Shortcuts to a problem like this? (making change)

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bazookaworm

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Nov 15, 2006
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There are various ways to make $207 using only $2 coins and $5 bills. One such way
is using one $2 coin and forty-one $5 bills. Including this way, in how many different
ways can $207 be made using only $2 coins and $5 bills?

Would I just work all the possibilities and jot them down, or is there a quicker method? A formula possibly? Just wondering as i'm only in eighth grade. :D

ps. sorry if this isn't the right section.
 
well starting from mostly twos, you have 101 x$2 + 1x$5, now if you want to have 2 $5 bills...you realize you can't! since 207 is odd and any amount of $2 coins is even, the amount of $5 bills possible has to be an odd number, so if you have 3 5$ bills, then its 96 2$ bills ...based on this concept, you should be able to figure it out
 
Because very other multiple of 5 can be made with a remainder divisible by 2.

1-5, 101-2
3-5, 96-2
5-5, 91-2
7-5, 86-2

and so on.

Because it's every other you divide by 10 instead of 5.
 
galactus said:
Because very other multiple of 5 can be made with a remainder divisible by 2.

1-5, 101-2
3-5, 96-2
5-5, 91-2
7-5, 86-2

and so on.

Because it's every other you divide by 10 instead of 5.
Click to expand...

Sorry I don't really understand :oops: thanks for helping though :D I don't get what you mean when you say:
1-5, 101-2
3-5, 96-2
5-5, 91-2
7-5, 86-2
Click to expand...
 
bazookaworm said:
galactus said:
Because very other multiple of 5 can be made with a remainder divisible by 2.

1-5, 101-2
3-5, 96-2
5-5, 91-2
7-5, 86-2

and so on.

Because it's every other you divide by 10 instead of 5.
Click to expand...

Sorry I don't really understand :oops: thanks for helping though :D I don't get what you mean when you say:
1-5, 101-2 .............1 $5 and 101 $2 bills together makes $207
3-5, 96-2.............3 $5 and 96 $2 bills together makes $207
5-5, 91-2.............5 $5 and 91 $2 bills together makes $207
7-5, 86-2
Click to expand...
Click to expand...
 
Thanks for clearing that up but I still can't see why you would divide by 10 and round up. I can see that every second multiple of 5 with a remainder of 2's can make 207. You also said that every other you divide by 10 instead of 5?

What do you mean by this?
Thanks.

edit: Well I know that

In order to get to $207 from a multiple of 5 using only $2 coins, the multiple of $5 must end in a 5 because if it was a 0 then adding $2 coins will keep the number even and 207 is odd.

So I count the multiples ending in 5 up to 207, which is 21 and that is the answer.
 
you're not dividing by 20 (which is the closest $207/5) since you dont want every amount of $5 bills, you only want amounts of $5, so that it turns up odd, and since every other number will make it odd, you only need half the numbers...10...
 
ilaggoodly said:
you're not dividing by 20 (which is the closest $207/5) since you dont want every amount of $5 bills, you only want amounts of $5, so that it turns up odd, and since every other number will make it odd, you only need half the numbers...10...
Click to expand...

Isn't 207/5 = 41.4?

you dont want every amount of $5 bills, you only want odd amounts of $5
Click to expand...

What the heck? I think i'm a visual learner...I'm trying to get this but I just don't understand?

So you only want ODD amounts of $5 so it turns up odd, because 207 is odd?

EDIT: Ok well I know that

In order to get to $207 from a multiple of 5 using only $2 coins, the multiple of $5 must end in a 5. If it ended in a 0, adding $2 coins would still give an even number and we want an odd because $207 is odd right?

So I count the multiples of 5 ending in 5 up to 207, which is 21. Which is also the answer. I just want to know how he got "Divide 207 by 10 and round up", (did you work it out like I did?)

EDIT 2: Ok so maybe the CLOSEST multiple of 5 to the answer is 20? But you don't want ALL $5, you want some $2's too. And because 207 is odd you only want the odd amounts of $5, so you divide 207 by 10(AND NOT 20 because you only want every other $5) To arrive at 21?
 
you understand it, you might just not realize that yet :p you don't really need to worry about the twos in the problem since there are a) less 5's and b) 2's always give even numbers
 
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