Looking for simple "interesting" math problems that cannot be easily solved without algebra

Question

I often find students who dislike algebra. They prefer to work with numbers in solving problems. I believe there are many problems that are hard to solve without algebra. For example:

1. Finding the value of $x$ such that the volume of a box without lid reaches a maximum value.

2. A tank contains 40 gallons of a solution composed of 90% water and 10% alcohol. A second solution containing half water and half alcohol is added to the tank at the rate of 4 gallons per minute. At the same time, the tank is being drained at the rate of 4 gallons per minute, as shown in below. Assuming that the solution is stirred constantly, how much alcohol will be in the tank after 10 minutes?

   

3. Etc, etc.

Could you provide me with other simple "interesting and challenging" examples so I can introduce them to my students?

Answer 1

The box problem may be hard for students who aren't into algebra. I like using the silly number puzzles when I teach beginning algebra.

1. Pick a number between 1 and 25.
2. Add 9 to it.
3. Multiply the result by 3.
4. Subtract 6.
5. Divide by 3.
6. Subtract your original number.

Then you get to go around and quietly say, "You got 7, right?" to each one. Once they get it that everyone has 7, it becomes interesting to figure out why.

I got the idea to use this in algebra class from Harold Jacobs' Elementary Algebra.

Answer 2

I recently came across the riddle that $\frac{3}{16} - \frac{3}{19} =\frac{3}{16} \cdot \frac{3}{19}$, and thus the question what values of the variables give the remarkable coincidence $$ \frac{a}{b} - \frac{a}{d} =\frac{a}{b} \cdot \frac{a}{d} $$ The point is that algebra gives a simple explanation to a mystifying phenomenon.

Answer 3

Here's a nice way to square a number ending in a "5"; let's take 75 as an example. You cut off the "5"; what remains is 7. You then multiply it by its successor: 7×8=56. Then you write "25" after the "56" and get the result: 5625.

Show it to high-schoolers, and they'll shout: "It's magic!" (tested!). Now show (and explain) them this: $$(10a+5)^2=100a^2+100a+25=100a(a+1)+25$$ and they (hopefully) get the "aha! so algebra isn't useless after all..." moment.

Answer 4

I believe Diophantus' riddle is a good example.

God gave him his boyhood one-sixth of his life,
One twelfth more as youth while whiskers grew rife;
And then yet one-seventh ere marriage begun;
In five years there came a bouncing new son.
Alas, the dear child of master and sage
After attaining half the measure of his father's life chill fate took him.
After consoling his fate by the science of numbers for four years,
He ended his life.

Answer 5

Here is a great riddle:

A bottle of wine costs €20. The wine costs €19 more than the empty bottle. How much does the empty bottle cost?

Everyone will answer €1. But 1 + (19 + 1) = 21. Putting it into an equation will lead to the €0.50 answer.

Answer 6

The most striking example of this to me is a reasonably complicated first-degree equation.

What's that, you say? There are no complicated first-degree equations? They're all just $ax+b=0\implies x=-b/a$? That's because you know algebra - it's not at all obvious they can be put into that standard form.

$$3\cdot(2\cdot\_\_+5)-2\cdot(\_\_+5)=3\cdot\_\_+14$$

What number can go in the blanks to make this true?

The reason I find this so striking is because, if you don't know any algebra at all, the above looks intractably difficult, but with algebra, it's so easy you can do it in your head in under a minute, with a bit of practice.

I'd like to think that a simple "fill in the blank" puzzle is simple and interesting enough to motivate all but the most math-phobic students, as long as you start simple.

$$\_\_+5=12$$

$$7\cdot\_\_=42$$

Easy, especially once you work out you can get the answer without guessing by doing division and subtraction.

$$3\cdot\_\_+9=90$$

A little trickier, but it doesn't take formal algebra training to see that you can get this with a subtraction followed by a division. Note that the problem is now getting too hard to make guessing as viable. Then, once you get to something like what I posted above, the situation is hopeless if you don't know about distributivity and balancing equations.

I remember when I was about 10, I knew roughly what "solve for $x$" meant, and I wanted to come up with something to stump my dad. I need something simple enough that I myself could solve it. If I remember correctly, my best effort after about 15 minutes of brainwracking was something like $x+1=2x$, if it wasn't even simpler. He solved it in his head in about half a second, and I was stunned.

