The other day my friend impressed me with her knowledge of perfect squares. She had all the squares up to 30 committed to memory, whereas I struggle to calculate 162! As she listed the squares, I started to notice a pattern:


202 400
212 441
222 484
232 529
242 576


Can you see it? The last digit of the answer can be found by squaring the last digit of the question. For example, 22 = 4, so the last digit of 222 must be 4. This trick works for even large numbers. You may not know what 6232 is, but I guarantee you the last digit is 9.1

However, I found another pattern that allowed me to calculate squares in my head. Here is how it works (using 22).

First, subtract 10: 22 – 10 = 12

Second, multiply the number by 4: 12 x 4 = 48

Third, add a zero: 48 → 480

Fourth, add the square of the last digit of the question: 480 + 4 = 484

Cool eh?2 After practicing a few times I was able to perform this process in a few seconds. However, we need to be careful as we extend the strategy. Let’s do 29:

First, subtract 10: 29 – 10 = 19

Second, multiply the number by 4: 19 x 4 = 76

Third, add a zero: 76 → 760

Fourth, add the square of the last digit of the question: 760 + 81 = 841

In this case, 19 x 4 is a bit trickier to do mentally. In addition, 760 + 81 requires a mental carry, which takes some practice. After thinking about it for a while, I realized that this pattern only applies to the numbers 20-30. However, there is a way to extend the method to the numbers 10-20. Consider the following example 17:

First, subtract 10: 17 – 10 = 7

Second, multiply the number by 4: 7 x 4 = 28

Third, add a zero: 28 → 280

Fourth, add the square of the difference from 20 (20 – 17 = 3): 280 + 9 = 289

To recap, here is my method for calculating squares to 30.

  • For the squares from 1-10, simply remember your multiplication facts.
  • For the squares from 10-20, use the process, remembering to add the square of the difference from 20.
  • For the squares from 20-30, use the process, remembering to add the square of the last digit.3

My friend criticized my method; calling it overly complicated. She said that memorizing the squares was easy and I should do the same. While I agree that memorizing the squares is faster and more reliable than working through my, somewhat tedious method, I quite enjoyed finding the pattern. I hope the pattern will aid in your memorization. I know it has aided in mine.


1The reason that the trick works can be explained as follows.

Break the number up into two pieces (the last digit and the remaining number): 623 = 620 + 3

Square the number: (620 + 3)2 = (620 + 3)(620 + 3)

Expand the brackets using FOIL: (620)(620) + (620)(3) + (3)(620) + (3)(3)

Calculate each term: 384400 + 1860 + 1860 + 9

Observe that every term has a 0 in the 1’s place expect the last term. This will always be the case. Hence, the last digit of the final answer is always determined by the last term of FOIL. The last term of FOIL is calculated by square the last digit of the original question (3×3).


2Similar to the note above, the pattern works because of the distributive property. Consider 222. The multiply by 4 and add a zero step can be combined by simply multiplying by 40.

(22 – 10) x 40 + 22

= 12 x 40 + 22                                                  (simplifying inside the brackets)

= 10 x 40 + 2 x 40 + 22                                     (splitting off 2 groups of 40)

= 20 x 20 + 2 x 40 + 22                                     (rewriting the first multiplication)

= 20 x 20 + 20 x 4 + 22                                     (rewriting the second multiplication)

= 20 x 20 + 20 x 2 + 20 x 2 + 22                                 (splitting off 2 groups of 20)

= (20 + 2)2                                                       (backwards foil)

= (22)2                                                             (simplifying inside the brackets)


3I leave it as an exercise for the reader to show why the trick works from 10 – 20.

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Averages of Averages

Averages of Averages

My friend recently graduated from high school. After the ceremony, he was comparing his transcript with a fellow student and notice something strange.

Below are the results he saw (names changed):

Bob Eve
Sem 1 83 81
Sem 2 93 92.8


Both Bob and Eve struggled in the first semester but Bob came out with a 2 percent lead. In the second semester, they stepped up their game and scored in the 90s. Again, Bob beat Eve. From these results, one would expect Bob to have the higher overall average for the year, no questions asked. Here are the final marks:

Bob Eve
Final Average 88 89.8


What!? He had a higher average than her in both the first and second semester. However, her overall average for the year was higher than his. How could this possibly happen?? I’ll give you a minute to speculate…

Here are the marks for the individual courses Bob and Eve completed:

Bob Eve
English 83 80
Biology 85 83
History 81 80
Chemistry 80 99
Physics 86 94
Math 83 99
Band 100 90
Gym 93 93
Art 91 91
IT 94 81
Programming 95 95
Design 85 93


The bold marks are first semester and the non-bold marks are second semester. As you can see, Eve took fewer courses in the first semester. Hence, her poor initial performance, was not weighted as heavily in her overall average. On the other hand, Bob took equal semesters. Hence, his poor performance had a greater negative influence on his final average. This is called Simpson’s Paradox1. The paradox is just what my friend experienced. He “won” both the first and second semester individually. However, he “lost” in the aggregate; the overall year average.

