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# On planet Enigma, the residents use a currency called the confusion. There are only 2 confusion bills on Enigma, one worth 8 confusions and

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(a) On planet Enigma, the residents use a currency called the confusion. There are only 2 confusion bills on Enigma, one worth 8 confusions and the other worth 11 confusions. There are also some coins of smaller value, but each weighs over 10 kilograms, so they are difficult to carry around. In how many ways can a resident of Enigma use only bills to purchase a toaster that costs 96 confusions? (Everybody knows, of course, that vendors on Enigma do not give change, so residents must make their purchases with exact change.)

(B) (This is a continuation of the problem above. See part (a) for an explanation of the planet Enigma and its currency.) In how many ways can a resident of Enigma use only bills to purchase a gyroscope that costs 97 confusions?

(c) (This is a continuation of the problem above. See part (a) for an explanation of the planet Enigma and its currency.) In how many ways can a resident of Enigma use only bills to purchase a single math book page that costs 69 confusions? (A page is so expensive because the residents of Enigma clearly don't understand much math.)

Sorry guys!!! For these lengthy problems

Mellie  Jul 2, 2015

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b )   8x + 11y = 97      ....I'm with Melody, here......"trial and error" might be the shortest distance between two points, since the numbers are small

Note that  x can't be greater than 12 and y can't be greater than 8 ....  and y must be odd because 8x is always even.....and the only way to get an odd (97, in this case) is to add an odd to an even....

When y = 3, x = 8

And y can't be 1, 5 or 7

So.... 8 "8" confusions and 3 "11" confusions seem to work.......

CPhill  Jul 2, 2015
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Hi Mellie,

Now, we can do this the easy way .... or we can do it the hard way.

11x+8y=96

if x=0     8y=96    y=12          So   12 * 8 confusion notes will be good

if x=1

11+8y=96

8y=85     No integer solution

if x=2          22+8y=96        8y=74        No integer solution

if x=3           33+8y=96        no integer solution

Just keep doing this and you will find that the only other solution is

8*11confusions+1*8confusion      88+8=96

So there are only 2 ways.

You can do the other two similarly. :)

This method is time consuming but the numbers are all quite small so it is quite doable :)

Melody  Jul 2, 2015
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The other way is to solve the diophantine equation more formally.

I have done the first one on a peice of paper in front of me but it takes even longer to get to your answer than the easier way that I just used, so do it the easy way :))

Melody  Jul 2, 2015
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A: I can only come up with two:    Twelve   8  Confusions bills

One 8 Confusions bill and eight 11 Confusions bills

Take 96 and remove multiples of 11 from it succsessively    96  85  74  63  52  41  30  19  8     only one of which is divisible by 8......That is the only combo of 11s and 8s that will work...

Likewise with the 8s   :  96 88 80 72 64 56 48  40  32 24  16  8    only 88 is divisible by 11 .....

B:  Do the same thing for 97 :  97  86  75  64  53  42  31  20  9  1       I find one way of buying something for 97 confusions with only 8s and 11s....    eight 8s  and three 11s

C.  For 69     69  58  47  36  25  14  3  (nothing divisible by 8)   I find no way to pay 69 using only 8s and 11s

69  61  53  45  37  29  21  13  5   (nothing evenly divisible by 11)       I think you will get some 10kg bitcoins!!

Guest Jul 2, 2015
#4
+80931
+15

b )   8x + 11y = 97      ....I'm with Melody, here......"trial and error" might be the shortest distance between two points, since the numbers are small

Note that  x can't be greater than 12 and y can't be greater than 8 ....  and y must be odd because 8x is always even.....and the only way to get an odd (97, in this case) is to add an odd to an even....

When y = 3, x = 8

And y can't be 1, 5 or 7

So.... 8 "8" confusions and 3 "11" confusions seem to work.......

CPhill  Jul 2, 2015
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Thanks Chris and anon - I think we might need change too :((

Mellie this isn't really for you - I am just practicing doing diophantine equations  :))

Part C

8A+11B=69

The FCF of 8 and 11 is 1    and 1 goes into 69 so this does have integral solutions  :/

 Euclidian Algorithm (1) 11=8*1+3 (6) 1=-8+3*(11-8*1) 1=-8-3*8+3*11 1=-4*8+3*11 69=8*-4*69+11*3*69 69=8*-276+11*207 (2) 8=3*2+2 (5) 1=3-(8-3*2) 1=-8+3*3 (3) 3=2*1+1 (4) 1=3-2 Extended Euclidean Algorithm

8*-276+11*207=69

8A+11B=69

So

A=-276+11N          B=207-8N

276/11=25 and a bit

A=-276+11*26=10         B=207-8*26=-1

Mmm       There are not any solutions!

Melody  Jul 3, 2015
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More practice

Part B

11A+8B=94

Largest common factor of 11 and 8 is 1.  One is a factor of 94.  So this does have integral solution.

 Euclidean Algorithm 11A+8B=94 (1) 11=8*1+3 (6) 1=-8+3*(11-8) 1=-8-3*8+3*11 1=-4*8+3*11 11*3+8*-4=1 11*3*94+8*-4*94=94 11*282+8*-376=94 so one integer solution is A=282 and B=-376 The general integer solution is A=282-8N                   B=-376+11N 282/8=35 and a bit A=282-8*35=2           B=-376+11*35=9       This is one sloution A=2+8=10                 B=9-11=-2  this will never do! So   2 elevens and 9 eights confusions is the only solution. (2) 8=3*2+2 (5) 1=3-(8-3*2) 1=3+2*3-8 1=3*3-8 1=-8+3*3 (3) 3=2*1+1 (4) 1=3-2 Extended Eurclidean Algorithm

2 elevens and 9 eights confusions is the only solution.

Melody  Jul 3, 2015

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