Fred G. Harwood

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

It's time for #MathStratChat Rules: post your favorite or a clever solution! It's also fun to comment on other's strategies. Tell us about your reasoning. Like/share so others can see! #MathIsFigureOutAble #MathStratChat #MTBoS #ITeachMath #MathEd

@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.

@gerryvarty @mATH_e_matics @ATMMathematics @howie_hua @ybgoi @Math @NCTM @sonukg4india @dment37 @HarMath @jamestanton Thank you, Gerry. My goal is to share my thoughts about the importance of the visual and the concrete preceding general form. I intentionally wrote it that way in the beginning before I would write 3^2 + 2•3 + 1^2. Then I would have the students repeat for a 4 x 4 to a 5x5

3x3 + 3 + 3 + 1 Your thoughts?

@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.

@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'.

The 3rd triangular # is T(3) = 1+2+3 = 6. Value of (T(1000))^2 - (T(999))^2? Mentally? Visually? Algebraically? @mATH_e_matics @howie_hua @HarMath @jamestanton @dment37 @RaminKhosravi4 @sonukg4india @maanow @NCTM

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

It's time for #MathStratChat! Rules: post your favorite or a clever solution! It's also fun to comment on other's strategies. Tell us about your reasoning. #MathIsFigureOutAble #MathsStratChat #MTBoS #ITeachMath #MathEd

@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?

@dmarain David can you give me some insight into why the (8-x+1)*(8 - y +1) formula gives me the number of x by y rectangles on the chessboard?

In Calc BC yesterday I posted a picture of a chessboard and asked for how many squares there would be. We worked our way through to the answer. I then posed the question of how many x by y rectangles there would be. One student proposed (8-x+1)*(8-y+1) 1/2

@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

@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.

Solve.

@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

@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.

What is the missing number?

@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.

It's Mental Math Monday! How would you mentally calculate this: 75 is 30% of _____?

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

@pwharris I'll bite the double-and-halve bait: 150×24=300×12=3600. Or 100×24=2400. Half again as much: 2400+1200=3600. Russian Peasant for kicks: Halve & floor: 24, 12, 6, 3, 1. Double: 150, 300, 600, 1200, 2400. First list mod 2, times second list, then summed: 0+0+0+1200+2400=3600.

It's time for #MathStratChat! Rules: post your favorite or a clever solution! It's also fun to comment on other's strategies. Tell us about your reasoning. Like/Retweet so others can see! #MathIsFigureOutAble #MathsStratChat #MTBoS #ITeachMath #MathEd

@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.

@pwharris 150 x 24 Think money! 150 is 6 quarters & 4 quarters is 1 dollar. 6 x 25 x 4 x 6 6 x 100 x 6 3600 #MathStratChat

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

@pwharris Here are some of my strategies for this week's #MathStratChat. Being able to play with various strategies seems to be getting easier with my commitment to try the weekly problems 👍 #MathIsFigureOutAble #MTBoS #ITeachMath #MathEd #abed

@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%

@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!

GR8+ A sequence a(n) is defined recursively as a(1)=5, a(n+1)=(a(n))^2, n+1,2,3,... Write an expression for a(n) in terms of n. @HarMath @mrdardy @decompwlj @RaminKhosravi4 @mATH_e_matics @MathCurmudgeon @amtnj @johngaccione @asitnof @DonFraser9 @dment37 @ybgoi

@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?

Consider the identity ((n+1)/2)^2 - ((n-1)/2)^2 = n (1) Verify algebraically (2) Replace n by 1-10. (3) What if n is even? Odd? An odd prime? Make at least 3 conjectures and attempt to prove them. @HarMath @howie_hua @decompwlj @RaminKhosravi4 @ybgoi @mATH_e_matics @jamestanton

@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'.

Gr5+ From 12:00 to 12:59, remove the colon from each digital time (e.g., 12:34 → 1234). How many of these four-digit numbers leave a remainder of 4 when divided by 9? List them and explain your method. @HarMath @howie_hua @RaminKhosravi4 @mATH_e_matics @ybgoi

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

If the difference of 2 consecutive triangular numbers is 100, what is their sum? (a) non-algebra strategies (b) algebraic approach @HarMath @howie_hua @RaminKhosravi4 @mATH_e_matics @ybgoi @DonFraser9 @mrdardy @jamestanton

@dmarain @howie_hua @RaminKhosravi4 @ybgoi @mATH_e_matics I like the way you've arranged the problem for younger grades. Always adding 1/2 in new ways by subdividing the previous. It's Halloween and the Blue Jays are in it! I think an interesting visual pattern might emerge but haven't sought it ... yet. Odd terms are whole. Ninth = n

TRICK OR TREAT a(1)=1 a(2)=1+1/2 = 3/2, a(3)=1+1/2+1/4+1/4 = 2, a(4)=1+1/2+1/4+1/4+1/8+1/8+1/8+1/8 = 5/2 (a) Continue pattern; write out next sum and compute it. (b) If a(n)=5 then n=? (c) Research this series. @HarMath @howie_hua @RaminKhosravi4 @ybgoi @mATH_e_matics