maihuna wrote:If p+q is a multiple of 35, where p and q are positive integers x=?

1. when p is divided by 5, the remainder is 1.
2. when q is divided by 7, the remainder is 3.

Answer : C

Please be careful when you post; there's clearly information missing from this question, since x is completely unconnected to p, q and the two statements.

Stuart Kovinsky wrote:

maihuna wrote:If p+q is a multiple of 35, where p and q are positive integers x=?

1. when p is divided by 5, the remainder is 1.
2. when q is divided by 7, the remainder is 3.

Answer : C

Please be careful when you post; there's clearly information missing from this question, since x is completely unconnected to p, q and the two statements.

Ohh it was a typo Stuart, I have changed x to p the intended question. Now proceed

maihuna wrote:If p+q is a multiple of 35, where p and q are positive integers p=?

1. when p is divided by 5, the remainder is 1.
2. when q is divided by 7, the remainder is 3.

Answer : C

any taker for this...

P+Q=35,70

I. possible values of P = 6,11,16,21,26,31,36 not Sufficient

II. possible values of Q= 3,10,17,24,30,38 Not Sufficient

I&II, Since the sum of P+Q must be a multiple of 35 The only value of P and Q that Add up to 35 is P= 11 and Q=24...

So I choose C

Giorgio wrote:P+Q=35,70

I. possible values of P = 6,11,16,21,26,31,36 not Sufficient

II. possible values of Q= 3,10,17,24,30,38 Not Sufficient

I&II, Since the sum of P+Q must be a multiple of 35 The only value of P and Q that Add up to 35 is P= 11 and Q=24...

So I choose C

There are several other pairs with P = 11 itseld, (11, 24) (11,59) (11, 94)

How I can be sure there is no other pairing possible with P other than 11.?

I highly doubt there is some subtle math tricks involve to close it within 2-minutes in an exam scenario.

maihuna wrote:

Giorgio wrote:P+Q=35,70

I. possible values of P = 6,11,16,21,26,31,36 not Sufficient

II. possible values of Q= 3,10,17,24,30,38 Not Sufficient

I&II, Since the sum of P+Q must be a multiple of 35 The only value of P and Q that Add up to 35 is P= 11 and Q=24...

So I choose C

There are several other pairs with P = 11 itseld, (11, 24) (11,59) (11, 94)

How I can be sure there is no other pairing possible with P other than 11.?

I highly doubt there is some subtle math tricks involve to close it within 2-minutes in an exam scenario.

Not sure where the question is from, but the answer is definitely E, not C.

If you understand remainders, it's actually a very quick question (under 30 seconds).

Remainders go in cycles, all the way up to infinity. Since we have every positive number in the universe at our disposal, there will be an infinite number of solutions, even when we take both statements into account.

So, as soon as we see that p+q fits a cycle (35, 70, 105, 140, ....), we know that there are an infinite number of possible values for p+q.

1) p = {1, 6, 11, 16, 21, ...} we have an infinite number of possible values for p, don't know anything about q: insufficient.

2) q = {3, 10, 17, 24, 31, ...} we have an infinite number of possible values for q, don't know anything about p: insufficient.

Combined: we have an infinite number of possible values for p AND an infinite number of possible values for q AND an infinite number of possible values for p+q; there will be an infinite number of points of intersection: insufficient, choose E.

Stuart Kovinsky wrote:

maihuna wrote:

Giorgio wrote:P+Q=35,70

I. possible values of P = 6,11,16,21,26,31,36 not Sufficient

II. possible values of Q= 3,10,17,24,30,38 Not Sufficient

I&II, Since the sum of P+Q must be a multiple of 35 The only value of P and Q that Add up to 35 is P= 11 and Q=24...

So I choose C

There are several other pairs with P = 11 itseld, (11, 24) (11,59) (11, 94)

How I can be sure there is no other pairing possible with P other than 11.?

I highly doubt there is some subtle math tricks involve to close it within 2-minutes in an exam scenario.

Not sure where the question is from, but the answer is definitely E, not C.

If you understand remainders, it's actually a very quick question (under 30 seconds).

Remainders go in cycles, all the way up to infinity. Since we have every positive number in the universe at our disposal, there will be an infinite number of solutions, even when we take both statements into account.

So, as soon as we see that p+q fits a cycle (35, 70, 105, 140, ....), we know that there are an infinite number of possible values for p+q.

