer value. We can conclude that sqrt(m^2) is an integer only if we already know that m itself is an integer.

As Jared points out, we are told that m and n are both integers. When we look at the statements, we must include this information in our analysis (we must never ignore information from the question stem itself).

(1) is clearly insufficient and if the algebra doesn't jump out to you, then the easiest way to see that (2) is sufficient is to pick some numbers, and to be organized in your scratchwork. Let's do that here:

2) n=m(m+1)

Let m = 1. Then n = 1*2 = 2. Is sqrt(2-1) an integer? Yes.

Let m = 2. Then, n = 2*3 =6. Is sqrt(6-2) an integer? Yes.

Let m = 3. Then, n = 3*4 = 12. Is sqrt(12-3) an integer? Yes.

By now we've convinced ourselves that (2) will always yield an integer value.
Kaplan Teacher in Toronto
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 22
Joined: Mon Jun 28, 2010 6:03 am
Thanked: 1 times
GMAT Score:680

by globalcitizen » Sun Jul 11, 2010 5:54 am
Thank you very much for your responses and for making me realize that I completely overlooked the information given in the question stem! This is something I find myself doing all too frequently. I'm going to try talking out loud in my head (if that makes any sense) to help me better focus...

Thanks again for the help!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1