Question 132459: Please help me solve this problem: Sanjay spent a long time searching for just the right birthday card for his mother. He narrowed his choice down to two different cards, a humorous one and a serious one. He finally decided to buy both for a total of $3.34. He gave the clerk $4.00. How many possible combinations of coins can you find for Sanjay's change?
I know that the change is $.66 and i can also use the fifty cent piece. But i have no clue what to do next... please help me..
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! I don't have to know the exact amount of each combination, just
how many combinations there are, so I can ignore the pennies, and
just be aware that I'll need pennies to fill in the shortfalls
when I have less that 66c
nickels
--------
(0-13)
I can have no nickels up to 13 nickels. 14 nickels would make
 c which exceeds the change
How many dimes can I possibly have with each of (0-13) nickels?
and not exceed 66c?
nickels dimes subtotal
------- ----- --------
0 (1-6) 6
1 (1-6) 12
2 (1-5) 17
3 (1-5) 22
4 (1-4) 26
5 (1-4) 30
6 (1-3) 33
7 (1-3) 36
8 (1-2) 38
9 (1-2) 40
10 (1) 41
11 (1) 42
12 (0)
13 (0)
---------------
To tally up what I have so far,
(1) (0 nickels, 66 pennies)
(2) (1 nickel, 61 pennies)
. . . .
(14) (13 nickels, 1 penny)
(15) (0 nickels, 1 dime, 56 pennies)
. . . .
(56) (11 nickels, 1 dime, 1 penny)
---------------
Now I have to take into account the
combinations that use quarters
---------------
nickels dimes: (1) (2) (3) (4)
------- ------------------------------
0 quarters: (1-2) (1) (1) (1)
1 (1-2) (1) (1) (0)
2 (1-2) (1) (1)
3 (1) (1) (0)
4 (1) (1)
5 (1) (0)
6 (1)
7 (0)
This table adds 19 more possibilities, bringing the
subtotal to 75
---------------
Now for the 50c as part of the change:
---------------
nickels dimes: (0) (1)
------- -----------------
0 50c: (1) (1)
1 (1) (1)
2 (1) (0)
3 (1)
4 (0)
---------------
This table sadds 6 more combinations.
The final grand total is 81, unless I
messed up somewhere.
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