1.4 Multiply whole numbers  (Page 3/7)

Solution

Write the numbers so each place lines up vertically.
Start by multiplying 7 by 62. Multiply 7 by the digit in the tens place of 62. $7\cdot 6=42.$ Write the 4 in the ones place of the product and carry the 1 to the tens place.
Multiply 7 by the digit in the tens place of 62. $7\cdot 6=42.$ Add the 1 ten we carried. $42+1=43$ . Write the 3 in the tens place of the product and the 4 in the hundreds place.
The first partial product is 434.
Now, write a 0 under the 4 in the ones place of the next partial product as a placeholder since we now multiply the digit in the tens place of 87 by 62. Multiply 8 by the digit in the ones place of 62. $8\cdot 2=16.$ Write the 6 in the next place of the product, which is the tens place. Carry the 1 to the tens place.
Multiply 8 by 6, the digit in the tens place of 62, then add the 1 ten we carried to get 49. Write the 9 in the hundreds place of the product and the 4 in the thousands place.
The second partial product is 4960. Add the partial products.

The product is $\mathrm{5,394}.$

Solution

• ⓐ
  $\begin{array}{c}\hfill 47\\ \hfill \underset{\text{\_\_\_}}{\times 10}\\ \hfill 00\\ \hfill \underset{\text{\_\_\_}}{470}\\ \hfill 470\end{array}$
• ⓑ
  $\begin{array}{c}\hfill 47\\ \hfill \underset{\text{\_\_\_\_\_}}{\times 100}\\ \hfill 00\\ \hfill \underset{\text{\_\_\_\_\_}}{\begin{array}{c}\hfill 000\\ \hfill 4700\end{array}}\\ \hfill \mathrm{4,700}\end{array}$

  When we multiplied $47$ times $10,$ the product was $470.$ Notice that $10$ has one zero, and we put one zero after $47$ to get the product. When we multiplied $47$ times $100,$ the product was $\mathrm{4,700}.$ Notice that $100$ has two zeros and we put two zeros after $47$ to get the product.

  Do you see the pattern? If we multiplied $47$ times $\mathrm{10,000},$ which has four zeros, we would put four zeros after $47$ to get the product $\mathrm{470,000}.$

Solution

There are three digits in the factors so there will be $3$ partial products. We do not have to write the $0$ as a placeholder as long as we write each partial product in the correct place.

Solution

There should be $3$ partial products. The second partial product will be the result of multiplying $896$ by $0.$

Notice that the second partial product of all zeros doesn’t really affect the result. We can place a zero as a placeholder in the tens place and then proceed directly to multiplying by the $2$ in the hundreds place, as shown.

Multiply by $10,$ but insert only one zero as a placeholder in the tens place. Multiply by $200,$ putting the $2$ from the $12.$ $2·6=12$ in the hundreds place.