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divisibility

Expert replies
by Gurpinder » Mon Jun 20, 2011 8:03 am
please explain this one
Attachments
divisibility5-a.png
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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Source: — Data Sufficiency |

by Frankenstein » Mon Jun 20, 2011 8:11 am
Hi,
k/6 + m/4 =t/12.
So, t = 2k+3m
From(1):k is a multiple of 3.
So, k = 3p where p is an integer.
So, t = 2(3p)+3m = 3(2p+m)
So, t is a multiple of 3. So, t and 12 have a common factor of 3.
Sufficient
From(2):m is a multiple of 3
if k=1 and m = 3, then t=11. common factor of t and 12 is 1
if k=1 and m = 6, then t=20. common factor of t and 12 is 4
Not sufficient to tell whether the common factor is 1 or greater than 1

Hence, A
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Things are not what they appear to be... nor are they otherwise
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by Gurpinder » Mon Jun 20, 2011 8:33 am
thanks buddy!


to make a note...

y = any number.

m and p are factors of y

so can i make a note that: factor m + or - factor p = x

and that as long as M and P are factors of Y, then x and Y will aways have factors greater than 1.
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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by Frankenstein » Mon Jun 20, 2011 8:56 am
Hi,
I don't understand your question precisely.
If you meant to say m and p are factors of y
and m + p =x, then x and y will always have factors more than 1.. this is incorrect
Consider y=12
let m=3, p=4
x = m+p = 7
x(7) and y(12) have only '1' as common factor.
Cheers!

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by Gurpinder » Mon Jun 20, 2011 9:20 am
yeah that is what i meant....but i guess that wrong....

so whats the generalization from the question?

or did you simply plugin numbers?
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
Join the discussion