Q: An eccentric casino owner decides that his casino should only use chips in $5 and $7 denominations. Which of the following amount cannot be paid out using these chips?
A. $31
B. $29
C. $26
D. $23
E. $21
OA is D, but I am looking for some easiest and fastest technique. Please help!
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smackmartine
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This problem tests you on how quickly you can sum up the multiples of 5 and 7 to get the values in the option
A. $31 = 7.3+5.2
B. $29 = 7.2+5.3
C. $26 = 7.3+5.1
D. $23 CANNOT
E. $21 = 7.3
A. $31 = 7.3+5.2
B. $29 = 7.2+5.3
C. $26 = 7.3+5.1
D. $23 CANNOT
E. $21 = 7.3
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GmatMathPro
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One thing you can do is just start subtracting 5's from the answer choices until you hit a multiple of 7. if you don't hit one before you get to zero, it's impossible.
A. 31,26,21 POSSIBLE
B. 29,24,19,14 POSSIBLE
C. 26,21 POSSIBLE
D. 23,18,13,8,3,-2 IMPOSSIBLE
E. 21 POSSIBLE
A. 31,26,21 POSSIBLE
B. 29,24,19,14 POSSIBLE
C. 26,21 POSSIBLE
D. 23,18,13,8,3,-2 IMPOSSIBLE
E. 21 POSSIBLE
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GmatMathPro
Thats really a fast approach to solve the question.
Thats really a fast approach to solve the question.
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smackmartine
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Yes!Ahmed MS wrote:Ok. I believe you meant (7.3+5.2), . indicates multiplication (7*3+5*2), right? Thanks.
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rohit_gmat
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just check the last digit or last digit - 5
and see if we have that last digit in any multiple of 7
A. $31 - 1 (yes 7 x 3 = 21)
B. $29 - 9 (7 x 7 - 49... too big) ... 9-5= 4 (yes 7 x 2 = 14)
C. $26 - 6 (no) ... 6-5 = 1 (yes 7x3=21)
D. $23
E. $21 - 1 (yes 7x3=21)
and see if we have that last digit in any multiple of 7
A. $31 - 1 (yes 7 x 3 = 21)
B. $29 - 9 (7 x 7 - 49... too big) ... 9-5= 4 (yes 7 x 2 = 14)
C. $26 - 6 (no) ... 6-5 = 1 (yes 7x3=21)
D. $23
E. $21 - 1 (yes 7x3=21)
