How many pies can be made out of 10 pieces of pastry?

[maxew]

1 piece of pastry can make 3 bottoms, and another piece of pastry can make 4 tops. (Tops are smaller than bottoms).

This is not a homework problem lol. I can work out "10" by guesswork (ie answer is 17 pies - 4 pieces x 4 tops = 16 tops, 6 pieces x 3 bottoms = 18 bottoms - the last piece of pastry has 2 bottoms & 1 top, with bit left over)...

What is the equation for this? What if I had, say, 557 pieces of pastry?

This is troubling me because I can't work it out & I think it is simple algebra?

 

[Ishuda]

1 piece of pastry can make 3 bottoms, and another piece of pastry can make 4 tops. (Tops are smaller than bottoms).

This is not a homework problem lol. I can work out "10" by guesswork (ie answer is 17 pies - 4 pieces x 4 tops = 16 tops, 6 pieces x 3 bottoms = 18 bottoms - the last piece of pastry has 2 bottoms & 1 top, with bit left over)...

What is the equation for this? What if I had, say, 557 pieces of pastry?

This is troubling me because I can't work it out & I think it is simple algebra?

Well not quite simple algebra if you want complete pies. This kind of problem comes under the branch of math call Number Theory. Suppose neither number was a multiple of the number like your problem. Then, to come out even you would need the same number of tops and bottoms so, to use a slightly different example, suppose you needed 1 pastry for 2 bottoms and 1 pastry for 5 tops [a really deep dish pie]. If you had 7 pastries you could make 10 bottoms (5 pastries) and 10 tops (2 pastries). That is 10=2*5 and 7 is 2+5. So, if you want to come out even you have to make multiples of 10 pies and use multiples of 7 pastries. Of course this assumes that you don't break up a pastry and make 1 pie with some left over.

Notice that 10 is a multiple of both 2 and 5. In fact, it is the smallest number which is a multiple of 2 and 5 and is called the least common multiple. That is the number you would need for any two (or more) numbers t and b required for tops and bottoms respectively. You find the least common multiple (a number which is a multiple of both t and b and is the smallest such number). Call that number L. Then the smallest number of pastries needed to complete a whole number of pies is [L/t+L/b].

For another example, suppose it were 4 tops per pastry and 2 bottoms [not quite as deep dish as the other]. The least common multiple of 2 and 4 is 4 and [4/4+4/2]=3. So to come out even we need 3 pastries to make 4 tops (=4/4 pastries) and 4 bottoms (4/2 pastries). The number of pies we need to make to come out even is a multiple of 4 and the number of pastries needed for that is a multiple of 3.

So how many do you need for your problem?

 

[maxew]

Hey thanks a lot you guys!