# Pocket Change 3

Directions: You have \$1.00 in change in your pocket. You have 15 coins. What coins do you have?

### Hint

Modifications: Provide coin manipulatives to make the problem more concrete.

Questions: Have you drawn out the coin possibilities?

Solutions may vary (e.g. 5 dimes and 10 nickels)

Source: Andrew Gael

## Product Close to 1,000

Directions: Using the digits 1 to 9 at most one time each, place a digit …

1. An example would be 5 dimes and 10 nickels

10 pennies, 1 nickel, 1 dime, 3 quarters
1 dime, 13 nickels, 1 quarter
9 dimes, 1 nickel, 5 pennies

3. Another example us 1 quarter, 13 nickels, 1 dime

4. Quarters:1
Nickles:12
Dimes:4

5. Quarters:2
Nickles:5
Dimes:3
Pennies:5

6. Isabella Mendoza

9 dimes 1 nickel and 5 pennies

7. 1 quarter 3 dimes 8 nickels 5 pennies = \$1.00

8. There are exactly six solutions. Let (pennies, nickels, dimes, quarters) represent the number of each type of coin. For example, (0,10,5,0) represents no pennies, 10 nickels, 5 dimes, and no quarters. The six solutions, therefore, are:
(0,10,5,0)
(0,13,1,1)
(5,1,9,0)
(5,4,5,1)
(5,7,1,2)
(10,1,1,3)

I knew that the number of pennies had to be a multiple of 5. I also knew the number of quarters would greatly affect the total. I combined these two ideas with the method of solving systems of equations. I found more solutions, but they have negative numbers in them, so they obviously don’t apply to this situation, but they did reveal a cool pattern.

First, I started with no pennies and no quarters. This gave the solution (0,10,5,0). Then, I examined the case of no pennies and one quarter. This resulted in the solution (0,13,1,1). Then, I considered no pennies and two quarters. The resulted in (0,16,-3,2), which is not possible. Note however that it does result in 15 coins and \$1. The next case was (0,19,-7,3). Notice that each time I add a quarter, the number of nickels goes up by 3 and the number of dimes goes down by 4. This makes sense because adding 3 nickels (15 cents) and subtracting 4 dimes (-40 cents) results in a decrease of 25 cents, which is offset by the increase of the quarter. Notice also that adding 1 quarter and 3 nickels and subtracting 4 dimes results in 4 added coins and 4 removed coins, keeping the number of coins the same.

Displaying the results in a table(ish) form, further illuminates the pattern:

no pennies
0, 10, 5, 0
0, 13, 1, 1
0, 16, -3, 2
0, 19, -7, 3
0, 22, -11, 4

five pennies
5, 1, 9, 0
5, 4, 5, 1
5, 7, 1, 2
5, 10, -3, 3
5, 13, -7, 4

ten pennies
10, -8, 13,0
10, -5, 9, 1
10, -2, 5, 2
10, 1, 1, 3
10, 4, -3, 4

9. I know one answer it is 5 dimes and 10 nickels

10. here you go all answers 0,10,5,0.
0,13,1,1.
5,1,9,0.
5,4,5,1.
5,7,1,2.
10,1,1,3.