$745 \times 51$

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Grade 6

Multiply numbers ending in zeros

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745 \times 51

Set up problem: Set up the multiplication problem in the standard format. $\newline$ Write $745$ above $51$ , with the larger number on top, and draw a line underneath them to prepare for multiplication.
Multiply ones place: Multiply the digit in the ones place of the bottom number ( $1$ ) by each digit of the top number ( $745$ ). $\newline$ $1 \times 5 = 5$ , write $5$ below the line under the ones place. $\newline$ $1 \times 4 = 4$ , write $4$ next to the $5$ , under the tens place. $\newline$ $1 \times 7 = 7$ , write $7$ next to the $4$ , under the hundreds place. $\newline$ The partial product is $745$ .
Multiply tens place: Multiply the digit in the tens place of the bottom number ( $5$ ) by each digit of the top number ( $745$ ), remembering to add a zero (place holder) for the tens place before writing the numbers. $\newline$ $5 \times 5 = 25$ , write $5$ and carry over $2$ . $\newline$ $5 \times 4 = 20$ plus the carried over $2$ equals $22$ , write $2$ and carry over another $2$ . $\newline$ $745$ $0$ plus the carried over $2$ equals $745$ $2$ , write $745$ $2$ . $\newline$ The partial product is $745$ $4$ , but remember to add a zero at the end because we are multiplying by the tens place, making it $745$ $5$ .
Add partial products: Add the two partial products together to find the final product. $\newline$ Add $745$ and $37250$ together. $\newline$ $5 + 0 = 5$ $\newline$ $4 + 5 = 9$ $\newline$ $7 + 2 = 9$ $\newline$ Since there are no more numbers to add in the thousands place, just bring down the $3$ from $37250$ . $\newline$ The sum is $37995$ .