All posts by mathconfidence

April 2020 Brain Puzzler Solution

SqRt Thingie

Using perfect squares:
This is super tricky as some of the smaller perfect squares that are 3 digits repeat digits (like 121, 144, and 225).  If we try 169 (whose square root is 13), then many of the two digit perfect squares cannot be used like 81, 64, 49, and 36).

For example, sqrt (169) – sqrt(25) – sqrt(4) would give 6.

If we bump up to 256 (whose square root is 16), then we can use 81 for the two digit and 9 for the single digit making it 16 – 9 – 3 = 4.

Using non-perfect squares:
Sqrt(145) – sqrt(92) – sqrt(6) = 0.00044178938368

Credit and thanks to openmiddle  (if you use these in the classroom, try to modify the instructions to make them less Google-able)

Math Promotes Calm and Flow

Our hats are off to the health care and essential workers.

Math can be therapeutic! I know there are dozens of studies re: Math anxiety but I propose that Math can bring on flow and thereby calmness.

I have the great fortune of working with students and teachers on this great thing we call Math.  In Math, we have the opportunity to find some comfort in the known and knowable with consistency and dependability.  This week, I observed a lesson on the Exterior Angle Theorem and the teacher did an excellent job of drawing 8th graders into the intrigue and puzzlement of Math and the ‘aha’/’Eureka’ of understanding.  This promotes a flow state and a sense of accomplishment and calm.

The teacher used this excellent Geogebra:
Geogebra Exterior Angle Applet
Geogebra Ext Angle ThmTriangle



March 2020 Brain Puzzler Solution

Q: How many three digit perfect squares are the sum of eighteen consecutive positive integers?

A: There are 3 three digit perfect square that are the sum of eighteen consecutive positive integers: 225, 441 and 729.

If there are 18 numbers in a row, there will be 9 pairs of numbers that will add up to the same sum.  Therefore we only need to check three digit perfect squares that are divisible by 9.  Another condition is that the two middle numbers in the list have to add up to that perfect square divided by 9.

144 (12^2) is even and divided by 9 = 16 each pair would have to add up to 16.  There are no two consecutive numbers that add to 16.

225 (15^2) divided by 9 is 25 so the two middle numbers are 12 and 13.
4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 has 12 and 13 for the middle numbers and every pair going outwards 11 and 14  10 and 15 etc adds up to 25.  This list would make 9 pairs of 25 or 225.

324 (18^2) is also even like 144.  324 divided by 9 is 36 and there are no two consecutive integers that add to 36.

441 (21^2) divided by 9 is 49 so the two middle numbers are 24 and 25.
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

576 (24^2) is also even.  576 divided by 9 is 64 and there are no two consecutive integers that add to 64.

729 (27^2) divided by 9 is 81 so the middle two numbers are 40 and 41.
32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

900 (30^2) is also even.  900 divided by 9 is 100 and there are no two consecutive integers that add to 100.

Notice each number divisible by 9 is a multiple of 3 squared.  The next multiple of 3 is 33 and 33^2 >1000.  Therefore, the only three numbers are:

225, 441 and 729.





Midterm “Exam” Suggestions from Students

Hope you all are well.  This week was back to school for my college students.   The form below is from the end of class Monday (our very first virtual class!)

Our college, The College of Mount Saint Vincent, has professors post midterm grades.  These grades do not go on transcripts but allow for transparency between teachers and students.  I spent much of the planning time for Monday’s lesson and Wednesday’s exam focused on the student experience and creating modules within the 85 minutes.

On Wednesday the midterm for my College Algebra students was a blend of Delta Math and a Google Form.  I chatted with students in real time and held virtual office hours.

I extended the Delta Math due date til Friday morning and also created a non Delta Math assignment on a Word document to provide students with a choice.   At this point in time, helping students complete assignments is not as important as building community and agency for our students.

Midterm Sugeestions

February 2020 Brain Puzzler Solution

Q: In an economy pack of 12 highlighters, 3 are defective.

If you purchase 2 highlighters (picked randomly), what is the probability that neither highlighter is defective?

A: 6/11
Since there are 3 defective highlighters out of 12, the probability of selecting the first non-defective highlighter is 9/12.
The probability of selecting the second non-defective highlighter is 8/11 as 11 highlighters remain and only 8 of them are good.
Therefore, the probability that a customer buys 2 non-defective highlighters is:
= 9/12 x 8/11
= 3/4 x 8/11
= 24/44
= 6/11.