This paradox teaches a valuable lesson: you cannot blindly average averages. Eve cannot calculate her final mark by averaging 81 and 92.8 to get 86.9. Instead, she must go back to her individual marks and calculate her average from the original data.

1 https://en.wikipedia.org/wiki/Simpson’s_paradox

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Polynomial Area

I was doodling the other day when a thought occurred to me: “what is the area of a polynomial?” In particular, consider the following two graphs:

poly1  poly3

In the first graph, the blue line is straight. Algebraically, this is a first degree polynomial.

In the second graph, the blue line is curved. Algebraically, this is a parabola; a second degree polynomial.

The area of the first shape is common knowledge:

\frac{1}{2} * base * height = \frac{1}{2} * 1 * 1 = \frac{1}{2}

The area of the second shape can be calucated using calculus and is:

\frac{2}{3} * base * height = \frac{2}{3} * 1 * 1 = \frac{2}{3}

What if we considered a more complicated polynomial? A third degree polynomial looks like this:


The new curve bends even more and takes up more area. Using calculus, we find that the area is  \frac{3}{4}

Below are the graphs of 10th, 20th, and 100th degree polynomials:




The larger and larger degree polynomials start to look like a square! This means, that we would expect the area to get closer and closer to 1. This is exactly what I found. The area of a polynomial with degree 100 is 0.99. In general, the area of a polynomial with degree n is:


Taking the limit, we find that:

\lim_{x \to \infty} \frac{n}{n+1} = 1

I found it very interesting that a large degree polynomial becomes so similar to a square. Click the link to view an animation displaying this fact https://www.desmos.com/calculator/afd21wuv0a

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Divisibility Problem

Is 59136 divisible by 2?

Of course it is. And you know this because the last digit is a 6.

Is 558162 divisible by 11? Not so easy. If you your schooling was like mine, you never learned a quick trick to check for divisibility for 11. Check this out:


5+8+6 = 19

5+1+2 = 8

198 = 11

Since 11 is divisible by 11, 558162 is divisible by 11. Go ahead, check your calculator!

Here is a challenge problem using the new for 11. What digit could you insert in the number 43160523 to make it divisible by 11? And where would you insert it?

First, we should ensure that the number is not already divisible by 11.


4+1+0+2 = 7

3+6+5+3 = 17

717 = -10

-10 is not divisible by 11. The nearest multiples of 11 are -11 and 0. We need to insert a digit somewhere to modify the number to get a total of -11 or 0. To get to -11, we could increase the sum of 17 to 18. One way to do this would be to insert a 1 at the beginning of the number. Hence, 143160523 would be our new number. You can verify that it is divisible by 11. If you came up with a different solution, please post it in the comments!


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Brothers Meets Physics

I was playing a game called “Brothers” when I noticed a sweet physics problem. Check out the video below:

My question was: is it possible for the little brother to hold on while his big brother makes those crazy swings? Here is what I found.

First, we should make some assumptions about the lengths involved. It looks to me like the rope is about 3 times as long as the big brother. Assuming the big brother is 5 feet tall (they are young boys), this gives us a rope length of 4.5m (see below).

physics problem 2

To determine the force exerted on the little brother, we have the following factors. First, we have the weight of the big brother. Second, we have the centripetal motion of the big brother. As a formula, it would look like this:

F = F_{g} + F_{c}

F = mg + \frac{mv^{2}}{r}

We know the radius of motion is the length of the rope. We won’t specify the mass of the big brother right now. Hence, we need to determine the velocity of the big brother. To do this, we can use energy considerations.

When the big brother is grabbing onto the wall, all of his energy is gravitational potential energy. When he is at the bottom of his swing, all of his energy is kinetic energy.

