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11 May 2015, 06:07
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (02:38) correct
67%
(02:40)
wrong
based on 275
sessions
History
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Time
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Not Attempted Yet
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.
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.
18 May 2015, 08:56
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Bunuel
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).
General Discussion
12 May 2015, 01:08
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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.
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.
