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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???
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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
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by HPengineer » Tue Dec 07, 2010 2:45 am
you are correct night reader... but im missing your logic..
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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.
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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.
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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.
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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..
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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.
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by HPengineer » Tue Dec 07, 2010 3:16 am
there we go thanks mitch...
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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.
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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
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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
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