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LCM (m06q26)

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Intern
Intern
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Joined: 12 Feb 2014
Posts: 7
Location: India
Concentration: General Management, International Business
GMAT 1: 660 Q47 V35
GPA: 3.1
WE: Consulting (Computer Software)
Followers: 0

Kudos [?]: 0 [0], given: 10

Re: LCM (m06q26) [#permalink] New post 25 Jul 2014, 04:26
The best way to solve this question is to simply explore the statement given in the question and modify it according to the answer options.
Condition: m>0 and m is integer
\sqrt{m}>25.

From this, we can deduce that m>625.

St.1: m is divisible by 50.
i.e., 50 is a multiple of m. m can be anything from 50, 100, 150 to 600, 650, 700 and so on.
Since there are answers which don't give a unique case that satisfies m>625, A is rejected.

St.2: m is divisible by 52.
Extend the approach used in st.1 here too. That is, multiples of 52 (potential values of m) can be <625 as well as >625. Hence, B is rejected too.

St.1 and St.2: m is divisible by 50 AND 52.
Since none of the common multiples (importantly, even the LCM) are greater than 625, i.e. \sqrt{25},
we have a unique answer to the question( yes, sqrt(m) >25. Hence, the answer is C.

Hope this helps.







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LCM (m06q26) [#permalink] New post 25 Jul 2014, 07:46
TallJTinChina wrote:
I make similar mistakes to raghavsp

I miss things in the first part of the question (I think it is officially called the stem) after I get to working at the problem. Recently I have started rewriting the sentences in inequalities.

So we take a look at the first part of the question.

bmwhype2 wrote:
If m is a positive integer, is \sqrt{m} \gt 25 ?


Because it is written "....m is a positive integer..."
I write out "m>0" on my whiteboard

Then we continue on.

bmwhype2 wrote:
1. m is divisible by 50


Based on (1) m is the set
M {0,50,100,150.....n50}

It is key to remember that divisible means when m is divided by 50 there is no remainder. So M could equal 0. Except in this situation the stem tells us M>0.

bmwhype2 wrote:
2. m is divisible by 52


As others have said (2) alone is not enough.

I took some time to calculate the smallest common denominator of \frac{m}{50} and \frac{m}{52}. I came up with 1300.

Does anyone know a quick way to do this? I had to do trial and error until I found 1300 (25x52)




1. M=50k1
2. M=52k2
Combining 1 and 2
50k1=50k2 or
k1/k2 =26/25;
since k1 and k2 are both positive integers, min value the can take are 26a nd 25 respectively
so M can be expressed as 25*26n, where ncan be any natural number.
Clearly, sqrt(M)>25………..ans(C)
LCM (m06q26)   [#permalink] 25 Jul 2014, 07:46
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