## week 15

Ss1 —
1. 32 x 3 = 96
2. I think of 0.75 as 3/4, and 1/4 of 32 is 8, so 3/8 is 24
3. 96 + 24 = 120
Ss2 —
1. I rounded 3.75 to 4, and did 32 x 4 = 128
2. Since it’s 1/4 less, and 1/4 of 32 is 8, I subtracted 8 from 128 to get 120
Ss3 —
1. I changed the problem by multiplying 3.75 by 4 and divide 32 by 4.
2. 3.75 x 4 = 3(4) + 0.75(2) + 0.75(2) = 12 + 1.5 + 1.5 = 15
3. 32 ÷ 4 = 8
4. 15 x 8 = 120
Ss4 —
1. 3 x 32 = 96
2. To multiply 0.75 and 32, I kept doubling 0.75 until it’s easy. So, 0.75 doubled is 1.5, and add another pair of 0.75 would give me 3
3. Since it took 4 sets of 0.75 to make 3, then there are 8 of these sets in 32
4. 3 x 8 = 24
5. Add this 24 to 96, I get 120

I shared this strategy with them as no one had mentioned it:

1. 3/8 = 0.375
2. So 3.75 is 3/8 of 10
3. 32 x 10 = 320
4. 3/8 of 320 = 120
Another strategy, similar to Ss3‘s:
1. 3.75 x 2 = 7.5 and 32 ÷ 2 = 16
2. 7.5 x 2 = 15 and 16 ÷ 2 = 8
3. 15 x 2 = 30 and 8 ÷ 2 = 4
4. 30 x 4 = 120 What is the equation for the number of squares in this pattern?

Ss1 — I see 2 squares and 1 rectangle: S = 2(n^2) + n(n+1). Ss2 — I see a full rectangle, then subtract the empty space: S = 2n(2n+1) – n(n+1). Ss3 — I see 2 rectangles: S = n(2n) + n(n+1). Ss4 — I move the whole left portion down to bottom of the right portion to have 1 rectangle: S = n(3n+1). What is 37.5 x 16?
Ss1 —
1. 37.5 x 10 = 375
2. 37.5 + 37.5 = 75
3. 75 x 3 = 225
4. 375 + 225 = 600
Ss2 —
1. 37.5 x 2 = 75
2. 16 ÷ 2 = 8
3. 75 x 8 = 600
Ss3 —
1. 40 x 16 = 640 (I did 4 times 16, then add the 0)
2. I went over by 2.5, so 2.5 x 16 = 2(16) + 0.5(16) = 32 + 8 = 40
3. 640 – 40 = 600
Ss4 —
1. 37.5 x 2 = 75
2. 16 ÷ 2 = 8
3. 75 x 2 = 150
4. 8 ÷ 2 = 4
5. 150 x 4 = 600 What is the equation for the number of squares in this pattern?

Ss1 — I see the middle rectangle separately from the 2 sides. Then I move the left side pieces to join the right side pieces to form 1 rectangle: S = n + (n^2) + n(n+1). Ss2 — I move the left pieces over to the right to form one large rectangle: S = 2n(n+1). Ss3 — I see a Gauss equation. I add the top and bottom row, which is (n+3n). The number of pairs of with this 4n sum is the height of the pattern divided by 2: S = 4n[(n+1)/2]. ## week 14

You need to cook a 24-pound turkey. The cookbook says it takes 15 minutes to cook every pound of turkey weight. How long will take to cook the turkey?

I stole this question from David Coffey. Actually, I just asked the kids what 24 x 15 is.

Ss1 — I broke the 15 into 3 chunks of 5.
1. 24 x 5 = 120
2. 3 x 120 = 360
Ss2 — I changed the problem to make the numbers nicer.
1. 24 ÷ 4 = 6
2. 15 x 4 = 60
3. 6 x 60 = 360
When some of my 6th graders weren’t sure why Ss2’s strategy would work, I showed them this area model. Ss3 — I just saw 15 on the clock face as a quarter. It takes 4 of these to make 1 whole. So I did 24 divided by 4 to get 6. Then 6 times 60 (full clock) to get 360.

Ss4 — I changed the problem.
1. 24 ÷ 2 = 12
2. 15 x 2 = 30
3. 12 x 30 = 360
Ss5 — I broke up 24 into 20 and 4.
1. 20 x 15 = 300
2. 4 x 15 = 60
3. 300 + 60 = 360
Ss6 — I broke up the numbers.
1. 24 x 10 = 240
2. 20 x 5 = 100
3. 4 x 5 = 20
4. 240 + 100 + 20 = 360
Ss7 —
1. 20 x 10 = 200
2. 20 x 5 = 100
3. 4 x 10 = 40
4. 4 x 5 = 20
5. Add them all up to get 360.
Ss8 — I divided both numbers by 3 to get easier numbers.
1. 24 ÷ 3 = 8
2. 15 ÷ 3 = 5
3. 8 x 5 = 40
4. Because I divided both numbers by 3, I have to multiply the 40 by 3, and then by 3 again. So 40 times 3 is 120, then 120 times 3 is 360. What is the equation for the number of hexagons in this pattern?

