Allen was at a neighborhood garage sale. He was standing at a table with all sorts of comic books divided into 3 piles. One pile was marked 10 cents, the second 5 cents and the third 1 cent. Allen had 26 cents. How many different combinations of comic books could Allen buy for 26 cents?

Question

anonymous · Answer

i think you have to grind it till you find it
1) all 1 cent
2) 1 nickel, 21 one cent
3) 2 nickels, 16 one cent
4) 3 nickels, 11 one cent
5) 4 nickels, 6 one cent
6) 5 nickels, 1 one cent
7) one dime, 16 one cent
8) etc

anonymous · Answer

I'm confused -_-

anonymous · Answer

list all the possible combinations
there are many but i just listed 7 of them

anonymous · Answer

would it take forever. I mean when could i stop doing thee pattern thing

anonymous · Answer

no it will not take forever
once you get out of the nickels there are fewer possibilities
maybe you will see a pattern, but i don't see it 
find all that you can do with one dime
7) one dimes, 16 1 cent
8) one dime, one nickel, 11 one cent
9) one dime, two nickels 6 one cent
10) one dime, three nickels, 1 one cent
that is all for one dime
now on to two dimes
11) two dimes, 6 one cent
12) two dimes, one nickel 1 one cent
that is all with two dimes

anonymous · Answer

10x+5y+z=26
If you could get this in y= form you could find how many x values return a y

anonymous · Answer

Okay so its 12. I can explain by making a chart right?

anonymous · Answer

It might not be exactly that, but I'm sure your math teacher doesn't want you to just list combinations. That would be pointless.

anonymous · Answer

i would start with all pennies, one way
then nickels and pennies
then one dime
then two dimes
that should give them all, unless i missed some
but i don't think so

anonymous · Answer

12 is not so big a number that you cannot count to it
i can't imagine what formula you would use for this
the problem with writing 
$$10x+5y+z=26$$ is two fold
a) it is one equation with three unknowns so there are an infinite number of solutions
b) in this context it is a "diophantine" equation, meaning the solutions have to be whole numbers, and i know of no way of saying in advance how many solutions there are to a diaphanine equation

anonymous · Answer

Thanks :)

anonymous · Answer

yw
make sure to check that we didn't miss any, but i think it is right

anonymous · Answer

:D