BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

800 score.com question..

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 165
Joined: Wed Mar 24, 2010 8:02 am
Thanked: 2 times
Followed by:1 members

800 score.com question..

by HPengineer » Tue Dec 07, 2010 2:10 am
The least common multiple of positive integer m and 3 digit integer n is 690. If n is not divisible by 3 and m is not divisible by 2 what is the value of n?

a.) 115
b.) 230
c.) 460
d.) 575
E.) 690

i always have trouble with these types of problems.. For me first thing i did was break down 690 into prime factors

2x3x5x23

from there i think about what is LCM... well its the highest order of each prime factors of the two numbers... I then attack the statement that N is not divisible by three... so i know that 3 came from M so when im calculating N i do not include three... here is where i get stuck...

im left with a 5 & 2 & 23 well it tells me that M is not divisible by 2 so i know that the 2 had to come from N... but what about the 5 and the 23?? couldnt they have come from either M or N???
Top
Source: — Problem Solving |
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pm
Thanked: 127 times
Followed by:10 members

by Night reader » Tue Dec 07, 2010 2:40 am
HPengineer wrote:The least common multiple of positive integer m and 3 digit integer n is 690. If n is not divisible by 3 and m is not divisible by 2 what is the value of n?

a.) 115
b.) 230
c.) 460
d.) 575
E.) 690

i always have trouble with these types of problems.. For me first thing i did was break down 690 into prime factors

2x3x5x23

from there i think about what is LCM... well its the highest order of each prime factors of the two numbers... I then attack the statement that N is not divisible by three... so i know that 3 came from M so when im calculating N i do not include three... here is where i get stuck...

im left with a 5 & 2 & 23 well it tells me that M is not divisible by 2 so i know that the 2 had to come from N... but what about the 5 and the 23?? couldnt they have come from either M or N???
IOM B 230

cancell E, sum of integers/3

test A, C, D => remainder

B => no remainder

multiple 690, factor 230
Top
Master | Next Rank: 500 Posts
Posts: 165
Joined: Wed Mar 24, 2010 8:02 am
Thanked: 2 times
Followed by:1 members

by HPengineer » Tue Dec 07, 2010 2:45 am
you are correct night reader... but im missing your logic..
Top
Master | Next Rank: 500 Posts
Posts: 165
Joined: Wed Mar 24, 2010 8:02 am
Thanked: 2 times
Followed by:1 members

by HPengineer » Tue Dec 07, 2010 2:48 am
ok the key to this problem was that stem mentioned N as three digit number...
Last edited by HPengineer on Tue Dec 07, 2010 2:55 am, edited 1 time in total.
Top
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Tue Dec 07, 2010 2:48 am
HPengineer wrote:The least common multiple of positive integer m and 3 digit integer n is 690. If n is not divisible by 3 and m is not divisible by 2 what is the value of n?

a.) 115
b.) 230
c.) 460
d.) 575
E.) 690

i always have trouble with these types of problems.. For me first thing i did was break down 690 into prime factors

2x3x5x23

from there i think about what is LCM... well its the highest order of each prime factors of the two numbers... I then attack the statement that N is not divisible by three... so i know that 3 came from M so when im calculating N i do not include three... here is where i get stuck...

im left with a 5 & 2 & 23 well it tells me that M is not divisible by 2 so i know that the 2 had to come from N... but what about the 5 and the 23?? couldnt they have come from either M or N???
690 = 2 * 3 * 5 * 23

AS M is not a multiple of 2, so N must contain 2, & N is not a multiple of 3 so M must contain 3

We know that M is multiple of 3 for sure and N is a multiple of 2 for sure,

Now it is also said that N is a Three digit Number,

Total Pairs Can be
{N = 2 * 5 * 23 , 2 * 23, 2 * 5 }

If we take the last two values it will be 2 digit Number so it has to be 2 * 5 * 23 = 230,

It does not matter Whether M does contain 5 , 23 Because the LCM in that case will still be 690.
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Tue May 11, 2010 5:46 am
Thanked: 3 times

by abhishekg21 » Tue Dec 07, 2010 3:12 am
goyalsau i thin k you forgot to consider value 115=23*5 as n.
though your final solution is still fine but we will have to consider cases of 115 so that we can neglect this value.
Top
Master | Next Rank: 500 Posts
Posts: 165
Joined: Wed Mar 24, 2010 8:02 am
Thanked: 2 times
Followed by:1 members

by HPengineer » Tue Dec 07, 2010 3:15 am
115 was my orignal answer... now im confused again.. since both 115 and 230 meet the requirement of 3 digit number and do not have a two.. how to determine which one to select as answer..
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Tue Dec 07, 2010 3:15 am
HPengineer wrote:The least common multiple of positive integer m and 3 digit integer n is 690. If n is not divisible by 3 and m is not divisible by 2 what is the value of n?

a.) 115
b.) 230
c.) 460
d.) 575
E.) 690
690 = 2*3*5*23

Now let's plug in the answer choices, which represent the value of n.

Answer choice A: n=115
If n = 115 = 5*23, then m = 2*3 = 6.
Doesn't work, because m cannot be divisible by 2. Eliminate A.

Answer choice B: n = 230.
If n = 230 = 2*5*23, then m = 3.
n is not divisible by 3, and m is not divisible by 2. Success!

The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Master | Next Rank: 500 Posts
Posts: 165
Joined: Wed Mar 24, 2010 8:02 am
Thanked: 2 times
Followed by:1 members

by HPengineer » Tue Dec 07, 2010 3:16 am
there we go thanks mitch...
Top
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Tue May 11, 2010 5:46 am
Thanked: 3 times

by abhishekg21 » Tue Dec 07, 2010 3:25 am
answer to your question "115 was my orignal answer... now im confused again.. since both 115 and 230 meet the requirement of 3 digit number and do not have a two.. how to determine which one to select as answer.."

if you are taking n as 115 then possible values of m can be 345,115 or 69

for all the possible values of m you can see that LCM is not 690.so you can ignore this value and hence n is 230.
Top
Master | Next Rank: 500 Posts
Posts: 165
Joined: Wed Mar 24, 2010 8:02 am
Thanked: 2 times
Followed by:1 members

by HPengineer » Tue Dec 07, 2010 3:27 am
yup finally got it now.. i see why 115 does nto work cause it breaks the requirement for M... thank you all for the help i have a few more for you to try as well. :)

Cheers

M
Top
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Fri Nov 05, 2010 5:49 am

by febus » Tue Dec 07, 2010 11:06 am
The least common multiple of positive integer m and 3 digit integer n is 690. If n is not divisible by 3 and m is not divisible by 2 what is the value of n?

a.) 115
b.) 230
c.) 460
d.) 575
E.) 690
I think the best way to solve this is getting the prime factors: 2x3x5x23 and then concluding

c.) 460 --> not possible for multiple reasons such as more than 690/2
d.) 575 --> same
e.) 690 --> not possible as divisible by 3

For the answers left, we need to see, that every factor of 690 has to be found at least once in either n or m (and remember m cannot include 2)

a.) 115 --> dont find factor 2, not possible (no 2 in n or m means there is no way getting 690 as multiple)
b.) 230 --> only answer left

Correct me if I'm wrong,

cheers
Top
Post new topic Post Reply
• Page 1 of 1