Author 
Message 
CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

LCM (m06q26) [#permalink]
Show Tags
12 Nov 2007, 09:56
This topic is locked. If you want to discuss this question please repost it in the respective forum. If \(m\) is a positive integer, is \(\sqrt{m} \gt 25\) ? 1. \(m\) is divisible by 50 2. \(m\) is divisible by 52 Source: GMAT Club Tests  hardest GMAT questions Please explain each step of the answer.



Senior Manager
Joined: 01 Sep 2006
Posts: 301
Location: Phoenix, AZ, USA

5
This post received KUDOS
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



Director
Joined: 25 Oct 2006
Posts: 635

2
This post received KUDOS
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



Director
Joined: 09 Aug 2006
Posts: 755

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.



SVP
Joined: 29 Aug 2007
Posts: 2473

Re: LCM (m06q26) [#permalink]
Show Tags
19 Dec 2009, 00:00
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.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



Intern
Joined: 23 Jul 2010
Posts: 17
Location: New York
Schools: Booth, Columbia, Ross, Kellogg

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 05:32
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.



Manager
Joined: 03 Aug 2010
Posts: 105
GMAT Date: 08082011

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 06:06
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.



Manager
Status: Waiting to hear from University of Texas at Austin
Joined: 24 May 2010
Posts: 76
Location: Changchun, China
Schools: University of Texas at Austin, Michigan State

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 06:20
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)



Manager
Joined: 03 Aug 2010
Posts: 105
GMAT Date: 08082011

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 06:33
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



Manager
Joined: 01 Apr 2010
Posts: 164

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 07:38
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.



Manager
Status: Waiting to hear from University of Texas at Austin
Joined: 24 May 2010
Posts: 76
Location: Changchun, China
Schools: University of Texas at Austin, Michigan State

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 07:42
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 \(\sqrt{m} \gt 25\) ?



Intern
Joined: 31 Jul 2010
Posts: 5

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



Manager
Joined: 03 Aug 2010
Posts: 105
GMAT Date: 08082011

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 08:34
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.



Intern
Joined: 09 Jun 2010
Posts: 4

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 09:51
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?



Math Expert
Joined: 02 Sep 2009
Posts: 39662

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 11:10
8
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 01 Apr 2010
Posts: 164

Re: LCM (m06q26) [#permalink]
Show Tags
04 Aug 2010, 19:44
Thanx Brunel for the detailed explanation with examples!!!!



Manager
Joined: 14 Mar 2011
Posts: 83

Re: LCM (m06q26) [#permalink]
Show Tags
17 Aug 2011, 02:44
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 Bingo C



Director
Status: Final Countdown
Joined: 17 Mar 2010
Posts: 537
Location: India
GPA: 3.82
WE: Account Management (Retail Banking)

Re: LCM (m06q26) [#permalink]
Show Tags
09 Aug 2012, 05:36
Thank you Bunuel Kudos +1 for you.
_________________
" Make more efforts " Press Kudos if you liked my post



Manager
Joined: 18 Jan 2012
Posts: 51
Location: United States

Re: LCM (m06q26) [#permalink]
Show Tags
09 Aug 2012, 11:01
2
This post received KUDOS
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
 IT TAKES TREMENDOUS EFFORT TO POST DETAILED ANSWERS. YOUR KUDOS IS VERY MUCH APPRECIATED.



Manager
Joined: 14 Jun 2012
Posts: 65

Re: LCM (m06q26) [#permalink]
Show Tags
29 Aug 2012, 08:46
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 squareroot of each is greater than 25. Hence I chose C.
_________________
My attempt to capture my BSchool Journey in a Blog : tranquilnomadgmat.blogspot.com
There are no shortcuts to any place worth going.







Go to page
1 2
Next
[ 22 posts ]




