IntMath Newsletter: hyperbolas, Pixar, PhotoMath

By Murray Bourne, 25 Nov 2016

25 Nov 2016

In this Newsletter:

1. Interactive hyperbola graphs
2. Pixar in a box
3. Resource: PhotoMath
4. Math movie: Teach for Mastery
5. Math puzzles
6. Final thought: beliefs

0. Gratitude

It's the season for giving thanks, so here goes. For me, the greatest thing to be grateful for is my well-rounded and fair education. It was very sad to read about the children of Mosul who were being indoctrinated with a corrupted curriculum while ISIS was in charge. Most kids stayed away, being denied a vital education.

On with the Newsletter.

1. Interactive hyperbola graphs

Interactive hyperbola graphs

This page has 4 different hyperbola graphs where you can explore the characteristics, such as what is meant by "path traced out by a point where the difference of the distances from 2 fixed points is a constant." A picture is worth a thousand words! See:

Interactive hyperbola graphs

2. Pixar in a box

There are a lot of interesting applications of mathematics to learn from Pixar in a Box. From the Khan Academy site:

Pixar in a Box is a behind-the-scenes look at how Pixar artists do their jobs. You will be able to animate bouncing balls, build a swarm of robots, and make virtual fireworks explode. The subjects you learn in school — math, science, computer science, and humanities — are used every day to create amazing movies at Pixar.

Pixar animations

In the Effects section, we learn how "artists create explosions, fire, and water by breaking them down into millions of tiny particles and controlling them using computer programming."

See: Pixar in a Box: Effects

3. Resource: PhotoMath

PhotoMath logo

PhotoMath is a clever iOS and Android app. You point your camera at an algebra question and it will solve it for you, including giving you the steps.

It reads (neat) handwriting, and in my tests, was able to handle questions involving integrals, square roots and trigonometry.

Search for PhotoMath on your phone or tablet in the App Store or Play Store. It's free until the end of November.

Disclaimer 1: IntMath has no connection with PhotoMath.

Disclaimer 2: PhotoMath, like all such computer-based math solving tools, should only be used to check your own work (which helps you to learn) - NOT to simply solve the homework for you (which usually results in your never knowing it).

4. Math movie: Salman Khan - Teach for Mastery

I once taught a class of pre-service primary school teachers. Many of them were not at all confident in mathematics generally, and certainly not at the level they were going to be teaching (decimals, fractions, money, units and so on).

I decided to use a mastery testing approach with this class. Students were given a series of quizzes throughout the semester, and they needed to obtain 80% to pass each one, and couldn't go on until they had "mastered" the earlier ones. They were given as many attempts as they needed. The test questions were taken from Year 5 and Year 6 primary school mathematics text books. While many students were initially unhappy about having to go through such a process, by the end of the semester they could see the point, and were much more confident (and competent) in the math they they would teach in the future.

Salman Khan Mastery in Math

This TED talk brought back strong memories of those wanna-be primary school student teachers. Salman Khan argues rightly that too often we allow math students to move on to the next level when many are not ready, guaranteeing frustration, boredom, and hatred for the subject.

See: Let's teach for mastery - not test scores

5. Math puzzles

The puzzle in the last IntMath Newsletter asked about the length of rope needed to allow a sheep to eat half of a circular pen of grass.

There was one correct solution and it was well done by Gerard. (I added some images so Gerard's answer would be easier for you to follow.) There was no "nice" algebraic way to solve the final equation involving an angle (which had the variable both inside and outside trigonometry expressions). Two methods are shown: (1) Solving by Newton's Method, and (2) Solving graphically.

New math puzzle

If the number ABC is written "in decimal form", it means it has value A × 100 + B × 10 + C × 1.

Given that identical letters stand for identical digits, and different letters are different digits, what are A, B and C if the following numbers are written in decimal form and:

ABC + AB + A = 300?

You can leave your responses here.

6. Final thought: beliefs

Considering Trump's astonishing announcements that he's appointed Stephen K. Bannon (former manager of the right wing Breitbart News) as a strategist, and that he's going to kill NASA's climate research budget, it's worth contemplating Voltaire's statement:

Those who can make you believe absurdities can make you commit atrocities. [Voltaire, 1694-1778]

Until next time, enjoy whatever you learn.

See the 32 Comments below.

