Question 885630: There are two of each of the following coins:5c,10c,20c 50c. How many different totals are possible?
Found 2 solutions by KMST, richwmiller:
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Obviously, since all the coin values are multiples of 5c, we can only make totals that are multiples of 5c.
Besides, 2 of each coin makes up a total of 170c, so we cannot make any amount larger than that.
There will be more than one way to make some amounts,
but as long as there is at least one way, we do not care.
We just need to know if there is we can make up totals equal to each and every multiple of 5c up to 170c,
or if there is any of those multiples of 5c that we cannot make.
I like using the most smaller value coins,
because that way I get rid of more coins and my wallet gets lighter and less bulky.
Using just the 5c coins, I could make 5c and 10c.
Adding to that one or both 10c coins, I could make any multiple of 5c from 15c to 10c+2(10c)=10c+20c=30c.
For somewhat larger amounts I would use the same combinations of 5c and 10c coins, adding one or both of the 20c coins.
For example, since I used a 5c coin and a 10c coin to make 15c, to make 35c=15c+20c, I would use the same coins plus a 20c coin,
and to make 55c=15c+40c, I would use the same coins as for 15c plus both 20c coins.
So, with just the 5c, 10c, and 20c coins I could make all multiples of 5c up to 30c+2(20c)=30c+40c=70c.
Making amounts larger than 70c would require using one or both of the 50c coins.
Adding one 50c coin to each of the combinations considered before would allow me to make up to 70c+50c=120c.
Adding the second 50c coin would allow me to get to 120c+50c=170c, and that would take all the coins we have.
However, for values over 100c, I would probably use both 50c coins first,
so as not to count that many coins.
So, maybe for 105c I would rather use 50c+50c+5c=105c rather than
5c+10c+20c+20c+50c=105c.
plus whatever combination I used to make the difference amount.
 <--->
All the multiples of 5c from  to are  .
I would say there are different possible totals.
If it is a mean trick question, we can also make 0c, using 0 coins of each type, and that would make 35 different total amounts from 0c to 170c, counting in 5c intervals.
Answer by richwmiller(17219)  (Show Source):
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