Idiot\'s Guides Basic Math and Pre-Algebra

Chapter 2: Arithmetic 25

1 hundred that’s been waiting and you’ll have 11 hundreds. Of course, 11 hundreds form 1 group
of 10 hundreds plus 1 more hundred, or 1 thousand and 1 hundred. Our multiplication ends up
looking like this.

When all the multiplying and regrouping is done, 594 v 2 =1,188.

MATH TRAP

Remember that any carrying you do in multiplication happens after you do a multi-

plication. The digit you carry is part of the result of the multiplication. It has already

been through the multiplication process, so make sure you wait and add it on after

the next multiplication is done. Don’t let it get into the multiplication again.

So our plan for multiplication of a larger number by a single digit is to start from the right,
multiply each digit in the larger number by the one-digit multiplier, and carry when the result of
a multiplication is more than one digit. That will work nicely when you want to multiply a larger
number by one digit, but what if both of the numbers have more than one digit? What if you need
to multiply 594 by 32 (instead of by just 2)?

For a problem like 594 v 32, you don’t have to invent a new method, but you do have to adapt the
method a little. You’ll still start from the right and multiply each digit of 594 by 2, carrying when
you need to. Then you’ll multiply each digit of 594 by 3, again starting from the right. But here’s
the catch: that 3 you’re multiplying by is 3 tens, not 3 ones. You have to work that change in place
value into your multiplication.

The problem is fairly easy to solve. Just look at the first time you multiply by the 3 tens: 3 v 4.
It’s really 3 tens times 4, and that should give you not 12, but 12 tens or 120. Multiplying by 3
tens instead of just 3 simply adds a zero.

So here’s how you’ll tackle the problem. First you’ll multiply 594 by the 2, just as you did before.

594

8

594 594

11

22

88

2

1188



594

32

1188

3