SOLUTION: What is the least posititve number of coins that is IMPOSSIBLW to give as change for a dollar? I am understanding this to mean what is the smallest number of coins that can't co

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Question 119198This question is from textbook Pre Alge ra
: What is the least posititve number of coins that is IMPOSSIBLW to give as change for a dollar?
I am understanding this to mean what is the smallest number of coins that can't combine to make $1. We charted it out starting at 1 (a dollar coin) 2 (two half dollars) 3 (hald dollar and 2 quarters) etc but are in the late 20s and still going. Is there a formula we could be using? THANKS!!
This question is from textbook Pre Alge ra

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
What is the least positive number of coins that is
IMPOSSIBLE to give as change for a dollar?

I'm pretty sure that's a trick question!

It's obviously 1.  You can't give 1 coin as change for a dollar.
                    J
Of course you could argue than swapping a paper dollar bill for one 
of those gold colored dollar coins the stamp vending machines in 
post offices give you when you put in a $5 or more - could be called 
"changing a dollar using only one coin".  If you count that as a way 
to change a dollar, then the smallest number of coins you could not 
have in change for a dollar would be 77.  But I only found that by 
writing a computer program, not by a formula.  Your teacher 
couldn't require you to get that answer unless you were taking a 
computer programming course.

Other than 1, here are the only numbers of coins you could NOT
POSSIBLY have if you had change for a dollar, using half dollars.

77, 81, 85, 86, 89, 90, 93, 94, 95, 97, 98, 99, 101, and all larger
integers. 

But half dollars are no longer in circulation, so if you
count swapping a paper dollar for a metal one as a way to
change a dollar, and don't use half dollars, then here are the only 
numbers of coins you could NOT POSSIBLY have if you had change for 
a dollar.

2, 3, 5, 77, 81, 85, 86, 89, 90, 93, 94, 95, 97, 98, 99, 101 and
all larger integers.
  
So in that case the answer would be 2.  But I'll bet anything it
was a trick question and the answer is simply 1.

                    J

Edwin

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