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November 27, 2014

November 27, 2014

Posted by **r. anthony** on Sunday, February 17, 2013 at 10:44pm.

she have?

- math -
**Reiny**, Monday, February 18, 2013 at 12:26amnickels --- x

dimes --- y

quarters --- 16-x-y

5x + 10y + 25(16-x-y) = 185

5x + 10y + 400 - 25x - 25y = 185

-20x - 15y = -215

divide by -5

4x + 3y = 43

but x and y must be positive integers

we know the x-intercept is < 11

and the y-intercept < 14

y = (43 - 4x)/3

by inspection, if x = 4 , y = 9 is the only combination that will work**so she has 4 nickels, 9 dimes and 16-4-9 or 3 quarters.**

check:

does she have 16 coins ?

4+9+3 = 16 , yes

does she have $1.85 ?

4(5) + 9(10) + 25(3) = 185 , yes!!

- math -
**Jeff**, Monday, February 18, 2013 at 12:33amShe could actualy do it several different combinations.

5 quarters, 1 dime, and 10 nickels

4 quarters, 5 dimes, and 7 nickels

3 quarters, 9 dimes, and 4 nickels

2 quarters, 13 dimes, and 1 nickel

- Jeff is right- math -
**Reiny**, Monday, February 18, 2013 at 2:04amShould not have stopped after I found one.

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