Answer 7

You can find a lot of such problems (freely available on the internet) among the Arithmetic Problems in the early issues of the American Mathematical Monthly and in 1800s algebra textbooks. Below are three such examples.

Solution to Arithmetic Problem #116, American Mathematical Monthly 6 #10 (October, 1899), 238-239. [The problem also appears on p. 120 of Advanced Algebra by Joseph Victor Collins (1918).]

Problem Statement: Two candles are of the same length. The one is consumed uniformly in $4$ hours, and the other in $5$ hours. If the candles are lighted at the same time, when will one be three times as long as the other?

Joseph Ray, Elements of Algebra, 1865.

Problem 24 on page 117 A tank is supplied with water from three pumps. The first and second will fill it in $30$ hours, the first and third in $40$ hours, and the second and third in $50$ hours. In what time can each fill it separately?

Horatio Nelson Robinson, Elementary Treatise on Algebra, 1846.

Problem 24 on p. 64 (variables changed to numerical values by me): A person engaged to work $24$ days on these conditions: For each day he worked he was to receive $25$ cents, for each day he was idle he was to forfeit $15$ cents. At the end of $24$ days he received $320$ cents. How many days was he idle?

Answer 8

Solve for a required grade at the end of a course. A common question that many of us get from students is, "What do I need to score on the final exam to get an A in the course?" or somesuch. My response is now always, "You've just asked me an algebra question, you should be able to solve that yourself."

I set up my grading formula specifically to support this exercise: $W = 15\%Q + 50\%T + 35\%F$, where W = weighted total for the course, Q = quiz average, T = test average, F = final exam score. A course "A" grade requires at least W = 90, "B" at least W = 80, etc. In the last week, if the student has a particular target grade then that dictates W, and Q and T are known, so the only unknown is F.

In my elementary algebra class, I spend an hour on this as its own exercise around mid-semester. In higher-level courses, if a student asks the question in the last week, then I'll help remind/set up the grading formula, and let them solve it themselves. Sometimes this gets a pretty intense "wow!" reaction, like it's the first time when a problem they personally instigated is being solved by algebra; sometimes phones come out and they take photos of it, etc.

Answer 9

This type of problem is almost impossible without algebra:

If John paints a house in 3 hours and Jane paints a house in 2 hours, how long does it take them to paint 5 houses together?

Answer 10

And here's my another, but unrelated answer. I sometimes did (more or less) the following: I showed high-schoolers the quadratic formula (i.e., the formula for the discriminant and solutions for a quadratic equation) and told them something along the lines: "Here's a formula that gives a way to solve this equation. Now try to explain the way to solve it without algebra." Then, I open up the Wikipedia page with the formula for the solutions for the quartic and say "And good luck with this one!". Maybe it's not interesting (for them), but I find it convincing that algebra can actually solve problems, not create them (in this case, the problem of concisely and precisely writing down the way of doing some computation).

Answer 11

"Alright class: what do you get when you divide 1 by 0?"

Almost inevitably you will get the response: "Infinity!"

But this is wrong, and you can use algebra to show it: The question itself is algebra in disguise. It is asking to solve for x: $$0*x = 1$$

At this point, you might write on the board the most basic of algebra equations: $$Solve\quad for\quad x:\quad\quad\quad ax = b$$

"This, class, is an algebra equation. It can also be written like this..." $$x=b/a$$

"In our problem, then, b=1 and a=0." $$0*x = 1$$

"Now if I plug in x = 1 million, will this solve the equation?"

Of course not: $0*(10^{6})=0\ne 1$

"How about if I plug in x = 1 billion?"

Still "no": $0*(10^{9})=0\ne 1$

"That's right, even if we plug in 1 trillion billion zillion +1, multiplying it by 0 gives us 0, which is not equal to 1."

The proper answer in the purely algebraic context is that x is indeterminate: there is no real number that will solve this equation. In fact, the answer "infinity" can only arise in the pre-calc/calc context of limits:

$$x = \lim_{a\downarrow 0}\frac{1}{a}=\infty$$

But this is the answer to "What is the limit of 1/a as a goes to zero," Which is not the same question as what we originally asked, which is "What number x do we get when we divide 1 by 0."