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What is the value of r? [#permalink]
lavya07
I'm not sure whether it can be ranked as 750+, but % of correct responses shows that it is definitely not an easy one :)
I think if I had encountered such a question on real GMAT I'd have spent more than 2 min to solve it...
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Re: What is the value of r? [#permalink]
How did you know 128 was a possible outcome too? I follow you until m=2 and from there thought q=2, k=4 and therefore i could solve. Didn't realize k could = 8 as well

niks18 wrote:
Alexey1989x wrote:
The sum of \(q\) and \(m\) equals to some positive odd number, where \(q\) and \(m\) are prime numbers.
When \(30\) is divided by \(q^k\) it leaves the remainder \(r\), where \(k\) is a positive integer. What is the value of \(r\) ?
(1) \(15 < q^k < 40\)
(2) \(q < m\) and \(q^k\) leaves \(2\) as a remainder when divided by \(7\)


Self-made

Analyzing question stem we can see that either q or m is 2 because q and m are prime numbers and their sum is odd. Since E+O = O and except 2 all other prime numbers are odd.
Now 30 = q^k*(Quotient) +r. this implies that q^k<30 because if q^k>=30 then either there will be no remainder (if q^k=30) or the remainder will be 30 itself.

Statement 1: if q=2 then from this statement q^k could be 16 or 32 i.e. powers of 2 but if q is not equal to 2 then it could be 27, 25. Hence the statement is not sufficient

Statement 2: As q<m this means q has to be 2 and m could be any prime no greater than 2. Also from this statement we get to know that q^k = 7n+2 (where "n" is the quotient). so we have 2^k = 7n+2

if n = 2; q^k = 16 and if n = 18; q^k = 128. Hence this statement is not sufficient.

Combining Statement 1 & 2 we get q^k = 16.
So option C
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Re: What is the value of r? [#permalink]
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mdacosta wrote:
How did you know 128 was a possible outcome too? I follow you until m=2 and from there thought q=2, k=4 and therefore i could solve. Didn't realize k could = 8 as well

niks18 wrote:
Alexey1989x wrote:
The sum of \(q\) and \(m\) equals to some positive odd number, where \(q\) and \(m\) are prime numbers.
When \(30\) is divided by \(q^k\) it leaves the remainder \(r\), where \(k\) is a positive integer. What is the value of \(r\) ?
(1) \(15 < q^k < 40\)
(2) \(q < m\) and \(q^k\) leaves \(2\) as a remainder when divided by \(7\)


Self-made

Analyzing question stem we can see that either q or m is 2 because q and m are prime numbers and their sum is odd. Since E+O = O and except 2 all other prime numbers are odd.
Now 30 = q^k*(Quotient) +r. this implies that q^k<30 because if q^k>=30 then either there will be no remainder (if q^k=30) or the remainder will be 30 itself.

Statement 1: if q=2 then from this statement q^k could be 16 or 32 i.e. powers of 2 but if q is not equal to 2 then it could be 27, 25. Hence the statement is not sufficient

Statement 2: As q<m this means q has to be 2 and m could be any prime no greater than 2. Also from this statement we get to know that q^k = 7n+2 (where "n" is the quotient). so we have 2^k = 7n+2

if n = 2; q^k = 16 and if n = 18; q^k = 128. Hence this statement is not sufficient.

Combining Statement 1 & 2 we get q^k = 16.
So option C


Hi mdacosta

My thought process was simple - as 2^k = 7n+2 and since 2^k is even so 7n has to be even and that is possible for any even value of "n" so there was high probability that this expression will yield another value equivalent to power of 2. From hereon you can either find the number by putting in even values of "n" or make a smart guess.

I made a smart guess - since 2^6 = 64 & 7*9=63, it is clear that there is only a difference of 1 (and my intention was to find out a remainder of 2) between these two numbers. so I checked the divisibility of next power of 2 with 7
i.e. 2^7 / 7 = 128/7......... and this was the answer that I was looking for.

I am not sure whether a mathematical calculation based approach will directly provide you the value of n=18 for the above equation. even if it does then I believe it will be time consuming. Substitution works best for these type of problems.
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Re: What is the value of r? [#permalink]
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