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A certain casino produces only two denominations of chips: those worth $6 and those worth $11.
Clearly it is impossible to possess certain sums of money, such as $10. Prove, inductively, that it
is possible to have any value greater than or equal to $50.

2007-09-13 08:16:52 · 3 answers · asked by shakirh 1 in Science & Mathematics Mathematics

3 answers

We first show that each of 50,51,...55 can be written as a linear combination of 6 and 11. We have
50 = 4(11) + 1(6)
51 = 3(11) + 3(6)
52 = 2(11) + 5(6)
53 = 1(11) + 7(6)
54 = 0(11) + 9(6)
55 = 5(11) + 0(6).

Now suppose all the numbers up to n >= 55 are linear combinations of 11 and 6, and consider n+1 >= 56. Look at (n + 1) - 6 >= 50, and say this is a(11) + b(6) = n - 5. Then a(11) + (b+1)(6) = n + 1, so by induction, all n >= 50 are combinations of 6 and 11.

2007-09-13 08:48:19 · answer #1 · answered by Tony 7 · 0 0

Tony's the name of a pizza man! Italianoooo pepperoniooooooooooo!

2007-09-13 09:09:13 · answer #2 · answered by Tameeka 1 · 0 0

11*4 +6*1=50
11*3+6*3=51
11*2+6*5=52
11*1+6*7=53
11*0+6*9=54
11*5+6*0=55
11*4+6*2=56
11*3+6*4=57
11*2+6*6=58
11*1+6*8=59
11*0+6*10=60

etc......

2007-09-13 08:25:52 · answer #3 · answered by BadHigh 2 · 0 0

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