42 Chapter 1 Whole Numbers
EXAMPLE (^2) Multiply: a. b. c.
Strategy For each multiplication, we will identify the factor that ends in zeros and
count the number of zeros that it contains.
WHY Each product can then be found by attaching that number of zeros to the
other factor.
Solution
a. Since 1,000 has three zeros, attach three 0’s after 6.
b. Since 100 has two zeros, attach two 0’s after 45.
c.912(10,000)9,120,000 Since 10,000 has four zeros, attach four 0’s after 912.
45 100 4,500
6 1,0006,000
6 1,000 45 100 912(10,000)
Self Check 2
Multiply:
a.
b.
c.
Now TryProblems 23 and 25
875(1,000)
25 100
9 1,000
We can use an approach similar to that of Example 2 for multiplication involving
any whole numbers that end in zeros. For example, to find , we have
Write 2,000 as 2 1,000.
Working left to right, multiply 67 and 2 to get 134.
Since 1,000 has three zeros, attach three 0’s
after 134.
This example suggests that to find we simply multiply 67 and 2 and
attach three zeros to that product. This method can be extended to find products of
two factors that both end in zeros.
67 2,000
134,000
134 1,000
67 2,000 67 2 1,000
67 2,000
EXAMPLE (^3) Multiply: a. b.
Strategy We will multiply the nonzero leading digits of each factor. To that
product, we will attach the sum of the number of trailing zeros in the factors.
WHY This method is faster than the standard vertical form multiplication of
factors that contain many zeros.
Solution
a. The factor 3 00 has two trailing zeros.
Attach two 0’s after 42.
Multiply 14 and 3
to get 42.
b. The factors 3,5 00 and 5 0 , 000 have a total of six trailing zeros.
Attach six 0’s after 175.
Multiply 35 and 5
to get 175.
3,50050,000175,000,000
14 300 4,200
14 300 3,50050,000
Self Check 3
Multiply:
a.
b.
Now TryProblems 29 and 33
3,1007,000
15 900
Success Tip Calculations that you cannot perform in your head should be
shown outside the steps of your solution.
3 Multiply whole numbers by two- (or more) digit numbers.
EXAMPLE (^4) Multiply:
Strategy We will write the multiplication in vertical form. Then we will multiply
436 by 3 and by 20, and add those products.
WHY Since , we can multiply 436 by 3 and by 20, and add those
products.
23 3 20
23 436
Self Check 4
Multiply:
Now TryProblem 37
36 334
1
1
4
3
42
3
2
5
5
(^175)