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Use the distributive property to solve the following multiplication problems: 12 imes 13, 10 imes 21, and 22 imes 14.

Mathematics

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The method shown in the example 11×1211 \times 12 involves breaking down each number into its tens and ones components and then using the distributive property (also known as the FOIL method or partial products).
For 11×1211 \times 12:
11=10+111 = 10 + 1
12=10+212 = 10 + 2
So, 11×12=(10+1)×(10+2)11 \times 12 = (10+1) \times (10+2).
This expands to (10×10)+(10×2)+(1×10)+(1×2)(10 \times 10) + (10 \times 2) + (1 \times 10) + (1 \times 2).
Calculating each part: 100+20+10+2100 + 20 + 10 + 2.
Adding these partial products: 100+20+10+2=132100 + 20 + 10 + 2 = 132.

We will apply this method to the given problems.

  1. Solve 12×1312 \times 13.

    We break down the numbers: 12=10+212 = 10 + 2 and 13=10+313 = 10 + 3.

    12×13=(10+2)×(10+3)=(10×10)+(10×3)+(2×10)+(2×3)=100+30+20+6=156\begin{align*} 12 \times 13 &= (10+2) \times (10+3) \\ &= (10 \times 10) + (10 \times 3) + (2 \times 10) + (2 \times 3) \\ &= 100 + 30 + 20 + 6 \\ &= 156 \end{align*}

    The product is 156\boxed{\text{156}}.

  2. Solve 10×2110 \times 21.

    We break down the numbers: 10=10+010 = 10 + 0 and 21=20+121 = 20 + 1.

    10×21=(10+0)×(20+1)=(10×20)+(10×1)+(0×20)+(0×1)=200+10+0+0=210\begin{align*} 10 \times 21 &= (10+0) \times (20+1) \\ &= (10 \times 20) + (10 \times 1) + (0 \times 20) + (0 \times 1) \\ &= 200 + 10 + 0 + 0 \\ &= 210 \end{align*}

    The product is 210\boxed{\text{210}}.

  3. Solve 22×1422 \times 14.

    We break down the numbers: 22=20+222 = 20 + 2 and 14=10+414 = 10 + 4.

    22×14=(20+2)×(10+4)=(20×10)+(20×4)+(2×10)+(2×4)=200+80+20+8=308\begin{align*} 22 \times 14 &= (20+2) \times (10+4) \\ &= (20 \times 10) + (20 \times 4) + (2 \times 10) + (2 \times 4) \\ &= 200 + 80 + 20 + 8 \\ &= 308 \end{align*}

    The product is 308\boxed{\text{308}}.

  4. Solve 23×1723 \times 17.

    We break down the numbers: 23=20+323 = 20 + 3 and 17=10+717 = 10 + 7.

    23×17=(20+3)×(10+7)=(20×10)+(20×7)+(3×10)+(3×7)=200+140+30+21=391\begin{align*} 23 \times 17 &= (20+3) \times (10+7) \\ &= (20 \times 10) + (20 \times 7) + (3 \times 10) + (3 \times 7) \\ &= 200 + 140 + 30 + 21 \\ &= 391 \end{align*}

    The product is 391\boxed{\text{391}}.

  5. Solve 19×2219 \times 22.

    We break down the numbers: 19=10+919 = 10 + 9 and 22=20+222 = 20 + 2.

    19×22=(10+9)×(20+2)=(10×20)+(10×2)+(9×20)+(9×2)=200+20+180+18=418\begin{align*} 19 \times 22 &= (10+9) \times (20+2) \\ &= (10 \times 20) + (10 \times 2) + (9 \times 20) + (9 \times 2) \\ &= 200 + 20 + 180 + 18 \\ &= 418 \end{align*}

    The product is 418\boxed{\text{418}}.

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