32 Comments on “IntMath Newsletter: hyperbolas, Pixar, PhotoMath”

  1. Chris Moon says:


    Just solve A then B and finally C so



    Ans: A=2, B=7, C=1

    ABC+AB+A = 300
    we can also write it as AAA+BB+C = 300 -- (i)
    by following logic based on fundamental summation procedure
    1) A+B+C=10*x
    2) A+B+x = 10*y
    3) A+y=3
    therefore, A>1 --(iii)
    therefore, A=2
    then BB+C = 300-222 = 78 -> B=7, C=1

  3. Γιώργος Βαρελάς says:

    MAXIMUM{A}=2, because: A=(300-11B-C)/111<300/11<3
    If A = 1, then: 11B+C = 189, but MAXIMUM{11b+c}=99+9=108, because:
    b,c < 10.
    therefore: A = 2, then:
    B=(78-C)11<78/11<8, therefore B<8
    also: C78-10=68, B>68/11>6, therefore B=7, then:
    A = 2, B = 7, C = 1

  4. Francis says:

    Working from the given observation that A × 100 + B × 10 + C × 1, we can state that
    111 * A + 11 * B + C = 300
    Clearly A cannot be greater or less than 2
    111 * 2 + 11 * B + C = 300
    11 * B + C = 300 - 222
    11 * B + C = 78 Where B cannot be greater than 7. If C is to be a single digit, B can also be no less than 7.
    11 * 7 + C = 78
    This results in C = 1
    A = 2, B = 7, C = 1

  5. Bill Harvey says:

    Given: ABC + AB + A = 300
    Rewrite in decimal form:
    100A + 10B + C + 10A + B + A = 300 or 111A + 11B + C = 300
    Assuming that we are dealing only with positive integers, then A must be 2; substituting we then would have 222 + 11B + C = 300 or
    11B + C = 78. B must equal 7 (if larger, then C would be negative, if smaller, then C would be a two digit number which isn't allowed). Then C = 1.

    Summary: A= 2, B=7, C=1

  6. GuidoV says:


    A=1 or A=2
    If A=1 ; carry must be 2, which is impossible since C+B+A<20
    So A=2 ; carry=1 ; B+2+1=10 ;B=7
    And C+7+2=10 ; C=1
    solution: A=2;B=7;C=1

  7. Eamon Day says:

    Solution to puzzle

    Therefore 100A+10B+C+10A+B+A=300
    Collecting terms, 111A+11B+C=300
    A must be equal to 1 or 2, otherwise total would exceed 300
    If A=1 then 11B+C=300-111=189
    This has no solutions for B & C between 1 and 9. Therefore A=2
    Hence 11B+C=300-222=78
    This has a solution B=7 and C=1

    Hence A=2, B=7, C=1

    Check: ABC+AB+A = 271+27+2=300

  8. Henri R. Carpentier says:

    This is in reference to item 6 (Final thought: beliefs) of the latest IntMath e-mail: Stay out of politics with this newsletter. It is supposed to be for math only. I resent you trying to influence my political beliefs and imply that I can be duped to commit atrocities.

  9. Murray says:

    @Henri: Trump said during his campaign he would re-introduce torture.

    I do not apologise for inviting people to think carefully about the world-wide implications of a Trump presidency.

  10. Nick Allen says:

    Maths might be an abstract subject, but there is nothing to exclude human opinion and experience in a newsletter. Murray is perfectly entitled to make a gentle comment about current events. There is nothing controversial in stating that Trump is controversial - controversy is part of his continued self referencing.

  11. Hamid.zandi says:

    A=1 or 2
    C,B= 0.,1,...,9.

  12. Nick Allen says:

    A+B+C=10 (write 0, carry 1)
    A+B=9 (total 10, write 0, carry 1
    A=2 (total 3)
    So, A=2, B=7, C=1

  13. Forrest says:

    PhotoMath is a great app for students to cheat on math tests with their cell phone. You should not be publicizing this app.

  14. Murray says:

    @Forrest: I agree with your sentiments, but can there are already many similar products out there which can be used to help learning.

    I have amended the post and added a second disclaimer (which I should have had in the first place). Thanks for pointing it out!

  15. Hannelie says:

    A=2 B=7 C=1

  16. Bruce Tivendale says:

    A cannot be 0 or 3, since the sum is 300.
    So A is either 1 or 2.
    C+B+A must be 10 (zero ones in 300).
    B+A must be 9 (tens) (zero tens in 300).
    If A = 1, then B = 8, which makes C = 1 also – not allowed.
    If A = 2, then B = 7, which means C = 1.
    SOLUTION: 271 + 27 + 2 = 300

  17. Maria Miller says:

    This was easy...
    271 + 27 + 2 = 300

  18. Suraj Kumar says:

    Expanding the given equation gives
    This can be solved by simple iteration or a computer program.
    This gives the answer as A=2, B=7, C=1

  19. Roger Evans says:

    Hello Murray,
    Loved the Pixar in a Box, have sent it to many friends. Kind regards, Roger

  20. Francis says:

    While reading your final thought it occurred to me that Mr. Trump is a business person. What kind of data is he willing to accept? What kind of data does he trust? What sources does he look to and when does he look? What kinds of risk is he unwilling to take?

