I chose to write about this problem because I love it! I love:
- the variety of possible solution methods: arithmetic, algebraic, pictorial, mathematical, and representational;
- the humor inherent in the problem;
- the potential fun for students and their joy in working it out;
- the gender switch on Farmer Montague;
- seeing the different ways students draw the chickens and goats;
- being surprised at who solves the problem and how;
- seeing the light come on in students’ eyes when they arrive at the solution.
This problem pushes students to think outside the box. It is non-discriminatory in that I can, have, and did give this and a similar problem to students learning at the most basic level and those taking the HSE exam in the same week.
How I Solved It
To solve this problem, I used an algebraic system of two equations, letting c represent the number of chickens and g represent the number of goats.
c + g = 22 and 2c + 4g = 56
First, I solved for one variable in terms of the other in the first equation.
c + g = 22
c = 22 – g
Then I substituted my new value for c into the second equation.
2c + 4g = 56
2(22 – g) + 4g = 56
44 – 2g + 4g = 56
44 + 2g = 56
44 + 2g = 56
2g = 12
g = 6
Once I had the value for g, I used it to solve for c in the first equation.
c + g = 22
c + 6 = 22
c = 16
I found that there must be 6 goats and 16 chickens. Finally, I checked my values by substituting both into the second equation.
2(16) + 4(6) = 56
32 + 24 = 56
56 = 56
Anticipating Student Approaches
Students could create a table. Start with a guesstimated number for the two animals. I might divide 22 by 2 and use 11 + 11 as a jumping-off point. Using this info, I could create a six-column table.
Once my columns are in place I can see that I have 10 feet too many. I can then work backwards to my solution. I begin by deducting one goat, adding one chicken, and reconciling the number of feet row by row.
If I’m using this method to show a possible solution pathway, this would be the point at which I ask the students if they see a pattern developing. I complete my table, systematically decreasing the number of feet by decreasing goats and adding chickens, while maintaining the necessary number of animals.
There are lots of other ways that students could solve this problem.
- Draw pictures representing 4-footed goats and 2-footed chickens. Start with a guess and check, then add or subtract animals and corresponding feet until you arrive at numbers that satisfy both conditions of the problem.
- Create a system using columns of 4’s and 2’s.
- Use hash marks. Mark off 4 marks for each goat and 2 for each chicken. Track your work using subtotals.
- Work backwards from 56 and 22, by adding and subtracting as needed.
- Select an arbitrary number of animals of each kind. Multiply to calculate the total number of feet (guess and check) until you arrive at the correct totals.
Supporting Productive Struggle
I believe that students will:
- try to add the animals and the feet together, not understanding that their sum will represent neither;
- probably divide the number of animals by 2, then not know what to do with the answer they get;
- in some cases, give up before really trying anything because the answer won’t be obvious or workable using a “standard” pathway, or they are paralyzed by the fear of being wrong;
- work out a solution that satisfies one condition but not the other;
- be upset because I won’t automatically give them the answer, but instead require them to think their way through by asking them questions.
To support students’ efforts and nudge them toward a solution, I might ask any or all of the following, as well as any others that would facilitate a student’s progress.
- What does the sum of 22 and 56 represent?
- Can the feet be added to the animals?
- How could you maintain your total of 22 animals but reduce the number of feet you got?
- What do you know about chickens? About goats?
- How are they different?
- How might you represent them on paper?
- Is there a pattern you see developing in this table/chart?
- Can you tell me the pattern you see in this column?
- What total, in the table, must change/has changed? What must remain/has remained the same?
- How would the next two cells/rows of the table look?
- Can you write an equation to represent the number of animals? What would it look like?
- Could you write an equation to represent the number of animal feet? What wold it look like?
- How might you use these two equations to find a solution?
Ioka is in my Level 1 class. She initially tried to draw the two sets of animals and count their legs but wasn’t successful. With each attempt, she was becoming increasingly frustrated. At one point she wanted to use her calculator but decided not to.
In the end, she added columns of 2’s and 4’s (not shown) and solved the problem. I chose to talk about Ioka’s paper because she worked with two different methods, maybe even three. When the first didn’t work, she thought of an alternate approach and was successful with that one. Both of her methods were very typical of the students in both classes.
Janice is one of the strongest students in this Level 1 class. She was also the first student to solve the problem.
Initially, she divided the 56 feet by the 22 animals, but realized that the quotient didn’t represent anything usable. Next she divided the 22 animals by 2–representing goats and chickens–then used multiplication and addition to guess and check until she reached the solution.
I was interested in Janice’s work because hers was not a typical solution strategy in either class.
Daniel is a student in my HSE class. Like Ioka, he used columns of 2’s and 4’s. However, unlike Ioka, it was his first-choice method, and he solved the problem rather quickly using this method. I chose to talk about his work for this reason.
