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When the positive integer n is divided by 15...

by The Jock » Mon Jul 12, 2010 3:29 am
When the positive integer n is divided by 15, the remainder is 4.
What is the value of n?
(1) When n is divided by 10, the remainder is 9
(2) n < 50

I used this approach:
n=15Q+4
from option 1
n=10Q+9
from above given equations...
Q =1 and n = 19
So my question is "isn't it enough to say that option 1 is sufficient"

Thanks in advance...
Thanks and Regards,
Varun
https://mbayogi.wordpress.com/
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by 4GMAT_Mumbai » Mon Jul 12, 2010 3:40 am
So my question is "isn't it enough to say that option 1 is sufficient"
Hi,

I am afraid that is not enough ... The question that you have to ask yourself is

"Does this enable me to give one single value as an answer to the question 'What is the value of n?'?"

If you ask yourself this question, you will realize that the answer is a NO.

When Q = 1, n = 19; when Q = 3, n = 49; so on and so forth.

The moment you get a second value of 'n', you know that the statement is not valid.

Try using this method for the second statement ...

Hope this helps. Thanks.
Naveenan Ramachandran
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by gmatmachoman » Mon Jul 12, 2010 3:42 am
The Jock wrote:When the positive integer n is divided by 15, the remainder is 4.
What is the value of n?
(1) When n is divided by 10, the remainder is 9
(2) n < 50

I used this approach:
n=15Q+4
from option 1
n=10Q+9
from above given equations...
Q =1 and n = 19
So my question is "isn't it enough to say that option 1 is sufficient"

Thanks in advance...
Varun, i see a glitch in ur method:

Your method:
n=15Q+4 ;

from option 1
n=10Q+9

My query is How did u assume both wil be Q???


It should be like:

n = 15y +4

n = 10K +9

Now equate for "n"

15y +4 = 10K +9

15 y = 10K +5

3y = 2k +1

This will be satisfied only when y = 1;K =1

So by that we find n = 19.

St 1 is sufficient

Pick A.

This question comes under the category " Remainders with one equation to find "n""
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by 4GMAT_Mumbai » Mon Jul 12, 2010 3:59 am
gmatmachoman,

Beg to differ.

Is 1 sufficient ?!?!? n could be one of 19, 49, 79, 109 ...
Naveenan Ramachandran
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by gmatmachoman » Mon Jul 12, 2010 4:09 am
4GMAT_Mumbai wrote:gmatmachoman,

Beg to differ.

Is 1 sufficient ?!?!? n could be one of 19, 49, 79, 109 ...

So naveen Ji, how to solve this kind of scenario?
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by amising6 » Mon Jul 12, 2010 4:12 am
The Jock wrote:When the positive integer n is divided by 15, the remainder is 4.
What is the value of n?
(1) When n is divided by 10, the remainder is 9
(2) n < 50
...
our number is 15n+4
1)When n is divided by 10, the remainder is 9
so number will be of form 10n+9
so first number satisfying this will be 19
lcm of 15,10 is 30 keep on addding 30 to 19 and you will get consecutive numbers
so n can be 19,49 or 79 0r 0r 109,139

2)n < 50
it doesnt say anything n can be anything like 19,49
not sufficient

combining again we are can see n can have 2 values so E
Ideation without execution is delusion
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by gmatmachoman » Mon Jul 12, 2010 4:14 am
My bad, i should have used that LCM concept!! Yuck!
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