Q&A for How to Do Double Digit Multiplication

Yes, if it helps you keep your place. Lots of people don't bother with the zero.

The zero doesn't have to be there. It's just a good reminder that the second partial product has to be offset one column to the left compared to the first partial product. Otherwise, you'll get the wrong answer.

Yes, the process is the same. If the multiplier is a three-digit number, you wind up with three partial products instead of two before adding up the answer.

First multiply 8 x 95 = 760 and 4 x 95 = 380. Write those two partial products on separate lines below the multiplication bar. First you write 760. Below that you write 380 by placing the zero immediately below the 6 in 760. Place the 8 below the 7. Place the 3 to the left of the 8 (below nothing). Then add the partial products together starting with the right-hand column, which consists of one zero. The next column consists of a 6 and a zero. The next column consists of a 7 and an 8. The left-hand column has only a 3. Add the partial products together that way to get 4560. That's your answer.

Yes, if the multiplier has at least two digits, you will have at least two partial products which must be added together.

They're called "partial products."

You just multiply the two-digit number by various whole numbers such as 2, 3, 4, 5, etc. For example, multiples of 25 are 50, 75, 100, 125, etc.

The first partial product results from multiplying the first 72 by the digit 2 of the second 72. The first partial product is 144, written so that "44" is directly below the second 72. The second partial product is 504, written so that "04" is directly below the "14" of the first partial product. There are now four columns among the two partial products. Add the columns beginning with the right-hand column, 4+0=4. In the next column to the left, 4+4=8. In the next column, 1+0=1. In the left-hand column, 0+5=5. The final product is 5,184.

Do the same thing you did in step 2, just multiply the digit in the tens place (instead of the ones place) of the bottom number by the top number. And write the result in front of the 0 you placed in step 3. For example, if you're multiplying 15 x 23, in step 4 you would multiply 5 x 2 and 1 x 2.

Count the number of places there are to the right of the decimal point in the multiplicand and/or the multiplier. That number is the number of places to the right of the decimal point in the product. For example, if the multiplicand has one place to the right of the decimal, and the multiplier has two places to the right of the decimal point, the product will have three places to the right of the decimal point.

In multiplication there is never a remainder. Remainders occur only in division, because in many cases one number will not divide evenly into another number.

Yes, it works for numbers of any size.

Just one zero. Any number multiplied by zero is the single digit number zero (0).

85 x 63. 3 x 5 = 15, so put a 5 below the 3 and carry the 1. 3 x 8 = 24 + 1 = 25. Write the 25 to form the first partial-product: 255. Next, 6 x 5 = 30, so put a 0 directly below the middle digit in 255, and carry the 3. 6 x 8 = 48 + 3 = 51. Form the second partial-product (510) by placing a 1 directly below the 2 in the first partial-product. Add the partial-products, noting that the second partial-product is offset one column to the left of the first partial-product. In the final product, the first column on the right just has a 5, followed in turn by (moving to the left) 5, 3 and 5. The final product is 5,355.

82 times 35. 5 times 2 is 10. Carry the 1. 5 times 8 is 40, plus 1 is 41. So the first partial product is 410. 3 times 2 is 6. 3 times 8 is 24. That makes the second partial product 246. Place that partial product below the first partial product offset one column to the left, so that the 6 in 246 is directly below the 1 in 410. Now add the partial products together, being careful to add only the numbers in the same column. Thus, starting from the right, bring the 0 straight down from the 410 to the final product line. In the next column, 1 plus 6 is 7. The 7 goes next to the 0 in the final product line. In the next column, 4 plus 4 is 8. The 8 goes next to the 7. Finally, bring the 2 straight down from the 246 to the final product line. The final product is 2870.

When you're multiplying double-digit numbers, the number in the tens place is that number times 10, so you need a zero when you're doing the second version.

Since 8 x 9 = 72, put the 2 down and carry the 7 above the 4 in 48.

36 x 15. Here’s a way to do it mentally: You’re doing 36 x 5 and 36 x 10 and adding them together. This is 180 + 360, which is 50, therefore your final answer is 540.

29 x 18. 9 x 8 = 72, so put the 2 and carry the 7. 8 x 2 = 16 + 7 = 23. So you have 232. Then drop a 0. 9 x 1 = 9, 2 x 1 = 2. 232 + 290 = 522.

You're probably asking about the factors of 124. You find them by dividing small numbers into 124 and seeing which of them will divide evenly without a remainder. The small numbers which divide evenly into 124 are 1, 2 and 4. When you divide each of those numbers into 124, you get quotients of 124, 62 and 31, respectively. That means the factors (divisors) of 124 are 1, 2, 4, 31, 62 and 124. (Remember that a number's factors (divisors) always include 1 and the number itself.)

Let the three-digit number be the multiplicand (on the top) and the two-digit number be the multiplier (on the bottom) so that you'll have only two partial products to add together.

Mostly because there are tenths and ones, and carrying helps you putting digits in their place where they belong.

Maybe, in my opinion, use a calculator which would be way easier! just type up 18 times 18 and it will come! For example, 18 times 18 is 324.