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LCM (m06q26)

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LCM (m06q26) [#permalink] New post 12 Nov 2007, 08:56
If m is a positive integer, is \sqrt{m} \gt 25 ?

1. m is divisible by 50
2. m is divisible by 52

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C

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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 10:10
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Is \sqrt{m}>25? Or is m>625?

(1) m is divisible by 50 --> m=50p, where p is a positive integer --> the least value of m is 50 (as m is positive). Max value of m is not limited. Not sufficient.
(2) m is divisible by 52 --> m=52q, where q is a positive integer --> the least value of m is 52 (as m is positive). Max value of m is not limited. Not sufficient.

(1)+(2) The least value of m would be LCM of the least values from (1) and (2): 50=2*5^2 and 52=2^2*13 --> m_{min}=LCM(50,52)=2^2*5^2*13=1300>625. Sufficient.

Answer: C.

srivicool wrote:
Hi YalePhd,

Thanks for the explanation. but i am not clear..

square root of a positive integer can be +ve or -ve. for eg. sq.rt of 4 = +2, -2.
so sqrt of m can be +ve or -ve..

am i missing something.

thanx in advance.


aero07 wrote:
I guess I'm confused as to why we know the sqrt(M) is positive. M is positive as stated in the stem but how does that mean that the sqrt(M) is positive. If M were 4 wouldn't the sqrt(M) be -2 or 2? The stem only specifies that M be positive, not the sqrt(M). What am I not seeing?


GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root.

That is, \sqrt{25}=5, NOT +5 or -5. In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \sqrt[3]{125} =5 and \sqrt[3]{-64} =-4.

Some other notes on this issue:
\sqrt[{even}]{positive}=positive - \sqrt{25}=5. Even roots have only a positive value on the GMAT.

\sqrt[{even}]{negative}=undefined - \sqrt{-25}=undefined. Even roots from negative number is undefined on the GMAT.

\sqrt[{odd}]{positive}=positive and \sqrt[{odd}]{negative}=negative - \sqrt[3]{125} =5 and \sqrt[3]{-64} =-4. Odd roots will have the same sign as the base of the root.

For our question we have: m_{min}=1300, question is \sqrt{m}>25? As \sqrt{1300} has only positive value, then \sqrt{m_{min}}=\sqrt{1300}\approx{36}>25.

For more please check Number Theory chapter of Math Book (link in my signature).

Hope it's clear.
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Re: LCM [#permalink] New post 12 Nov 2007, 09:10
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bmwhype2 wrote:
If M is a positive integer, is sqrt(M) > 25?

1. M is divisible by 50
2. M is divisible by 52

Please explain each step of the answer.


Using 1 M could be 50,100,150 etc
choose lowest i.e. 50 root 50 is 5*root(2) i.e less than 25
choose 2500 then root 2500 is 25 so eaual
choose 75000 the n root 7500 is greater than 25

insuff


Same logic for 2. u can prove it could be les equal ot more than 25

togather find LCM of 50 and 52

2*26*25 = M

The root will be greater than 25 beacuse root of 25 is 5 and root of 26 is greater than 5


Answer C
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[#permalink] New post 12 Nov 2007, 09:45
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A) M/50 = divisible; So M can be 50 or 900. sqrt(M) can or can’t be >25
B) M/52= divisible; So M can be 52 or 2704. sqrt(M) can or can’t be >25
Applying both A and B: 50 = 2 * 5 * 5; and 52 = 2 * 2 * 13
So Lowest value of M is: 5 * 5 * 2 * 2 * 13 to be divisible by 50 and 52
Hence Sqrt(M) = 5 * 2 * 3.61 = greater than 30
So, SUFF

Ans: C
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Re: LCM (m06q26) [#permalink] New post 09 Aug 2012, 10:01
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This question can be solved with a little bit of number picking and analysis.
I believe everybody can easily come to the conclusion that each statement on its own is not sufficient.
Lets combine both statements

In Data sufficiency questions, we need to look at the question stem to try to rephrase the question to something that is easier to understand.
Its like, Ok, i understand what you are asking .. But what are you REALLY asking ?
Let me deviate a bit. Here is a conversation between between a guy and a gal on a first time date.

