Question

Find using distributive property: (a) \[728 \times 101\] (b) \[5437 \times 1001\] (c) \[824 \times 25\]
 (d) \[4275 \times 125\] (e) \[504 \times 35\]

Answer 1

Hint: The distributive property allows us to rewrite the term in parentheses as a sum of two different terms, or to say that the term in parentheses is distributed into two terms. We use the distributive property to simplify calculations so that larger terms can be easily multiplied with each other.

Complete step by step answer:
\[\left( a \right)728 \times 101\]
We can write 101 as (101+1)
Therefore,
\[ = 728 \times \left( {100 + 1} \right)\]
\[ = \;728 \times 100 + 728 \times 1\]
\[ = \;728 \times 100 + 728 \times 1\]
\[ = \;73528\]
\[\left( b \right)5437 \times 1001\]
We can write 1001 as (1000+1)
Therefore,
\[ = 5437 \times \left( {1000 + 1} \right)\]
\[ = 5437 \times 1000 + 5437 \times 1\]
\[ = 5437000 + 5437\]
\[ = 5442437\]
\[c)824 \times 25\]
We can write 25 as (20+5)
Therefore,
\[ = 824 \times \left( {20 + 5} \right)\]
\[ = 824 \times 20 + 824 \times 5\]
\[ = {\text{ }}824 \times 2 \times 10 + 824 \times 5\]
\[ = 16480 + 4120\]
\[ = 20600\]
\[\left( d \right)4275 \times 125\]
We can write 125 as (100+25)
\[ = 4275 \times \left( {100 + 25} \right)\]
\[ = 4275 \times 100 + 4275 \times 25\]
\[ = 427500 + 106875\]
\[ = 534375\]
\[\left( e \right)504 \times 35\]
We can write 504 as (500+4)
\[ = 35 \times \left( {500 + 4} \right)\]
\[ = 35 \times 500 + 35 \times 4\]
\[ = 17500{\text{ + }}140\]
\[ = 17640\]

Note:
We have written this note as an example. Here, 101 can be distributed in a variety of ways, but we know that multiplying a number by a multiple of ten is the simplest, so we rewrite the number with this in mind. Because 100 is a multiple of ten, we divided the number 101 into the sum of 100 and one.