# Matt intends to buy precisely the amount he needs and to spend as little money as possible. How many packets of each kind does he have to buy? Find all solutions to the problem below.Matt is in...

Matt intends to buy precisely the amount he needs and to spend as little money as possible. How many packets of each kind does he have to buy? Find all solutions to the problem below.

Matt is in charge of organizing a drawing competition. The main part is, of course, the ceremony where he will announce the winners and give away prizes. Matt knows that he needs exactly 833 lollies for the ceremony and the three most popular kinds of lollies are 'Smarties', 'Jelly Beans' and 'Jelly Snakes'.

He has learnt that he can buy packets of 'Smarties' with 63 lollies for the price of $1.28, packets of 'Jelly Beans' wih 91 lollies for the price of $0.81 and packets of 'Jelly Snakes' with 44 lollies for the price of $0.60.

Matt intends to buy precisely the amount he needs and to spend as little money as possible. How many packets of each kind does he have to buy? Find all solutions.

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The key to a question like this is to set up a good table so that you do not miss any possible solutions.

Let x=# packs of Jelly Snakes, y=#packs of Smarties and z=# packs of Jelly Beans.

Then 44x+63y+91z=833 with `x,y,z >=0`

In a systematic way, try each possible combination.

Let x=0:

y=1,z=1 63+91<833

...

y=1,z=9 63+9(91)>833

y=2,z=,1 126+91<833

...

y=2,z=8 126+91(8)>833

y=3,z=1 189+91<833

...

y=3,z=7 189+7(91)>833

...

Let x=1:

y=1,z=1 44+63+91<833

y=1,z=2 44+63+182<833

y=1,z=3 44+63+3(91)<833

y=1,z=4 44+63+4(91)<833

y=1,z=5 44+63+5(91)<833

y=1,z=6 44+63+6(91)<833

y=1,z=7 44+63+7(91)<833

y=1,z=8 44+63+8(91)=835>833

y=2,z=1 44+126+91<833

...

y=2,z=8 44+126+8(91)>833

and so on.

Once you have exhausted the possibilities for x=1, try x=2.

The number of trials quicly reduces.

The only answer with all three nonnegative is 7x,4y,3z

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**The solution is 7 packets of Jelly Snakes, 4 packets of Smarties, and 3 packets of Jelly Beans. This results in 7(44)+4(63)+3(91)=833 lollies at a cost of 7(.6)+4(1.28)+3(.81)=11.75**

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A complete solution is obtainable. The solution is:

`x=7c_1,y=11c_1+13c_2+6,z=-11c_1-9c_2+5` where `c_1,c_2` are integers. The particular solution found was obtained when `c_1=1,c_2=-1` .

You will note that `c_1>=0` since `x>=0` . Also `c_2<=0` so that `z>=0` if you want x,y,z to all be nonnegative.