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Q134 in OG 11th Edition

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Q134 in OG 11th Edition

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.
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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
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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.
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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!
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