## week 12

March 9, 2014 § Leave a comment

What fraction of this square region is shaded? Stripes are equal in width, and the figure is drawn to scale. Ss1 — I don’t have the exact answer, but I know it’s more than 1/2. I visually drew in grid lines.

Ss2  mentally drew in grid lines like Ss1 did. It’s a 6×6 square, so 36 square units. Starting in the left corner, I saw 1 white square. Diagonally from there, I notice these are consecutive odd numbers. I just added the white parts to get 15. Subtract 15 from 36 gives 21. So, 21/36 is the answer. Ss 3 — I got 7/12. I split it up into smaller squares too and got 36 because it’s sides are 6 each. There are 21 shaded squares, 21/36 simplifies to 7/12. What is the equation for the number of green dots in this pattern?

Ss1 — I see the diagonal, then 3 groups of n on the outside, then the bottom group. My equation is D = (n+2) + 3n + (n+1) Ss2 — I have the exact same equation as Ss1, but I saw it differently, like this: Ss3 — Left and right, each is n. Top and bottom, each is (n+2). The leftover dot is 1 less than n. My equation: D = 2n + 2(n+2) + (n-1). Ss4 — I saw these 4 in the right corner in every step. My equation is D = 4 + 2n + 2(n-1) + (n+1). Can you figure out the rule: 2, 4, 8, …?
I had fun watching this video from Veritasium, so I posed the question to the kids. The quickest answer came from a 6th grader, and he even used the word “ascending” in his rule! What is the equation for the number of blue rods in this pattern?

Ss1 — I see top and bottom as 2 groups of (n-1). The middle has (n+1) by (n). My equation isR = 2(n-1) + n(n+1). Ss2 — I move one rod from the bottom row to the top to make it a complete “rectangle” of (n+2) by (n). The leftover rods on the bottom row is (n-2). My equation is R = n(n+2) + (n-2). Three corn dogs cost \$5.25. How much does one corn dog cost?

I asked this question because Justin Lanier shares how he and his friend had done this math problem mentally.

Ss1 — I just divided like I normally would on paper. Yea, I just set it up in my head because it’s not too bad.

Ss2 — I knew that if the price had been \$6, then each corn dog would cost \$2. But it costs 75 cents less than that, so 75 divided by 3 is 25. I subtracted 25 cents from \$2 to get \$1.75 for each.

Ss3 — I knew each must cost at least \$1. Then I split up the leftover \$2.25 into thirds: three 25-cents are not enough, three 50-cents are still not enough, so I tried three 75-cents, and that added up to \$2.25.

Ss4 — I used the divisibility rule to see if \$5.25 is divisible by 3: 5+2+5 = 12, and 12 is divisible by 3, so I knew the amount is divisible by 3. I just tried some numbers and \$1.75 works.

Ss5 — I changed \$5.25 to 21 quarters. 21 divided by 3 is 7 — 7 quarters is \$1.75.

Ss6 — I tried \$1.50 first, that got me \$4.50 for 3 corn dogs. I tried \$2, and that’s \$6 total. So I tried the middle number of \$1.75, and it worked.

When I shared with the students how Mr. Lanier had done the problem, they said they would never have guessed that! I said to them, “Math Munch people are just cool like that.” 🙂

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