Ss1 — I see the shape as 2 halves, top and bottom. For each half, I see n groups of 3. But the groups overlap, and the overlap is always (n-1). My equation is H = 2(3n) – 2(n-1). Ss2 — I see the two middle rows of hexagons of (n+1) each. Then there are leftover groups of n on top and bottom: H = 2(n+1) + 2n. Ss3 — I always see 2 hexagons at the front. Then I see n groups of 5, but that includes the white inside hexagons, so I have to subtract them — and there are always n white ones. My equation is H = 2 + 5n – n. Ss4 — I see n groups of 6. But there are (n-1) overlaps of 2 each. My equation is H = 6n – 2(n-1). Today is March 17. Write down as many equations as you can, each equaling to 17.

102 – 83 = 17

sqrt(289) = 17

(15 + 4) – 2 = 17

42 + 1 = 17

sqrt(9) + 14 = 17

abs(-16 – 1) = 17

25 – 15 = 17

52 – 8 = 17 What is the equation for the number of cubes in this pattern?

Ss1 — I moved the front pieces to the left side and the back pieces to the right side, so it’s one flat rectangle. The equation is C = (n+1)(2n+1). Ss2 — I see the center column as (n+1). Then there are 4 identical groups around the center, each one is a Gauss addition. My equation is C = (n+1) + 4[(n2+n)/2]. Ss3 — I see the center column as (n+1). Then I rearranged the front and back pieces together to form a rectangle. I do the same with the left and right pieces. My equation is C = (n+1) + n(n+1) + n(n+1)

Similar to how Ss1 sees it, but she creates 2 separate rectangles instead of 1. Ss4 — I take the left and right pieces (including the middle column and flatten them out into a rectangle like everyone else has done, but I see that I can move the pieces to form a square of (n+1) sides. So I have 2 of these because I do the same to the front and back pieces. Because I count the middle column of (n+1) for both squares, so I have to subtract it from my equation:C = 2(n+1)2 – (n+1). Today is March 21. Write down as many equations as you can, each equaling to 21.

620 ÷ 2 – 300 + 11 = 21

52 – 4 = 21

85 – 64 = 21

sqrt(36) × 3 + 3 = 21

sqrt(400) + 1 = 21

42 × 2 – 11 = 21

4! – 3 = 21

## week 13

A pair of jeans costs \$60. The sales tax is 8.25%. What is the total cost for the pair of jeans?

Ss1 — I rounded 8.25% to 10%, 10% of \$60 is 6, so the total cost is \$66. But I know the answer is slightly lower because I’d rounded up.

Ss2 — I knew 10% would be \$6, so the tax is less than that. Then I just ignored the % sign and the 0 in 60, so that the math I did mentally was 825 times 6. I got 4950. Since I knew the answer was less than 6, I knew the number I wanted was \$4.95. Add this to \$60, that’s \$64.95.

Ss3 —
1. 8% of \$60 is \$4.80
2. .25% of \$60 is 15 cents
3. Add above to get \$4.95
Ss4 — I set up the problem as 2 fractions being multiplied: (60/1) × (8/100) = 480/100 = 4.8. Add 60 to 4.8, that’s \$64.8, but since I rounded the 8.25% to 8%, I think the answer is closer to \$65. What is the equation for the number of dots in this pattern?

Ss1 — I always see 2 groups of (n+1). The leftover dots are (n-1). My equation is D = 2(n+1) + (n-1). Ss2 — I see the bottom dot separately. The rest are n groups of 3. So, it’s D = 3n + 1. Ss3 — I saw step 1 in every step, so a constant of 4. Then I see (n-1) groups of 3. My equation is D = 4 + 3(n-1). Ss4 — I saw (n+1) groups of 3 dots. But there are always 2 [red] dots missing. So, D = 3(n+1) – 2. Ss5 — I see 4 groups of n. But there’ll be overlaps to subtract. That overlap is (n-1). My equation is D = 4n – (n-1). Ss6 — I always see two groups of 2 on the outside. The middle dots are (n-1) groups of 3. My equation is D = 2(2) + 3(n-1). How would you find the area of the shaded? 4-leaf clover was created by drawing 4 semicircles of radius equal to 1/2 of side of square

Ss1 —
1. Find area of square
2. Find area of circle
3. Subtract circle from square, this leaves area of the 2 white parts
4. Subtract 2 white parts from circle gives only the shaded parts Ss2 —
1. Find area of square
2. Find area of 1 semicircle
3. Multiply this by 4
4. Subtract area of square from area of 4 semicircles gives you the shaded part only
Ss3 — (I drew in the 2 segments to hint students at this strategy.)
1. Find area of 1/4 of circle
2. Find area of right triangle within the quarter circle
3. Subtract to get half the leaf part
4. Multiply this by 8  What is the equation for the area of this pattern?

Ss1 — I see a square of side n. The leftover is 2 groups of n with overlap of 1. My equation is A = n2 + 2n – 1. Ss2 — I see 2 overlapping squares. The overlapped region is also a square. My equation is A = 2n2 – (n-1)2. Ss3 — I see a large square that’s always missing 2 pieces: A = (n+1)2 – 2. Ss4 — I see 2 groups of n on top and bottom. The middle is a rectangle of dimensions (n-1) and (n+1). My equation is A = 2n + (n-1)(n+1). ## week 12

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.” 🙂