Fred G. Harwood

Maths teacher, mentor teacher, ed consultant fascinated with discovery learning and inquiry, problem solving, assessment, visualization & making a difference.

[Fred G. Harwood @HarMath]

@dmarain A lovely inquiry that seems to generate its own furthering questions in response to the "Wait! What? Really?" Extensions: can it work with three different factors? (1/n = 1/a + 1/b + 1/c ) Does the n have to be a good composite number?

[Fred G. Harwood @HarMath]

@pwharris #MathStratChat How many 1/2's are in 9/2? Nine as 9/2 is nine halves.

[Fred G. Harwood @HarMath]

@dmarain @gerryvarty @mATH_e_matics @ATMMathematics @howie_hua @ybgoi @Math @NCTM @sonukg4india @dment37 @jamestanton If 50^2 is 2500 then 51^2 is two 50's and 1 in the corner bigger or 2601, then 52^2 is 103 bigger etc.

[Fred G. Harwood @HarMath]

@dmarain @gerryvarty @mATH_e_matics @ATMMathematics @howie_hua @ybgoi @Math @NCTM @sonukg4india @dment37 @jamestanton Or for 51^2 and 42^2. This problem was a centre piece of my "visualizing mathematics" workshop. The sum of consecutive odd numbers always being a square (when arranged as you've shown) leads to estimating the next square from a known square.

[Fred G. Harwood @HarMath]

@dmarain @mATH_e_matics @howie_hua @jamestanton @dment37 @RaminKhosravi4 @sonukg4india @maanow @NCTM Left X for a while: visually seeing the steps losing all but the highest stack is probably the easiest solving. Simple cases allowed me to do it mentally. Algebraically was the 'messiest'.

[Fred G. Harwood @HarMath]

@pwharris 24524–12875 --> 24649 – 13000 = 11,649
Solved by pushing each up 125 as a constant shift has an identical difference.

[Fred G. Harwood @HarMath]

@mrdardy @dmarain Pt. 2 Now I'm curious what either of your approaches might change by recording the data in this way for a specific rectangle. Would the emerging patterns lead to a generalization?

[Fred G. Harwood @HarMath]

@mrdardy @dmarain When I had extended the squares question on a checkerboard, I asked, "How many rectangles of any size is on an m x n grid?" The students started with simple cases, organized counting and then one had a breakthrough by organizing the data on a grid. A 6x4 has 210 rectangles Pt.1

[Fred G. Harwood @HarMath]

@MickeyCootes @MathGuyTFL This is the correct approach. Those stating PEMDAS need to see that - 4^2 has the 4^2 taking priority and then being multiplied by the -1.
(-4)^2 =16 but -4^2 = -16
Even the platforms AI knows this.

[Fred G. Harwood @HarMath]

@RHSTUDYZONE 5(s^2 – 25) = 0
(s^2 – 25) = 0
(s + 5)(s – 5) = 0 so either
s + 5 = 0 or s – 5 = 0
then s = -5 or 5

[Fred G. Harwood @HarMath]

@Matt_Pinner I like that some will do a x b + a.
Others will do a x (b+1).
Then the fight starts until equivalence is discovered.

[Fred G. Harwood @HarMath]

@howie_hua Reduce 30% to 3/10, then multiply both by 25 to get 75/250. 250 is the total that 75 is 30% of.

[Fred G. Harwood @HarMath]

@BradBMath @pwharris I haven't used this method in a while. Thanks for the reminder.

[Fred G. Harwood @HarMath]

@KarenCampe @pwharris I had just sent this partial factoring to @KarenCampe
and then continued looking through other answers. I love taking things they already know well and invoking them in the strategies.

[Fred G. Harwood @HarMath]

@ccampbel14 @pwharris Your partial factoring could be:
6x25 x 4 x 6 = 25x4 x 6x6 = 100 x 36
Love the richness of your solutions.

[Fred G. Harwood @HarMath]

@pwharris Missed the last few weeks with busy Wednesday nights.
150 x 24 --> I'd multiply by 100 and add half of that answer since 50 is half of 100.
2400 + 1200 = 3,600.
Double/halving works too.
I love doing percentages this way. 12% of something is 10% + 1% + 1%
15% is 10% + 5%

[Fred G. Harwood @HarMath]

@dmarain @mrdardy @decompwlj @RaminKhosravi4 @mATH_e_matics @MathCurmudgeon @amtnj @johngaccione @asitnof @DonFraser9 @dment37 @ybgoi I think I see where you were going. If students only did n=1, 2, 3 they might come up with a pattern of
a(n) = 5^(2(n-1)) <-- even numbered patterns of exponents, 0, 2, 4... from 5^0, 5^2, 5^4, . . .
This would break down on n=4
A 'powerful' pattern, David!

[Fred G. Harwood @HarMath]

@dmarain @howie_hua @decompwlj @RaminKhosravi4 @ybgoi @mATH_e_matics @jamestanton I used difference of squares to see (2n/2)(1/1) =n.
After verifying, I found it hard to do the 1-10 check. 3rd part made me. I didn't differentiate between odd & odd prime. 9 gave two parts of a pythagorean triples. Instead of conjectures: Q which other n are pythorean legs?

[Fred G. Harwood @HarMath]

@dmarain @howie_hua @RaminKhosravi4 @mATH_e_matics @ybgoi 1206 is the first multiple of 9 (by digit sum) so 1205 ÷ 9 has remainder 8 & 1207 has remainder 1. Using this data, count down to 1201 and up to 1210, we see to add 9 repeatedly; 1201, 1210, 1219, 1228, 1237, 1246, 1255.
I like the use of time as a 'time constraint'.

[Fred G. Harwood @HarMath]

@dmarain @howie_hua @RaminKhosravi4 @mATH_e_matics @ybgoi @DonFraser9 @mrdardy @jamestanton How about (c) a visual approach?