Gal : Where do you work and how long have you been working ? (Real Question : Give me a good estimate of how much you make ?)
Guy : How many boy friends have you had ? ( Real Question : Do you remain committed in relationships ? )


Sorry for puns, but i hope that i have been able to drive home the point that, when trying to solve GMAT questions, we need to peel back the layers and
ask, -OK guyz, what do you REALLY want to know ?

Lets rephrase the question a lil bit
IS SQRT (M) > 25

This can be rephrased to IS M > 625 ? ( Squaring both sides)

A) M is divisible by 50
What does this really mean ?
We need to identify the prime factors and their respective powers
M is divisible by 50, implies M is divisible by all prime factors of 50.
50 can be factorized as 5 x 5 x 2
This implies that M contains in its factors ATLEAST 2 FIVES and 1 TWO

A) M is divisible by 52
Following the same line of thinking that we followed above
52 = 13 x 2 x 2
This implies that M contains in its factors ATLEAST 1 THIRTEEN and 2 TWOs

Lets combine these statements. When combining these statements, lets ONLY focus on the highest powers of the prime factors.

50 = 5 x 5 x 2
52 = 13 x 2 x 2

We can establish these factors conclusively
a) M has as its factor AT LEAST 1 instance of 13
b) M has as its factors AT LEAST 2 instances of 5
c) M has as its factors AT LEAST 2 instances of 2
Hence M = K * 13 x 5 x 5 x 2 x 2 ( Where K is some integer .)
Hence M = K * 1300

Hence M is a multiple of 1300 and any multiple of 1300 is always greater than 625.
Hence We can conclusively establish that regardless of the value of K, M > 625

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Re: LCM [#permalink] New post 12 Nov 2007, 09:51
bmwhype2 wrote:
If M is a positive integer, is sqrt(M) > 25?

1. M is divisible by 50
2. M is divisible by 52

Please explain each step of the answer.


C.

Stat 1: M could be 50 or a much bigger number whose sqrt is greater than 25. Insuff.

Stat 2: M could be 52 or a much bigger number whose sqrt is greater than 25. Insuff.

Stat 1 & 2: The least value that M can take is 1300 (LCM of 50, 52). Therefore, sqrt (M) has to be greater than 25 the answer is a definite yes. Suff.
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Re: LCM (m06q26) [#permalink] New post 18 Dec 2009, 23:00
raghavsp wrote:
Isn't the answer E?
Since
Sqrt(m) where m is a positive number could be positive or negative number.
so it is insufficient to say Sqrt(m) > 25
Am I missing something here....


M has two roots:
1) Sqrt(m), which is always +ve.
2) -Sqrt(m), which is always -ve.
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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 04:32
Ans : C

1) M can be 50, 100, 150....
2) M can be 52, 104, 156 ...

Using both..

LCM of 52, 50 = 1300

Question :
Is sqrt(m) > 25
=> m > 625..

As LCM is 1300 , we get the one answer .

so Condition is sufficient when both the statements taken together.
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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 05:06
raghavsp wrote:
Isn't the answer E?
Since
Sqrt(m) where m is a positive number could be positive or negative number.
so it is insufficient to say Sqrt(m) > 25
Am I missing something here....


Remember that you cannot take the squre root of a negative number on the GMAT. Also, the stem reports that M is a positive interger.
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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 05:20
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)
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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 05:33
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)


You can use the prime factorization method to find the LCM for 50 and 52.
Prime factorization for 50 = 5, 5, 2 or 5^2 and 2
Prime factorization for 52 = 2, 2, 13 or 2^2 and 13.
Multiply the common factors using the one with the larger exponent and the factors unique to each number.
5^2 * 2^2 * 13 = 1300
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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 06:38
Hi YalePhd,

Thanks for the explanation. but i am not clear..

square root of a positive integer can be +ve or -ve. for eg. sq.rt of 4 = +2, -2.
so sqrt of m can be +ve or -ve..

am i missing something.

thanx in advance.
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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 06:42
srivicool wrote:
Hi YalePhd,

Thanks for the explanation. but i am not clear..

square root of a positive integer can be +ve or -ve. for eg. sq.rt of 4 = +2, -2.
so sqrt of m can be +ve or -ve..

am i missing something.

thanx in advance.