    If we who understand that climate change is real can start putting the facts and the risks in front of him in a place he will read them and in a way that encourages him to trust them, then we might have a chance of bringing him on board and getting his help. It should be remembered that the EPA was instituted by a Republican, Richard Nixon. There is always hope.

    Thanks for the excellent newsletters.
    Francis Kisner

  21. karam says:

    Hello Murray , The solution of the puzzle is A=2 , B = 7 , C = 1
    If the number ABC is written "in decimal form", it means it has value A × 100 + B × 10 + C × 1.

    Given that identical letters stand for identical digits, and different letters are different digits, what are A, B and C if the following numbers are written in decimal form and:
    ABC + AB + A = 300?
    Is this true solution

  22. Thomas A Buckley says:

    The answer is A=2, B=7, and C=1,
    because from:-
    ABC + AB + A = 300
    (100A +10B +C) + (10A+B) + A =300
    111A + 11B + C = 300
    By examination, 2 is a likely substitute for A, so;
    (222) +11B + C =300,
    Again by examination 7 is a likely substitute for B, so;
    222 + 77 + C = 300. So C = 1
    Proof check
    (2x100 + 7x10 + 1x1) + (2x10 +7x1) + 2x1 =300

    So A=2, B=7, and C=1

    Solutions by the Power of Number Crunching.

  23. Randy says:

    271 + 27 + 2 = 300

  24. karam says:

    Hello Murray

  25. ManyArrows says:

    Re: 6. Final thoughts: beliefs

    That quote cuts both ways...

    Whether climate change, brexit, gun control/rights, insert your own topic, if you follow the money, you will find selfish actors on both sides that seek to obfuscate the truth and push their own agendas of profit...

  26. Murray says:

    @Karam: Please ask such math questions in the IntMath Forum.

  27. Aaron says:

    Apologies if I'm misinterpreting the question; I'm new
    A = 2, B = 7, C = 1?
    ABC + AB + A
    271 + 27 + 2 = 300

  28. Gerard Grufferty says:

    We must assume that A, B and C are all natural numbers between 1 and 9 inclusive.
    Start with
    100 A + 10 B + C + 10A + B + A = 300
    That is
    111A + 11B + C = 300
    C = 300 - 111A - 11B
    If A = 3 then C = 300 - 333 - 11B which is negative.
    Then A must be less than 3 or A = 1 to A = 2.

    If A = 1 then C = 300 -111 - 11B = 189 - 11B
    Therefore 189 - 11B > 0
    11B < 189
    B 0
    11B < 78
    B < 7·90909.. < 8
    If B = 7 then C = 78 - 77 = 1 which looks good.
    If B = 6 then C = 78 - 66 = 12 which is not less than 9
    and any other value for B less than 6 would make C larger.
    Then B must be greater than 6 and so B = 7

    Then A = 2, B = 7 and C = 1

    271 + 27 + 2 = 300.

    Maybe the Diophantine equation could be solved by ancient methods but lief is too short for that.

  29. Peggy says:

    For the math puzzle: A = 2, B = 7, C = 1.

    The original sum works out to
    (100A + 10B + C) + (10A + B) + A = 111A + 11B + C = 300.

    Since B and C are single digits, the maximum value of 11B + C is 11(9) + 9 = 108. And 300 - 108 = 192. So 111A has to be equal to at least 192. The least value of the sum 11B + C is 0, and now we have 192 \<= 111A \<= 300. The only integer solution for this inequality is 2, and therefore A = 2.

    Now 222 + 11B + C = 300, so 11B + C = 78. In a similar process, C is on the interval [0, 9], so  69 \<= 11B \<= 78, and the only possible integer solution is B = 7.

    Finally, 222 + 77 + C = 300, and C = 1.

    I liked your puzzle.

  30. Rex says:

    A lie told often enough become the truth. Unknown

  31. Ipshita Bhuwania says:

    ABC-271, AB-27, A-2
    The sum of these three would add up to 300.

  32. ernest says:


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