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A certain game pays players in tokens, each of which is worth either m  [#permalink]

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Question Stats: 32% (02:24) correct 68% (02:36) wrong based on 117 sessions

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A certain game pays players in tokens, each of which is worth either m points or n points, where m and n are different positive integers whose greatest common factor is 1. In terms of m and n, what is the greatest possible sum, in points, that can be paid out with only one unique combination of these tokens? (For example, if m = 2 and n = 3, then a sum of 5 points can be created using only one combination, m + n, which is a unique combination. By contrast, a sum of 11 points can be created by 4m + n or by m + 3n. This solution does not represent a unique combination; two combinations are possible.)

A) 2mn

B) 2mn – m – n

C) 2mn – m – n – 1

D) mn + m + n – 1

E) mn – m – n

Kudos for a correct solution.

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Re: A certain game pays players in tokens, each of which is worth either m  [#permalink]

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Bunuel wrote:
A certain game pays players in tokens, each of which is worth either m points or n points, where m and n are different positive integers whose greatest common factor is 1. In terms of m and n, what is the greatest possible sum, in points, that can be paid out with only one unique combination of these tokens? (For example, if m = 2 and n = 3, then a sum of 5 points can be created using only one combination, m + n, which is a unique combination. By contrast, a sum of 11 points can be created by 4m + n or by m + 3n. This solution does not represent a unique combination; two combinations are possible.)

A) 2mn

B) 2mn – m – n

C) 2mn – m – n – 1

D) mn + m + n – 1

E) mn – m – n

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Multiple combinations are created when you can “exchange” a set of m tokens worth n points each for a set of n tokens worth m points each. In the example given in the problem, the first solution uses 4 m’s (worth 8 points total) and one n (worth 3 points) to reach 11 points. The second solution is created when 3 m’s (worth 2 points each, for a total of 6 points) are exchanged for 2 n’s (worth 3 points each, for a total of 6 points), leading to the new solution m + 3n.

The question asks for a sum with a unique solution, or a solution for which such an “exchange” is not possible. What must be true of this kind of solution?

The payout may include no more than (n – 1) m-point tokens. (If there were n of them, they could be exchanged to make a different combination.)
The payout may include no more than (m – 1) n-point tokens. (If there were m of them, they could be exchanged to make a different combination.)
The question asks for the greatest possible such sum, so use the maximum possible number of m and n tokens, as determined above:

(n – 1)(m) + (m – 1)(n)
= mn – m + mn – n
= 2mn – m – n.

Alternatively, try plugging in numbers. For instance, if m = 2 and n = 3, then you’re looking for the greatest possible sum that can be paid with a unique combination of 2- and 3-point tokens. That sum must be one of the answer choices, so check the sums in the answer choices and stop at the highest one with a unique payout combination.

With m = 2 and n = 3, the answer choices are (A) 12 points, (B) 7 points, (C) 6 points, (D) 10 points, (E) 1 point. Try them in decreasing order (because the problem asks for the greatest sum):

(A) A sum of 12 points can be paid out in three ways: six 2-point tokens; three 2-point tokens and two 3-point tokens; and four 3-point tokens. This answer is not correct.

(D) A sum of 10 points can be paid out in two ways: five 2-point tokens, or two 2-point tokens and two 3-point tokens. This answer is not correct.

(B) A sum of 7 points can be paid out only as two 2-point tokens and one 3-point token. This answer is the largest one left with a unique combination, so it is the correct answer.

The correct answer is (B).
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Re: A certain game pays players in tokens, each of which is worth either m  [#permalink]

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I wonder if thats an adequate way to solve this but I literally just plugged the very same numbers in the task as in m = 2 and n = 3. a and b ar variables corresponding to the amount of n and m-worth chips used
A)2*a + 3*b = 2*2*3-1 = 11 = 2*4+3*1 = 2*1+3*2 - fail
B)2*a+3*b = 2*2*3-2-3 = 7 = 2*2+3*1 - looks good, that means any answer below 7 is incorrect.
C)2*a+3*b = 6 = incorrect (less than 7)
D)2*a+3*b = 2*3+2+3-1 = 10 = 2*2+ 3*2 = 5*2 + 3*0 - fail
E)2*a+3*b = 6 - 2 -3 = 1 - fail

B that is then.
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Re: A certain game pays players in tokens, each of which is worth either m  [#permalink]

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Bunuel wrote:
A certain game pays players in tokens, each of which is worth either m points or n points, where m and n are different positive integers whose greatest common factor is 1. In terms of m and n, what is the greatest possible sum, in points, that can be paid out with only one unique combination of these tokens? (For example, if m = 2 and n = 3, then a sum of 5 points can be created using only one combination, m + n, which is a unique combination. By contrast, a sum of 11 points can be created by 4m + n or by m + 3n. This solution does not represent a unique combination; two combinations are possible.)

