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prime & divisibility

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prime & divisibility

by Gurpinder » Tue Aug 24, 2010 2:19 pm
If k, m, and t are positive integers and k/6+m/4=t/12 do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

if i rearrange the equation it works out to be 2k+3m=t

now stmt (1)

k is a multiple of 3, so that means that 2k has to be an EVEN multiple of 3 like 6,12,24

how is this statement sufficient when 2k could mean any of the above?

can someone also please explain: t and 12 have a common factor greater than 1 ?

Is this basically asking whether T is a multiple of 12 greater than 12????
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by thirst4edu » Tue Aug 24, 2010 6:20 pm
Gurpinder wrote:If k, m, and t are positive integers and k/6+m/4=t/12 do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

if i rearrange the equation it works out to be 2k+3m=t

now stmt (1)

k is a multiple of 3, so that means that 2k has to be an EVEN multiple of 3 like 6,12,24

how is this statement sufficient when 2k could mean any of the above?

can someone also please explain: t and 12 have a common factor greater than 1 ?

Is this basically asking whether T is a multiple of 12 greater than 12????
This is how I would interpret this - factors of 12 greater than 1 are 2, 3, 4, 6, 12.

In order for t to have *a* common factor with 12, t should at least have one of these (2, 3, 4, 6, 12).. Only statement 1 proves it has 3, which is one of the common factor with 12, hence sufficient.
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by Gurpinder » Tue Aug 24, 2010 6:49 pm
Alright now I get how A is sufficient. How is B not sufficient?

m is a multiple of 3. so 3m means that the number is a multiple of 9. so lets just take 9 for example, its factors are 3 and this is common with 12.

in fact any multiple of 9 will always be divisible by 3, so why isnt statement 2 sufficient.
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by sk818020 » Tue Aug 24, 2010 7:19 pm
This is a number properties question.

As noted above the equation simplifies to 2k+3m=t

One important number properties rule is;

A multiple of n plus a multiple of n will sum to another multiple of n. Algebraic example, 3x+3y = 3(x+y)

1) k is a multiple of 3.

3m is obviously a multiple of 3 and if k is also a multiple of 3, then t is a multiple of 3, because of the rule stated above.

Thus, A is sufficient. 3 is a common factor.

2) Tell us that m is a multiple of 3 which does not tell us anything new in regards to t because the only way we can determine any other factor of t is if 2k and 3m share a factor, which this answer does not tell us.

2 is insufficient.

A is the answer.

Please confirm OA.

Thanks,

Jared
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by thirst4edu » Tue Aug 24, 2010 7:30 pm
Gurpinder wrote:Alright now I get how A is sufficient. How is B not sufficient?

m is a multiple of 3. so 3m means that the number is a multiple of 9. so lets just take 9 for example, its factors are 3 and this is common with 12.

in fact any multiple of 9 will always be divisible by 3, so why isnt statement 2 sufficient.
The equation in question is 2k + 3m = t, in your example, we should be able to factor out 3 from the sum (2k+3m), not just 3m.
To understand this, lets go back to equation 1, why it works..

1) k is multiple of 3,
so, we can say value of k is such that k=3 * x (where x is integer)

so equation becomes => 2(3*x) + 3m = t
this can be written as => 3(2x+m) = t
Now, here we know 3 is factor of t.

but in case of statement 2, saying m is multiple of 3 is not enough to factor out 3 from the whole equation.

Hope that helps!
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by Gurpinder » Tue Aug 24, 2010 7:36 pm
thirst4edu wrote:
Gurpinder wrote:Alright now I get how A is sufficient. How is B not sufficient?

m is a multiple of 3. so 3m means that the number is a multiple of 9. so lets just take 9 for example, its factors are 3 and this is common with 12.

in fact any multiple of 9 will always be divisible by 3, so why isnt statement 2 sufficient.
The equation in question is 2k + 3m = t, in your example, we should be able to factor out 3 from the sum (2k+3m), not just 3m.
To understand this, lets go back to equation 1, why it works..

1) k is multiple of 3,
so, we can say value of k is such that k=3 * x (where x is integer)

so equation becomes => 2(3*x) + 3m = t
this can be written as => 3(2x+m) = t
Now, here we know 3 is factor of t.

but in case of statement 2, saying m is multiple of 3 is not enough to factor out 3 from the whole equation.

Hope that helps!
Ahh silly me!

