Fact Fluency Question

[Kiana]

Explain 3 different ways that you could use fact fluency to solve the following multiplication: 45x16.

 

[Steven G]

90*8 = 720
(9*5)(4*4) = (9*4)(5*4)= 36*20 = 720
45*16 = (9*5)*16= 9*(5*16) = 9*80 = 720

Finally a topic I really shine in.

 

[Romsek]

so you're fluent in fact fluency?

 

[Steven G]

so you're fluent in fact fluency?

Well I am certainly not fluent in English. In fact it was about 15 years ago that I realized that I do know a language. That language is mathematics.

When it comes to doing mental addition, subtraction or multiplication I hardly ever do the precise problem I need to figure out but rather I convert it an easier one and solve that one instead.

 

[Steven G]

Here's one. Compute 37^2 - 36^2 in your head. From previous post by Dr Peterson I know that he could do this. Lets hear from others.

 

[topsquark]

Here's one. Compute 37^2 - 36^2 in your head. From previous post by Dr Peterson I know that he could do this. Lets hear from others.

Is this what you are after? My internet search didn't really give me a straightforward idea.
I'd simply factor it: [math]37^2 - 36^2 = (37 - 36)(37 + 36) = 1 \cdot 73 = 73[/math].

FYI: I'm (con)fluent in English and Quantum Mechanics.

-Dan

 

[JeffM]

Here's one. Compute 37^2 - 36^2 in your head. From previous post by Dr Peterson I know that he could do this. Lets hear from others.

Difference of two squares I can do in my head. More complicated than that, I need tools.

 

[Farzin]

Here's one. Compute 37^2 - 36^2 in your head. From previous post by Dr Peterson I know that he could do this. Lets hear from others.

[MATH](a+1)^2-a^2=2a+1[/MATH]Thus the difference between two consecutive squares [MATH]a^2[/MATH] and [MATH](a+1)^2[/MATH] is equal to [MATH]2a+1[/MATH].
[MATH]37^2-36^2=2×36+1=73[/MATH]

 

[Steven G]

[MATH](a+1)^2-a^2=2a+1[/MATH]Thus the difference between two consecutive squares [MATH]a^2[/MATH] and [MATH](a+1)^2[/MATH] is equal to [MATH]2a+1[/MATH].
[MATH]37^2-36^2=2×36+1=73[/MATH]

Or 37+36 = 73.

 

[Steven G]

Can you generate pythagorean triples where two sides differ by 1? What is the algorithm?

 

[topsquark]

Can you generate pythagorean triples where two sides differ by 1? What is the algorithm?

There is only one where the sides differ by 1: 3, 4, 5.

For the rest, solve [math]a^2 + b^2 = (b + 1)^2 \implies b = \dfrac{a^2 - 1}{2}[/math]. For example, with a = 5 you get b = 12. (Note that a must be odd for this to work. I can't say if there are any triples for an even a.)

-Dan

 

[Steven G]

3^2 = 5^2 - 4^2----3, 4, 5
5^2 = 13^2 - 12^2----5, 12, 13
7^2 = 25^2-24^2-----7, 24, 25
9^2 = 41^2-40^2-----9, 40, 41
....
If n=2k+1, (2k+1)^2 = (k+1)^2-k^2.

If n is even, this can't be done.

 

[MarkFL]

Can you generate pythagorean triples where two sides differ by 1? What is the algorithm?

About 10 years ago, I thought of doing that in an effort to look for rational approximations for \(\sqrt{2}\) and wound up "discovering" the Pell sequence:

[MATH]P_{n}=2P_{n-1}+P_{n-2}[/MATH] where \(P_1=1,P_2=2\).

Then you can use any two consecutive members of the sequence to generate Pythagorean triples whose legs differ by 1.

 

[Steven G]

I guess that I took over this thread. Sorry. But we are having fun.
Since you mentioned sqt(2) I thought of how that is a constructible number. Does anyone know how to construct sqrt(n), where n is a positive integer, using a straight edge and compass?

 

[Steven G]

3^2 = 5^2 - 4^2----3, 4, 5
5^2 = 13^2 - 12^2----5, 12, 13
7^2 = 25^2-24^2-----7, 24, 25
9^2 = 41^2-40^2-----9, 40, 41
....
If n=2k+1, (2k+1)^2 = (k+1)^2-k^2.

If n is even, this can't be done.

If n=2k+1, (2k+1)^2 = (k+1)^2-k^2 is total nonsense!

If n=2k + 1, then (2k+1)^2 = 4k^2 + 4k + 1 = (2k^2 + 2k + 1)^2 - (2k^2 + 2k)^2