The operator asked Angie to deposit 55 cents to make a call. How many ways can Angie pay using quarter, dimes and nickles?

Question

 Need help with question 

Answer 1

The important ability that this problem teaches is called 'developing an algorithm' (that is, a systematic method for solving the problem).  Now, some algorithms are best for beginning math, some are best for computers, etc.  The goal is to understand the problem and to select a 'tool' (method) for solving it.

 

Let's do an exhaustive iteration.  What's that?  Well it examines all possibilities (using a computer) and checks to see whether the total is 55.  The units are 25, 10, and 5 and the total must be 25, and we will start with either 0 or 1 or 2 quarters; either 0,1,2,3,4, or 5 dimes; and from 0-11 nickels.  Then test to see whether the total value is 55.

 

For Q = 0 to 2

  For D = 0 to 5

    For N = 0 to 11

      If (25*Q + 10*D + 5*N) = 55 then Print Q,D,N

    Next N

  Next D

Next Q

 

The results:

 

Quarters   Dimes   Nickels

0             0             11
0             1             9
0             2             7
0             3             5
0             4             3
0             5             1

1             0             6
1             1             4
1             2             2
1             3             0

2             0             1

 

Angie has 11 possibilities.

Answer 2

Hi Misty,

 

An intriguing problem for sure !!

 

Let's begin by making a few guiding observations:

    1.  Each way will involve the use of nickels since the desired amount (0.55) ends in 5. Only exception is one quarter which can an add an amount ending in 0, which does NOT require the use of any nickels.

    2.  Next, we can observe that when an odd number of nickels is involved - dimes will also be involved, since their denomination ends in 0 and an odd number of nickels ends in 5.  Only exception is 2 quarters, which amount also ends in 0.

   3.  Finally, we can observe that when an even number of nickels is involved - quarters will be involved, since we need to go from amount ending in 5 and adding an amount ending in 0.  Dimes will also be involved when using one quarter in which cases dimes can also be used to add an amount ending in 0. 

 

So with the above information in hand, we can arrange our coin combinations in ascending order of the number of nickels required using these coin abbreviations (<Q>uarters/<D>imes/<N>ickels):

    0.  0N:  1Q/3D 

    1.  1N:  2Q/1N; 5D/1N

    2.  2N:  1Q/2D/2N

    3.  3N:  4D/3N

    4.  4N:  1Q/1D/4N

    5.  5N:  3D/5N

    6.  6N:  1Q/6N

    7.  7N:  2D/7N

    8.  8N:  no combinations

    9.  9N:  1D/9N

  10. 10N:  no combinations

  11. 11N:  11N

 

So from our above list, we can see that there are 11 combinations using these three different coins to reach an amount of 55 cents.

 

Looking  for observable number properties allowed us to come up with an orderly procedure for coming up with our list of different coin combinations.

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.