1) p = {1, 6, 11, 16, 21, ...} we have an infinite number of possible values for p, don't know anything about q: insufficient.

2) q = {3, 10, 17, 24, 31, ...} we have an infinite number of possible values for q, don't know anything about p: insufficient.

Combined: we have an infinite number of possible values for p AND an infinite number of possible values for q AND an infinite number of possible values for p+q; there will be an infinite number of points of intersection: insufficient, choose E.

Hi Stuart,
can u please provide a suitable mathematical framework to close such Q quickly.?

maihuna wrote: Hi Stuart,
can u please provide a suitable mathematical framework to close such Q quickly.?

There is no quick way to solve this algebraically; it's far far faster to either pick numbers to show that there's more than 1 possible answer or to understand the concepts to see that there are an infinite number of possibilities.

I suppose you could set up equations to show that you don't have enough information to solve the system:

p + q = 35k
p/5 = m + 1
q/7 = n + 3

in which k is a positive integer and m and n are non-negative integers.

We have 3 equations and 5 unknowns, so there's no way to solve for p.

Stuart Kovinsky wrote:

maihuna wrote: Hi Stuart,
can u please provide a suitable mathematical framework to close such Q quickly.?

There is no quick way to solve this algebraically; it's far far faster to either pick numbers to show that there's more than 1 possible answer or to understand the concepts to see that there are an infinite number of possibilities.

I suppose you could set up equations to show that you don't have enough information to solve the system:

p + q = 35k
p/5 = m + 1
q/7 = n + 3

in which k is a positive integer and m and n are non-negative integers.

We have 3 equations and 5 unknowns, so there's no way to solve for p.

Hi Stuart,
But the dangerous part is that remainders follow a pattern some time and most of the tough questions are solved just because they only allow a few options that match the given crietrion.

maihuna wrote:
Hi Stuart,
But the dangerous part is that remainders follow a pattern some time and most of the tough questions are solved just because they only allow a few options that match the given crietrion.

I'm not sure what you mean.

Remainders always follow a pattern. In problem solving, we can use those patterns to narrow down the choices; in data sufficiency, we can use those patterns to determine the sufficiency of the statements.

I'm not suggesting that we should always say "this is a remainder question, so the answer is E"; of course it depends on what the question is asking (which is why it's imperative to understand the question before evaluating the statements).

p=11 q=24 p+q=35
p=11 q=59 p+q=70

no unique answers , therefore E

Dear All...question asks for unique p...not for unique p and q....hence be careful about

p=11 q=24 p+q=35
p=11 q=59 p+q=70

no unique answers , therefore E


We have 3 equations and 5 unknowns hence unsolvable mathematically...as Stuart rightly pointed out...

p=11, q = 59.... p+q = 70

p = 81, q = 59... p+q = 140.... no unique value of p, hence E

viidyasagar wrote:Dear All...question asks for unique p...not for unique p and q....hence be careful about

p=11 q=24 p+q=35
p=11 q=59 p+q=70

no unique answers , therefore E


We have 3 equations and 5 unknowns hence unsolvable mathematically...as Stuart rightly pointed out...

p=11, q = 59.... p+q = 70

p = 81, q = 59... p+q = 140.... no unique value of p, hence E

I think stuart and others even did not got my core emphasis, reminders short of things are having certain structure and so do not require as many equation as that many variables, take for example, integer solutions for 3 variables can be constructed using one equation only, add some constraints and even unique values can be found,

Point that I was trying to make was: In exam I tried to make a table and coudn't see a combination other than p=11 for almost a dozen value, as we normally do assume GMAT pattern do not change much later, I assumed it is uniques,

So what we need is a formulatic version or a quick way to find values will not be or will be unique.

The variable/equation based conclusion, as suggested, is absurd, in case you have understood my concern.

P's possibilities, 1 6 11 16 21 26 31 36 41 46 51 56..

Q's possibilities: 3 10 17 24 31 38 45 52 59 66 73 ...

35 multiples, 35, 70, 105, 140 ...

Check all the P+Q 5 multiples.

11+24 = 35 YES

16+24 = 40 NO

24 + 26 = 50 NO

24+ 36 = 60 NO
.
.
.

46 + 59 = 105 YES

Since we are getting more than 2 values, the answer should be E.