SOLUTION: I know the answer to this problem, but I don't know how to set it up in an equation to show it. Here is the problem: Austin has 441 pennies that he finds very hard to carry aro

Question 116420: I know the answer to this problem, but I don't know how to set it up in an equation to show it. Here is the problem:
Austin has 441 pennies that he finds very hard to carry around. So he takes them to the bank to cash in for other coins (half-dollars, quarters, dimes, nickels and pennies). What is the fewest coins he would need to carry?
Thanks in Advance!
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
4.41/.50=8 half dollars with .41 remaining.
.41/.25=1 quarter with .16 remaining
.16/.10=1 dime with .06 remaining
.06/.05=1 nickel with .01 remaining
.01/.01=1
8+1+1+1+1=12 total coins.
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