E_{g} = E_{k}

mgh = \frac{1}{2} mv^{2}

Rearranging the fomula above for v2 gives:

v^{2} = 2gh

Substituting this info our equation for force, we have:

F = mg + \frac{m2gh}{r}

However, h (the height of the drop) and r (the radius of the swing) are the same. Therefore, our formula reduces to:

F = mg + 2mg

F = 3mg

Since we solved the above algebraically, we can be certain that no matter the length of the rope, the force on the little brother will be 3 times the weight of the big brother. Could the little brother hold 3 copies of his big brother dangling from a rope? I don’t know. It seems plausible. Regardless, the physics is quite interesting.

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Stuffing Sacks

I came across an interesting problem posted by a fellow teacher. Think about all those plastic bags you get from the grocery store. Say you shove them into in a single bag, and throw that bag under the sink. How many different ways can you store those bags?

For 1 bag, there is only 1 way.
For 2 bags, there is still only 1 way.
For 3 bags, there are 2 ways.

Here is a picture to illustrate:

stuffing sacks 1

How many ways for 4 bags? 5? How about 20 bags?

Give the problem a try for 4 bags. Once you think you have an answer, click here to confirm.

If you want, I encourage you to try to find the number of ways for 5 and 6 bags.

As I started to try to find all the combinations for 5 bags, I found that drawing the individual bags became cumbersome. I wanted a better method of diagramming. Ready and waiting was my trusty friend, graph theory.

I represented the bags as nodes and connected 2 nodes if one bag contained the other. Here is what the graph would look like for the 1, 2 and 3 bag cases:

Stuffing sacks 3

Here is the drawing for 4 bags, again showing that there are only 4 ways of stuffing the 4 sacks:

Stuffing sacks 4

Before we make the jump to 5 bags, we need to explore the graph theory representation a bit more. Consider these 2 different drawings, do they represent the same situation?

Stuffing sacks 5

On one hand, the drawings are clearly different. One has a branch going to the right and one has a branch going to the left. On the other hand, if we draw out the stuffing situation that they represent, the two situations seem identical:

Stuffing sacks 5

Indeed, the reason these 2 stuffing arrangements are the same, is because it doesn’t matter what order the 2 smallest bags are in. The only thing that matters is which bag they are contained in.

To visualize this with our graph, we are going to add a rule. You can transform the graph by dragging around the nodes, as long as you do not disconnect the lines. As you can see, I can easily transform the left graph into the right graph; showing they represent the same situation.


Using this rule, and trial and error, I was able to produce all 9 different arrangements for 5 bags:

Stuffing sacks 7

I challenge you to try to find all of the combinations for 6 bags. Hint: the answer is larger than 14 and less than 25.



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Flip and Multiply

What is \frac{2}{3} \div \frac{7}{5} ?

Some of you may remember the simple rule for dealing with fraction division. If you encounter a question about the division of fractions, simply flip the second fraction and change the division to multiplication. Flip and multiply.


Why does this work? To justify this technique, we need to remember a crucial rule in math. If we see a division, we can change the question into a fraction, or vise-versa. For example:

4\div3=\frac{4}{3} and \frac{11}{5}=11\div 5

Here is the argument:

\frac{2}{3}\div\frac{7}{5}=\cfrac{\frac{2}{3}}{\frac{7}{5}}    because division can be converted to a fraction

\cfrac{\frac{2}{3}}{\frac{7}{5}}=\cfrac{\frac{2}{3}\times\frac{5}{7}}{\frac{7}{5}\times\frac{5}{7}}    because we can multiply the numerator and the denominator of a fraction by the same quantity

\cfrac{\frac{2}{3}\times\frac{5}{7}}{\frac{7}{5}\times\frac{5}{7}}=\cfrac{\frac{2}{3}\times\frac{5}{7}}{\frac{35}{35}}    by multiplying out the denominator

\cfrac{\frac{2}{3}\times\frac{5}{7}}{\frac{35}{35}}=\cfrac{\frac{2}{3}\times\frac{5}{7}}{1}     by simplifying the denominator

\cfrac{\frac{2}{3}\times\frac{5}{7}}{1}=\frac{2}{3}\times\frac{5}{7}\div 1    because a fraction can be converted to division

\frac{2}{3}\times\frac{5}{7}\div 1=\frac{2}{3}\times\frac{5}{7}    because anything divided by 1 remains the same]

Thus \frac{2}{3}\div\frac{7}{5}=\frac{2}{3}\times\frac{5}{7}

The above technique works with any fraction division question. Hence, the flip and multiply shortcut will work for any fraction division question. QED

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