Unfortunately, he seems to have erased some of his work, but the column of 11’s on the back side of his paper shows that he was considering the number of 11’s in 56, maybe after having divided 22 by 2. He abandoned this approach when he realized that he would have either 1 foot left over, or a one-legged/footed chicken–a situation proposed by more than a few students!
I chose to include Andrea’s solution because her approach to counting 2’s and 4’s was unique. Instead of putting them into separate columns, she incorporated them into a table of sorts, roughly alternating between several of each number, and keeping running subtotals.
Alan was the only student who attempted an algebraic solution, as I expected he would.
He successfully created two equations from the information in the problem and what he knows about goats and chickens. But then he got stuck. He couldn’t figure out how to use these equations to find the solution. He also, out of ego, pride, or just plain not knowing, did not allow himself to seek a less sophisticated/mathematical solution. I sense that this problem may not be uncommon with more advanced students.
Elliott has passed all parts of the HSE exam except the math, which he struggles. I chose his work because he was one of the few students in either class who used has marks to count animals.
Although he arrived at a set of relevant numbers, he could not reason his way through to what the numbers represented–i.e., feet, and not animals.
I selected Sualdy’s work because I was so impressed by the amount of thought and work she put into finding a solution. Her initial attempt is on one side of another sheet, where she drew 22 individual chickens and the same number of goats, albeit each with two feet and two “hands,” which weren’t counted.
On the back of the sheet, she again drew 22 chickens and 22 goats, this time showing the goats with four legs each. Although her numbers still weren’t correct, it was an improvement over her first attempt, and showed that she had realized an error in her earlier thinking. Even though the numbers didn’t pan out, Sualdy didn’t give up. On the front of the problem sheet she drew a third set of goats and chickens, again taking the time to count and label each set, but with no change in the numbers.
Sualdy is among the least mathematically skilled students in my Level 1 class. However, her diligence, patience, and determination to get to a solution are evident in her work.
Here are some statistics from my experience with the Farmer Montague problem.
- Total number of students given the problem = 22
- Level 1/Basic Math students = 9
- Level 3/HSE students = 13
- Total correct solutions = 10
- Level 1 students = 4 correct = 44%
- HSE students = 6 = 46%
I learned that I should never underestimate students’ abilities and diligence. The Level 1 students worked just as hard as the upper-level students in their efforts to solve the problem. I saw striking similarities of method in both groups. Most of the upper-level students reverted to the same “simplistic” solution methods used by the lower-level students, even though they had access to a much wider range of mathematical skills.
I liked that once a student knew the answer, they assisted classmates with figuring out the solution without giving it away. Iespecially learned that students, if encouraged and guided to do so, will for the most part readily and gladly think, regardless of their level of skill.
What I Might Change
I’m sure I will use this, or a similar problem, again. For lower-level students I might introduce Farmer Montague as the second problem of this type, rather than the first. To prepare students, I might introduce a simpler problem of the same ilk. For example:
Joe’s Repair Shop fixes cars and motorcycles. Right now there are 9 vehicles in the shop. They have a total of 22 wheels. How many of each kind of vehicle are in the shop?
By the same toke, if students are struggling with Farmer Montague, I could give them Joe’s Repair Shop as an alternative.
I was surprised that the student who chose the algebraic method wasn’t able to solve the problem. What I learned is that he has difficulty abstracting information. He very good at rote memorization, but gets stuck if he has to think out of the box.
I believe the benefits were far-reaching.
- Students who:
- found the solution felt empowered in their math ability;
- almost got it were strengthened because they were on the right track;
- were stalled got a better understanding of process.
- understood that there can be more than one way to solve a complex problem;
- learned that solving a complex problem does not necessarily require advanced math skills;
- learned that simplistic solutions work well and sometimes work best, which was validation for many of the lower-level and struggling students;
- learned, through reviewing other student work/solutions, new ways to solve the problem that they might try next time around;
- learned that math can be fun, even silly.
I continually required/asked students to consider and tell me what the numbers they were finding represented. It is important for students to realize that they will always get some number when working on a math problem. The key to a successful solution is to know what that number represents, and to ascertain whether or not that number answers the question/s asked. I asked students to keep in mind the conditions the problem sets out (22 animals/56 feet) and to make sure their answer satisfies both conditions.
I also prodded students with questions like:
- What do you think you should do next?
- Can you add the feet and the animals together?
- What does your sum represent?
- How many feet does a goat have? A chicken?
- Can you put the information you have into a table?
- What would your table headings be?
Advice for Teachers
Discourage or even disallow students from using a calculator. Make it clear that brain power is enough. Allow students to work in pairs or small groups if they wish to. Try to ensure that abilities within the groups are balanced and that all students are participating and contributing. Give students enough time to work on the problem. I devoted 30 to 45 minutes for the work based on the class level, and another 15 to 25 minutes for discussion and solution review.
Do this problem with your students! I believe they’ll enjoy it.