A) 2mn – 1

B) 2mn – m – n

C) 2mn – m – n – 1

D) mn + m + n – 1

E) mn – m – n

Kudos for a correct solution.

Ans: (A)
I am not sure if I am right.
this is how i thought of it.
as question says highest common factor is 1 means these two do not have any common factor.
so lets say M=x*y and N=a*b
where x,y,a and b are different prime numbers.
so now putting these conditions in all the options
1. 2mn-1 = 2xyab-1 this can have any value but no two different combinations are possible for the same value.
2. 2mn – m – n = 2xyab-xy-ab = xy(2ab-ab-1) = xy(ab-1)
--------OR
2xyab-xy-ab = ab(2xy-xy-1) = ab(xy-1)
two different combinations are possible.
same way we can do for all options and
the Ans: A
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Re: A certain game pays players in tokens, each of which is worth either m  [#permalink]

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dkumar2012 wrote:
Bunuel wrote:
A certain game pays players in tokens, each of which is worth either m points or n points, where m and n are different positive integers whose greatest common factor is 1. In terms of m and n, what is the greatest possible sum, in points, that can be paid out with only one unique combination of these tokens? (For example, if m = 2 and n = 3, then a sum of 5 points can be created using only one combination, m + n, which is a unique combination. By contrast, a sum of 11 points can be created by 4m + n or by m + 3n. This solution does not represent a unique combination; two combinations are possible.)

A) 2mn – 1

B) 2mn – m – n

C) 2mn – m – n – 1

D) mn + m + n – 1

E) mn – m – n

Kudos for a correct solution.

Ans: (A)
I am not sure if I am right.
this is how i thought of it.
as question says highest common factor is 1 means these two do not have any common factor.
so lets say M=x*y and N=a*b
where x,y,a and b are different prime numbers.
so now putting these conditions in all the options
1. 2mn-1 = 2xyab-1 this can have any value but no two different combinations are possible for the same value.
2. 2mn – m – n = 2xyab-xy-ab = xy(2ab-ab-1) = xy(ab-1)
--------OR
2xyab-xy-ab = ab(2xy-xy-1) = ab(xy-1)
two different combinations are possible.
same way we can do for all options and
the Ans: A

If two numbers have a GCF of 1 this does not necessary imply that those numbers itself are prime. http://mathworld.wolfram.com/RelativelyPrime.html

They are relatively prime, just for your information. Not sure if that changes the approach.
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Re: A certain game pays players in tokens, each of which is worth either m  [#permalink]

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noTh1ng wrote:

If two numbers have a GCF of 1 this does not necessary imply that those numbers itself are prime. http://mathworld.wolfram.com/RelativelyPrime.html

They are relatively prime, just for your information. Not sure if that changes the approach.

I did not say that they are prime but they don't have any common factors with each other except 1. which is given.
i hope i am getting it right.
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Re: A certain game pays players in tokens, each of which is worth either m  [#permalink]

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i plugged in numbers for this m=3 and n=5 and i got C as the answer
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Joined: 06 Oct 2013
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Re: A certain game pays players in tokens, each of which is worth either m  [#permalink]

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Bunuel wrote:
Bunuel wrote:
A certain game pays players in tokens, each of which is worth either m points or n points, where m and n are different positive integers whose greatest common factor is 1. In terms of m and n, what is the greatest possible sum, in points, that can be paid out with only one unique combination of these tokens? (For example, if m = 2 and n = 3, then a sum of 5 points can be created using only one combination, m + n, which is a unique combination. By contrast, a sum of 11 points can be created by 4m + n or by m + 3n. This solution does not represent a unique combination; two combinations are possible.)

A) 2mn

B) 2mn – m – n

C) 2mn – m – n – 1

D) mn + m + n – 1

E) mn – m – n

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Multiple combinations are created when you can “exchange” a set of m tokens worth n points each for a set of n tokens worth m points each. In the example given in the problem, the first solution uses 4 m’s (worth 8 points total) and one n (worth 3 points) to reach 11 points. The second solution is created when 3 m’s (worth 2 points each, for a total of 6 points) are exchanged for 2 n’s (worth 3 points each, for a total of 6 points), leading to the new solution m + 3n.

The question asks for a sum with a unique solution, or a solution for which such an “exchange” is not possible. What must be true of this kind of solution?