Thanks man!
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by thebigkats » Thu Jan 06, 2011 11:50 pm
we need to figure out if t has some common factor as 12 which can be only 2, 3

STAT #1
2k + 3m = t
if k is multiple of 3 then k = 3x (where x is something)
2.3x + 3m = t
3 (2x+m) = t
means that t is a multiple of 3, same as 12 --> SUFF

STAT #2
2k + 3m = t
if m is multiple of 3 then m = 3x (where x is something)
2k + 3.3x = t

==> we can't get either 2 or 3 as common factor from this equation --> INSUFF
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by niazsna786 » Wed May 18, 2011 3:37 am
i approach to this problem as
k/6 + m/4 = t/12
1.stmt k is a multiple of 3. which means k can be 3, 6, 9, 12 ...
by using the above values we reduce expr to
x/2 + m/4 = t/12 or x + m/4 = t/12 ( where x is output of k/6)
so independent of value of m we can say denominator will be max 4. so we can say t and 12 have more than one common factor
2 stmt m is multiple of 3 which means m can be 3, 6, 9 , 12..
k/6+ 3x/4 = t/12 here the denominator will depend on value of k .. so we can say 2 is not sufficient
answer is A
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by factor26 » Tue Jun 21, 2011 7:51 pm
4K + 6M = 2T

STATEMENT 1:K IS A MULTIPLE OF 3

PLUG IN TO SATISFY STATEMENT A (MUST BE MULTIPLE OF 3; 3,6,9 ...)

PUT RANDOM POSITIVE INTEGERS IN TO SOLVE (SUGGEST EVEN AND ODD)

K=3
M=3

4(3)+ 6(3) = 2T
T=15 (COMMON FACTOR WITH 12 ...3) YES

K=3
M=6

T= 18 (COMMON FACTOR WITH 12 ...3) YES

TRY TO KEEP PLUGGING IN #'S FOR STATEMENT 1, MAKE SURE TO SATISFY CRITERIA ... ANSWER IS ALWAYS YES

ANSWER CAN NOT BE BCE



STATEMENT 2: M IS A MULTIPLE OF 3

TAKE THE SAME APPROACH AS DESCRIBED IN STATEMENT 1

4K + 6M = 2T

K=3
M=3

12 + 18 = 2T
T=15 (TRY TO RECYCLE ALREADY USED NUMBERS) .... YES

K=4
M=3

T=14 NO

CANT HAVE A YES AND A NO IN A DS QUESTION ... MUST BE ONE OR THE OTHER. ELIMINATE B

THE ANSWER IS A [/u][/b]
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by prashant misra » Sun Oct 30, 2011 12:01 am
statement 1 is sufficient to answer the question.we can simplify the question as (2k+3m)/12=t/12 when k is a multiple of 3 then we have m also as a multiple of 3 thus t and 12 will have a common factor greater than 1.
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by Rastis » Thu Dec 01, 2011 12:05 pm
Can someone explain this better please?
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by GMATGuruNY » Thu Dec 01, 2011 12:35 pm
Gurpinder wrote:If k, m, and t are positive integers and k/6+m/4=t/12 do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.
To clear the fractions, multiple the given equation by 12:
12(k/6 + m/4) = 12/(t/12)
2k + 3m = t.

The factors of 12 greater than 1 are 2,3,4,6, and 12.
Question rephrased: Is t a multiple of 2,3,4,6 or 12?

Statement 1: k is a multiple of 3
Thus:
t = 2(multiple of 3) + 3m.
t = multiple of 3 + multiple of 3.
Since the sum of two multiples of 3 must also be a multiple of 3, t is a multiple of 3.
SUFFICIENT.

Statement 2: m is a multiple of 3
If k=3 and m=3, then t = 2(3) + 3(3) = 15.
In this case, t is a multiple of 3.
If k=1 and m=3, then t = 2(1) + 3(3) = 11.
In this case, t is not a multiple of 2,3,4,6, or 12.
INSUFFICIENT.

The correct answer is A.
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by ArunangsuSahu » Mon Jan 09, 2012 10:49 am
Let me simplify the problem

k/6+m/4=t/12
or 2k+3m=t---(i)

Statement 1

Let k =3n
Then from equation(1) 2*(3n)+3m=t
or 3*(2n+m)=t

so t and 12 has a common factor 3.. Hence SUFFICIENT

Statement 2:

Let m=3n

or 2k +3*3n=t

No definite conclusion about factors. So Insufficient
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by jayoptimist » Wed Jan 18, 2012 8:38 pm
My Method same as Arunangsu.
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by ronnie1985 » Fri Mar 30, 2012 6:31 am
Given: k,m,t>0 and {k,m,t} belongs to N
k/6 + m/4 = t/12
=> 2k + 3m = t

To find if HCF of 12 and t is greater than 1

S1: k = 3a
6a+3m = t. 3 is HCF. Sufficient.

S2: m = 3b
2k+9b = t. Not sufficient

(A) is answer.
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