Page 24 - ArithBook5thEd ~ BCC

and then performing the addition

24
× 32
48
+ 7 2 0
76 8

Here’s another example.

Example 18. Find the product of 29 and 135.

Solution. We choose 29 as the multiplier since it has the fewest digits.

13 5
× 29

We use the 1-digit multiplier 9 to obtain the first partial product

{3}{4}
13 5
× 29
12 1 5

Notice that we put down 5 and carried 4 to the tens place, and also put down 1 and carried 3 to the
hundreds place. Next we use the 1-digit multiplier 2 (standing for 2 tens) to obtain the second partial
product, shifted left by putting a 0 in the ones place

{1}
13 5

× 29
121 5
270 0

(What carry did we perform?) Finally, we add the partial products to obtain the (total) product

13 5
× 29
12 15
+ 2 7 0 0
39 15

Note that the whole procedure is compactly recorded in the last step, which is all that you need to write
down.

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