Problem Solving Strategy of Working Backwards  
Problem Solving website prepared for Problem Solving Strategies for City College course EDUC 6200E, Dr. Moresh  Prepared
by Katie Swartchild, Adam Wagner, David Krulwich


What is Working Backwards in problem solving?  
The strategy of working backwards entails starting with the end results and reversing the steps you need to get those results, in order to figure out the answer to the problem.  
When do we use this strategy? What are real life examples?  
There are at least two different types of problems which can best be
solved by this strategy:
(1) When the goal is singular and there are a variety of alternative routes to take. In this situation, the strategy of working backwards allows us to ascertain which of the alternative routes was optimal. An example of this is when you are trying to figure out the best route to take to get from your house to a store. You would first look at what neighborhood the store is in and trace the optimal route backwards on a map to your home. (2) When end results are given or known in the problem and you're asked for the initial conditions. An example of this is when we are trying to figure out how much money we started with at the beginning of the day, if we know how much money we have at the end of the day and all of the transactions we made during the day.


Sample
Problem:
Joe forgot to check how much money he began the day with. During the day, he spent $8.00 on breakfast, withdrew $40.00 from the ATM, got his dry cleaning for $12.00, bought 5 shirts for $22.00 a piece (plus 8% sales tax). At the end of the day, he had $100.00, how much did he start the day with? 
Solution:
Rather than letting x = the initial amount and creating a long algebraic equation, if we use the working backwards strategy, the problem is more easily solved. $100 Initial amount + 1.08($22*5) = $218.80 (adding back shirt purchase) + $12 = $230.80 (adding back dry cleaning)  $40 = $190.80 (subtracting ATM withdrawal) + 8 = $198.80 (adding back breakfast) Joe began the day with $198.80. 

Below are more sample problems:  
Problem
#1
Joe gives Nick and Tom as many peanuts as each already has. Then Nick gives Joe and Tom as many peanuts as each of them then has. Finally, Tom gives Nick and Joe as many peanuts as each has. If at the end each has sixteen peanuts, how many peanuts did each have at the beginning?

Solution:
Working backwards, we can start with the end result and analyze each step: Joe Nick Tom Peanuts at end: 16 16 16 Previous round: 8 8 32 (because Tom must have given each half of their 16 peanuts, or 8 peanuts) Previous round: 4 28 16 (because Nick gave each half of their peanuts) Peanuts at start: 26 14 8 (because Joe gave each half of their peanuts) The answer, therefore, is that Joe began with 26 peanuts, Nick began with 14, and Tom with 8. 

Problem
#2
I have seven coins whose total value is $0.57. What coins do I have? And, how many of each coin do I have?

Solution:
In this problem, we can solve it most easily by working backwards  instead of starting with zero and adding the different combinations of coins towards a goal of $0.57, we can begin with the end result and work backwards towards $0.00. Step 1: There must be 2 pennies. (The only other option would be to use seven pennies, but that would use up all the coins prematurely.) Step 2: Now we need to figure out how to use the 5 remaining coins to make a total of $0.55. Because 5 dimes is less than $0.55, we must use at least one quarter. Step 3: Now we need to use 4 coins to make up the remaining $0.30. At this point, all the remaining coins must be dimes and nickels, and the only possible combination is to use 2 dimes and 2 nickels. 

Problem
#3
The grid represents a section of town with only 2way streets. You want to go from point A to point B without backtracking. How many different routes are possible? (You can only go north or east at each intersection.) B
A

Solution:
Starting at point B, we work backwards, one intersection at a time. In our solution table, the number at each intersection denotes the number of possible routes to B from that intersection. Each number is simply the sum of the two numbers north and east of it. Using this system, we can fill in all the numbers back to Point A, and we see that there are seventy possible routes.
