My mother gave me twenty cents to buy 20 candies but only 3 kinds. how many should I buy? Of a,bor c?

Question

if the cost of each kind in cents are as follows;
Kind a cost 5 cents
kind b cost 3 cents
kind c cost one half cents
No broken pieces.

Correction ;the price of kind 'a 'went down when i got to the store=the price is two cents.

Answer 1

You've given two cases:
Pricing scheme 1:
A = 5 cents, B = 3 cents, C = 0.5 cents

You could just buy all C, but that would result in 40 candies. This is no good.

However, you could exchange 6 pieces of C for 1 piece of B. This is a net change of -5 pieces. Thus it is readily apparent that {0,4,12} is one solution.

You can't do a straight trade of A pieces for C pieces because that is a net change of -9 pieces, and 20 isn't divisible by -9...

Let's try other combinations of A and B.
1A and 1B, 2 pieces for 8 cents = 16 pieces for 0.5 cent.
This is a net change of -14, so that won't work.

1A and 2B, 3 pieces for 11 cents = 22 pieces for 0.5 cent.
This is a net change of -19, so that won't work either

1A and 3B, 4 pieces for 14 cents = 28 pieces for 0.5 cent.
This is a net change of -24, so that's no good.

2A and 1B, 3 pieces for 13 cents = 26 pieces for 0.5 cent
This is -23 pieces, no good.

So there is no way with integer combinations to buy at least one of A, B and C.

The only solution to part 1 is:
0 pieces of A
4 pieces of B (12 cents)
16 pieces of C (8 cents)
--------------------------
20 pieces for 20 cents

Pricing scheme 2:
A = 2 cents, B = 3 cents, C = 0.5 cents

Again we can start with 40 pieces of C. We need to exchange 20 pieces.

We already know that {0,4,16} is one solution.

Let's try the other combinations:
1A + 1B: 2 pieces for 5 cents = 10 pieces for 0.5 cent
Net change of -8, not good.

1A + 2B: 3 pieces for 8 cents = 16 pieces for 0.5 cent
Net change of -13, not good

1A + 3B: 4 pieces for 11 cents = 22 pieces...
Net change -18, bad

1A + 4B: 5 pieces for 14 cents = 28 pieces...
Net change - 23, bad

2A + 1B: 3 pieces for 7 cents = 14 pieces...
Net: -11 bad

2A + 2B (same as 1A + 1B case) - bad

2A + 3B: 5 pieces for 13 cents = 26 pieces...
Net - 21 bad

3A + 1B: 4 pieces for 9 cents = 18 pieces...
Net -14 bad

3A + 2B: 5 pieces for 12 cents = 24 pieces...
Net -19 bad

3A + 3B, same as 1A + 1B - bad

4A + 1B: 5 pieces for 11 cents = 22 pieces...
Net -17 bad

4A + 2B, same as 2A + 1B - bad

4A + 3B: 7 pieces for 17 cents = 34 pieces...
Net -27 bad

5A + 1B: 6 pieces for 13 cents = 26 cents
Net -20 GOOD!

Finally another solution and one that allows us to buy one of each candy:

The two solutions to part 2 are:
0 pieces of A
4 pieces of B (12 cents)
16 pieces of C (8 cents)
--------------------------
20 pieces for 20 cents

5 pieces of A (10 cents)
1 pieces of B (3 cents)
14 pieces of C (7 cents)
--------------------------
20 pieces for 20 cents

Answer 2

You need:

None of kind A
4 pieces of kind B - 4 @ 3 cents = 12 cents
16 pieces of kind C - 16 @ .5 cents = 8 cents

20 pieces of candy for 20 cents!!

Answer 3

You cant buy all the three kinds, but you can buy four candies of three cents and the remaining of half cents.

But instead of going for this math, just buy the ones you like for those 20 cents ;-)

Answer 4

I want to know where to buy 20 candies for twenty cents!

Answer 5

u will end up wit 3 pieces of A, 4 pieces of B, and 4 pieces of C

Answer 6

1 piece of kind A
4 pieces of kind B
2 pieces of kind C

yeahhhh!!!

Answer 7

U should save your money n buy something more beneficial... or buy as many of your favurite ones as possible... or if u dont have enogh money ask your father for more...