The stem states that m is positive.

bmwhype2 wrote:
If m is a positive integer, is \sqrt{m} \gt 25 ?
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Re: LCM [#permalink] New post 04 Aug 2010, 06:47
Damager wrote:
bmwhype2 wrote:
If M is a positive integer, is sqrt(M) > 25?

1. M is divisible by 50
2. M is divisible by 52

Please explain each step of the answer.


Using 1 M could be 50,100,150 etc
choose lowest i.e. 50 root 50 is 5*root(2) i.e less than 25
choose 2500 then root 2500 is 25 so eaual
choose 75000 the n root 7500 is greater than 25

insuff


Same logic for 2. u can prove it could be les equal ot more than 25

togather find LCM of 50 and 52

2*26*25 = M

The root will be greater than 25 beacuse root of 25 is 5 and root of 26 is greater than 5


Answer C


I don't understand how M could be equal to 100 or 150 based on the information we are given in the first clue. The prime factorization only gives us 2*5*5 which at most equals 50.

The second clue gives us the prime factorization of 2*2*13. Using both clues we have a prime factorization of 2*2*5*5*13 which equals 1300.

So I do understand why the answer is C but your response to clue 1 though me off completely. Please xplain if you don't mind. Thanks, :) H
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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 07:34
TallJTinChina wrote:
srivicool wrote:
Hi YalePhd,

Thanks for the explanation. but i am not clear..

square root of a positive integer can be +ve or -ve. for eg. sq.rt of 4 = +2, -2.
so sqrt of m can be +ve or -ve..

am i missing something.

thanx in advance.


The stem states that m is positive.

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


I think you might be thinking about square and square root in the same way. Yes, -2 squared and 2 squared are both 4, but you can't take the square root of -4 on the GMAT, as the GMAT only deals with real numbers.

Think of it this way, what number multiplied by itself would result in a product of -4? None.

Also, don't forget that the stem explicitly states that M is a positive integer. With that, you can rule out all negative numbers.

N.B. the square root of -4 is the imaginary number 2i. But you will never work with imaginary numbers on the GMAT.
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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 08:51
I guess I'm confused as to why we know the sqrt(M) is positive. M is positive as stated in the stem but how does that mean that the sqrt(M) is positive. If M were 4 wouldn't the sqrt(M) be -2 or 2? The stem only specifies that M be positive, not the sqrt(M). What am I not seeing?
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Re: LCM (m06q26) [#permalink] New post 04 Aug 2010, 18:44
Thanx Brunel for the detailed explanation with examples!!!!
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Re: LCM (m06q26) [#permalink] New post 17 Aug 2011, 01:44
Clearly its C
we know sqrt M >25 means M >625
1. Insufficient because it can any multiple of 50
2. Insufficient because it again any multiple of 52
Now combining both M= 2X 25X 2 X 13 which greater than 50*26 = 1300 assuming 2 is common in 50 & 52
Sufficient
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Re: LCM (m06q26) [#permalink] New post 09 Aug 2012, 04:36
Thank you Bunuel Kudos +1 for you.
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Re: LCM (m06q26) [#permalink] New post 29 Aug 2012, 07:46
Came down to choosing between C and E.

Considering both together, the possible values for "m" by plugging in values (multiples of both 50 and 52) come down to 5200, 10400, etc, The square-root of each is greater than 25. Hence I chose C.
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Re: LCM (m06q26)   [#permalink] 29 Aug 2012, 07:46
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