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we will have to wait for the OA for this for one week

let us discuss it

here is my attempt at the problem

IMO A is the answer to this one

my line of reasoning is as follows

statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time

now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5

statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent

Accordingly, A is the answer for this

I hope i have taken the correct approach to solve the question and arrived at the correct answer

we will have to wait for the OA for this for one week

let us discuss it

here is my attempt at the problem

IMO A is the answer to this one

my line of reasoning is as follows

statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time

now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5

statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent

Accordingly, A is the answer for this

I hope i have taken the correct approach to solve the question and arrived at the correct answer P/Q = R/S PS=RQ (1) P is divisible by 140 140 k*s=RQ

now if k=2 and Q=140 and s=1 then r is not divisible by 5 Insuff

(2)Q = 7^x , where x is a positive integer

Insuff

combine 140 k*s=R 7^x

R=(140 k*s/(7^x))

since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5

Comments please

Many thanks

Ok i see i mistake in statement 1 (thanks for pointing this out) but i just have one question

u say that

Quote:

140 k*s=R 7^x

R=(140 k*s/(7^x))

since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5

agreed that (7^x) can never be a multiple of 10, but let us try some numbers then let us say x =2 and k and s = 1 what we get is R=140/49 which then clearly is not divisible by 5

we will have to wait for the OA for this for one week

let us discuss it

here is my attempt at the problem

IMO A is the answer to this one

my line of reasoning is as follows

statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time

now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5

statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent

Accordingly, A is the answer for this

I hope i have taken the correct approach to solve the question and arrived at the correct answer P/Q = R/S PS=RQ (1) P is divisible by 140 140 k*s=RQ

now if k=2 and Q=140 and s=1 then r is not divisible by 5 Insuff

(2)Q = 7^x , where x is a positive integer

Insuff

combine 140 k*s=R 7^x

R=(140 k*s/(7^x))

since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5

Comments please

Many thanks

Ok i see i mistake in statement 1 (thanks for pointing this out) but i just have one question

u say that

Quote:

140 k*s=R 7^x

R=(140 k*s/(7^x))

since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5

agreed that (7^x) can never be a multiple of 10, but let us try some numbers then let us say x =2 and k and s = 1 what we get is R=140/49 which then clearly is not divisible by 5

so then the answer to this should be E

what do u think

why are you people forgetting If P, Q, R, and S are positive integers

we will have to wait for the OA for this for one week

let us discuss it

here is my attempt at the problem

IMO A is the answer to this one

my line of reasoning is as follows

statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time

now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5

statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent

Accordingly, A is the answer for this

I hope i have taken the correct approach to solve the question and arrived at the correct answer P/Q = R/S PS=RQ (1) P is divisible by 140 140 k*s=RQ

now if k=2 and Q=140 and s=1 then r is not divisible by 5 Insuff

(2)Q = 7^x , where x is a positive integer

Insuff

combine 140 k*s=R 7^x

R=(140 k*s/(7^x))

since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5

Comments please

Many thanks[/quote][/quote]

Ok i see i mistake in statement 1 (thanks for pointing this out) but i just have one question

u say that

Quote:

140 k*s=R 7^x

R=(140 k*s/(7^x))

since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5

agreed that (7^x) can never be a multiple of 10, but let us try some numbers then let us say x =2 and k and s = 1 what we get is R=140/49 which then clearly is not divisible by 5

so then the answer to this should be E

what do u think[/quote]

why are you people forgetting If P, Q, R, and S are positive integers

Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R divisible [#permalink]

Show Tags

03 Jul 2008, 04:28

I get answer C too, following the same reasoning as vdhawan1 did

Please, post the official solution

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Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R divisible [#permalink]

Show Tags

03 Jul 2008, 06:26

I think the answer should be E.

vdhawan1 wrote:

If P, Q, R, and S are positive integers, and P/Q = R/S, is R divisible by 5 ?

(1) P is divisible by 140

(2)Q = 7^x , where x is a positive integer

Statement 1) First, all we know is P is divisible by 140, we don't know at all what Q is. The value of Q will help determine what value R can be.

If p = 140 (which is divisible by 140), we must pick a number for Q. We could pick 280 for Q, and so 140 / 280 = 1/2. R could be 1 and that is not divisible by 5. The question asked of us is essentially "Is R always or never divisible by 5?" If we can come to a conclusion of "always" or "never" - the statement is sufficient. Becuase we can't do that for #1, it is insufficient.

Statement 2) This fails for a similar reason to Statement 1. We know possible values of Q, but we don't know the value of P. If x = 2, then Q = 49 (7^2=49). We are not limited at all as to what q can be. Insufficient.

Together) This should prove to be sufficient. If P is a multiple of 140, and Q is 7^x, we need to see what values are available for R/S.

Factors of 140 will help us determine if (when reduced) the value will be divisible by 5.

1-2-4-5-10-14-28-35-70-140.

Some available numbers for Q: 7-49-343-2401...(I stopped doing the math, because there is a useful principle here).

The only factors of (7^x) will be the number itself and the numbers ahead of it in the progression of 7^x (i.e., factors of 7^5 = 7^1, 7^2, 7^3, 7^4,7^5) There is a pattern that emerges here with the units column (the column that tells us if it is divisible by 5).

You see the pattern. So no matter what P/Q is, it will never be reduced down to something that makes P divisible by 5. Any multiple of 140 divided by 7^x will not leave a 5 or 0 in the units column because any multiple of 140 will have a 0 in the units column and in order to divide anything with a 0 in the units column and result in a 0 or 5 in the units column you mus divide by multiples of 5 or 10 (and mulltiples of 2 [that are also not multiples of 6]).

After all of this analysis and I was ready to write the answer is C, the thought struck me.

If P/Q = R/S, and P=140 and Q = 7^x, P/Q could be 140/49. and if that = RS, there is nothing that would keep R/S from being 140/49 and R is then divisible by 5. The reasoning above shows that we can find plenty of situations that R is not divisible by 5, but because we found one where R is divisible by 5

Answer has to be E. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

we will have to wait for the OA for this for one week

let us discuss it

here is my attempt at the problem

IMO A is the answer to this one

my line of reasoning is as follows

statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time

now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5

statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent

Accordingly, A is the answer for this

I hope i have taken the correct approach to solve the question and arrived at the correct answer

Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R divisible [#permalink]

Show Tags

02 Mar 2015, 06:07

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If P, Q, R, and S are positive integers, and P/Q=R/S, is R divisible by 5?

Given: \(R=\frac{PS}{Q}\)

(1) P is divisible by 140 --> P is multiple of 5 --> now, if all 5-s from P and S (if there are any) are reduced by 5-s in Q then the answer will be No (for example P=140=5*28, S=1 and Q=5) but if the sum of powers of 5 in P and S is higher than the power of 5 in Q then not all 5-s will be reduced and R will be a multiple of 5 (for example P=140=5*28, S=1 and Q=1 or P=140*5=5^2*28, S=1 and Q=5). Not sufficient.

(2) Q= 7^x, where x is a positive integer. Clearly insufficient.

(1)+(2) Q is not a multiple of 5 at all thus 5-s in P won't be reduced so as \(R=\frac{PS}{Q}=5*integer\) then R is indeed a multiple of 5. Sufficient.

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