When multiplying multiple digit numbers, it is best to set the problem up much like an addition problem. Layout the problem in a vertical manner with one number on the top and one below that. It is easiest to set the problem with the highest amount of digits on top.
Example:
In this problem we have 45 multiplied by 7. The first step of this problem is perform multiplication of the ones column. In this case 5 x 7 = 35. Since we can only put one number in the ones column for the answer, much like addition, we carry over the 3 part of 35 which represents the number 30, leaving just 5 in the ones column.
We now must multiply the 10s column. Unlike addition, we do not just assume a zero and multiply 4 x 0. With multiplication we must multiply every digit in the first number by every digit in the second number. In this case we have multiplied the 5 x 7 in the ones column, now we must multiply 4 x 7 in the tens column to get an answer of 28. We must not forget to add that 3 we carried over from the ones column. 28 + 3 = 31. Since there are no more digits in the first number we can simply write 31 at the front of our answer to get our final answer of 315.
Breaking down the Process
This process doesn't seem to have much reason at a quick glance. Let's take a look at this problem a little more closely. When we break down the number 45 into place values it is actually 4 tens and 5 ones or 40 and 5. Since multiplication is in reality taking a number and adding it to itself a certain number of times this problem breaks down to 45 + 45 + 45 + 45 + 45 + 45 + 45. Since 45 = 40 + 5 we can now rewrite it:
(40 + 5) + (40 + 5) +(40 + 5) + (40 + 5) + (40 + 5) + (40 + 5) + (40 + 5)
Given the
Associative and Commutative Properties of Addition, we can rearrange this to be:
40 + 40 + 40 + 40 + 40 + 40 + 40 + 5 + 5 + 5 + 5 + 5 + 5 + 5.
Going back to multiplication, we have 7 40's and 7 5's so we can now rewrite it as
(40 x 7) + (5 x 7) which is equal to (40 + 5) x 7 via the
Distributive Property bringing us full circle to 45 x 7.
Going back a couple steps if we add up all of the 40's and all of the 5's we come up with 280 for the 40's and 35 for the 5's. Breaking it down into place values we get the following:
2 hundreds 8 tens 0 ones
+ 0 hundreds 3 tens 5 ones
2 hundreds 11 tens 5 ones
As you can see the 7 40's gave use 2 hundreds and 8 tens while the 7 5's gave us 3 tens and 5 ones. Since we now add the tens places to find our result we now know why we add that carry value of 3 to the result of 4 x 7.
Finally to simplify our answer since 11 tens = 1 hundred and 1 ten, we can rewrite our answer to be 3 hundreds 1 ten and 5 ones or 315.