The payout may include no more than (n – 1) m-point tokens. (If there were n of them, they could be exchanged to make a different combination.)
The payout may include no more than (m – 1) n-point tokens. (If there were m of them, they could be exchanged to make a different combination.)
The question asks for the greatest possible such sum, so use the maximum possible number of m and n tokens, as determined above:

(n – 1)(m) + (m – 1)(n)
= mn – m + mn – n
= 2mn – m – n.

Alternatively, try plugging in numbers. For instance, if m = 2 and n = 3, then you’re looking for the greatest possible sum that can be paid with a unique combination of 2- and 3-point tokens. That sum must be one of the answer choices, so check the sums in the answer choices and stop at the highest one with a unique payout combination.

With m = 2 and n = 3, the answer choices are (A) 12 points, (B) 7 points, (C) 6 points, (D) 10 points, (E) 1 point. Try them in decreasing order (because the problem asks for the greatest sum):

(A) A sum of 12 points can be paid out in three ways: six 2-point tokens; three 2-point tokens and two 3-point tokens; and four 3-point tokens. This answer is not correct.

(D) A sum of 10 points can be paid out in two ways: five 2-point tokens, or two 2-point tokens and two 3-point tokens. This answer is not correct.

(B) A sum of 7 points can be paid out only as two 2-point tokens and one 3-point token. This answer is the largest one left with a unique combination, so it is the correct answer.

The correct answer is (B).

Can anyone please explain the highlighted part. why (n-1) , why not (n-2) or (n-3)....

is it because greatest common factor 1.
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GMAT 1: 690 Q49 V35 Re: A certain game pays players in tokens, each of which is worth either m  [#permalink]

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Bunuel wrote:
A certain game pays players in tokens, each of which is worth either m points or n points, where m and n are different positive integers whose greatest common factor is 1. In terms of m and n, what is the greatest possible sum, in points, that can be paid out with only one unique combination of these tokens? (For example, if m = 2 and n = 3, then a sum of 5 points can be created using only one combination, m + n, which is a unique combination. By contrast, a sum of 11 points can be created by 4m + n or by m + 3n. This solution does not represent a unique combination; two combinations are possible.)

A) 2mn

B) 2mn – m – n

C) 2mn – m – n – 1

D) mn + m + n – 1

E) mn – m – n

Kudos for a correct solution.

Here is my take on this question. After looking at the question, I realized that solving it the typical algebrical way would be hard. So I rather moved backward from option choices. First and foremost take some sample numbers. Why not start with m=3 and n=2. Now lets get to option choices:

A) 2mn = 12, ahaa- you have two combinations right away- 3*4 or 2*6. So this can't be true.
B) 2mn - m- n : 12-2-3=7, now how can you get 7 using 2s and 3s? Only one way: 2*(2) + 1*(3). There is no other way to get this value. You can try it yourselves. Most likely answer, but lets check if any other option gives an even higher value using one unique combination as the question asks for the greatest possible sum.
C) 2mn-m-n-1= 6, now right away you have 2 possible combinations. One way is to do 2*(3) and the other way is to do 3*(2). Hence this is wrong too.
D) mn + m+ n-1= 10, One way is to do 2*(2) + 2*(3), and the other way is to simply do five times 2. Hence this is also wrong.
E) mn - m - n= 1, there is no way to get this. Hence wrong

CORRECT ANSWER IS B
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Re: A certain game pays players in tokens, each of which is worth either m  [#permalink]

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Bunuel wrote:
A certain game pays players in tokens, each of which is worth either m points or n points, where m and n are different positive integers whose greatest common factor is 1. In terms of m and n, what is the greatest possible sum, in points, that can be paid out with only one unique combination of these tokens? (For example, if m = 2 and n = 3, then a sum of 5 points can be created using only one combination, m + n, which is a unique combination. By contrast, a sum of 11 points can be created by 4m + n or by m + 3n. This solution does not represent a unique combination; two combinations are possible.)

A) 2mn

B) 2mn – m – n

C) 2mn – m – n – 1

D) mn + m + n – 1

E) mn – m – n

Kudos for a correct solution.

Like the example given in the problem, we can let m = 2 and n = 3 and analyze each answer choice:

A) 2mn = 2(2)(3) = 12

However, 12 = 6m + 0n or 0m + 4n.

B) 2mn - m - n = 2(2)(3) - 2 - 3 = 7

We see that 7 = 2m + n only. We can skip choices C and E since the value of those expressions will be less than that of choice B and we are looking for the greatest possible sum.

D) mn + m + n - 1 = 2(3) + 2 + 3 - 1 = 10

However, 10 = 5m + 0n or 2m + 2n.

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