Chapter 3 BEYOND
REAL NUMBERS
LESSON 1 A New Look at Digits of a Number
We
have learned during elementary about how to write expanded form using the place
value of the digit and how to express in exponential form using powers of 10.
Let
us consider the following example:
36 = 10 × 3 + 6 =
278 = 100 × 2 + 10 × 7 + 8
5
732 = 1 000 × 5 + 100 × 7 + 10 × 3 + 2
In
general, a two digit number having a and b respectively as tens
and units (ones) digit. Using the above notations, the number can be written as
10 × a + b
Let
us use the notation ab to denote this number.
ab
= 10 × a + b
REMEMBER
Any
two digit number can be written as ab, where a and b
are whole numbers taking values from 0 to 9 such that a ≠ 0. This is known as
the generalized form of a two digit number.
Similarly,,
if a,
b, c denote numbers taking values from 0 to 9, then abc,
where a ≠ 0, is the generalized form of a three digit number whose ones,
tens, and hundreds digits are c, b, and a, respectively.
In
general, any natural number can be written as …cba in which ones digit
is a,
tens digit is b, hundreds digit is c and so on. Here, dots mean that
there may be more digits to the left of c.
In
expanded form, we have
ab
= 10a + b
abc = 100a + 10b + c
and …cba = …
+ 100c + 10b + a
Let
us use the three digit number abc. If we are going to change the
order of these digits, we obtain numbers bca and cab.
We
can write this numbers in expanded form as
abc = 100a + 10b + c
bca
= 100b + 10c + a
cab = 100c + 10a + b
If
we add these three numbers, we get
abc
+
bca
+ cab
= (100a
+ 10b + c) + (100b + 10c + a) + (100c + 10a + b)
= 111a + 111b + 111c
= 111 (a + b + c)
= 3 × 37 × (a + b + c)
It
follows from the above expression that the sum abc + bca
+ cab
is exactly divisible by 3, 37, a + b + c, 111, 3(a
+ b
+ c),
37(a
+ b
+ c)
The
quotient in each case are listed below.
|
Number
|
Divisor
|
Quotient
|
|
abc + bca + cab
|
111
|
a + b + c
|
|
a + b + c
|
111
|
|
|
37
|
3(a + b + c)
|
|
|
3
|
37(a + b + c)
|
|
|
3(a + b + c)
|
37
|
|
|
37(a + b + c)
|
3
|
The
following divisors listed on the table are all fixed or constant to be used
when solving. Likewise, the quotients are fixed or constant to be used when
solving.
Example
1. Without performing actual addition and division, find the quotient when the
sum of 57 and 75 is
divided by (a) 12 (b)
11
Solution: We have
ab + ba =
11 × (a + b)
Here a = 5 and b
= 7.
(a)
Obviously, 57 and 75 are the numbers we can obtain by reversing the digits of
the other.
Therefore, when ab + ba is divided by the sum
of the digits (a + b), that is, 5 + 7 = 12, we can get answer of 11 as the quotient.
(b)
If we divide the sum of these two numbers 57 + 75 by 11, we can get answer of 5
+ 7 (sum of the digits) = 12 as the
quotient.
Example
2. Without performing actual computation,
find the quotient when 83 – 38 is divided by
(a) 9 (b) 5
Solution:
Note that
ab – ba = 9 (a
– b)
or ab
– ba
=
32 × (a – b)
Here, a = 8 and b
= 3.
(a) We
know that
ab
– ba
when divided by 9, the quotient is (a – b).
Therefore, 83 – 38 when divided by 9, the quotient
is 8 – 3 = 5.
(b) We
know that
ab
– ba
when divided by (a – b), the quotient is 9.
Therefore, 83 – 38 when divided by 8 – 3 = 5, the quotient is 9.
Example
3. Without performing actual addition, find
the quotient when 528 +285 + 852 is divided by
(a) 111 (b) 15 (c)
37 (d) 3 (e) 45 (f) 555
Solution: If the digits 5, 2, and 8 are arranged in
cyclic order, we get three numbers, 528, 285, and 852. Let us
use the following given on the table.
The sum of 528 +285 + 852 is divided by
(a) 111,
the quotient is 5 + 2 + 8 = 15
(b) 15,
that is, 5 + 2 + 8 gives the quotient 111
(c) 37,
the quotients is 3(5 + 2 + 8) = 45
(d) 3,
the quotient is 37(5 + 2 + 8) = 555
(e) 45,
the quotient is 37
(f) 555,
the quotient is 3
Example
4. Find the quotients when the
difference of 634 and the number obtained by interchanging its ones and
hundreds digits is divided by (a) 99 (b) 2 (c) 33 (d) 3 (e) 22
Solution:
Note that
abc – cba = 99 (a
– c)
or abc
– cba
= 32×
11 × (a – c) = 3 × 33 × (a – c)
On the given, a = 6, b
= 3, and c = 4.
Therefore, 634 – 436 = 32
× 11 × 2
(a) We
know that
abc
– cba
when divided by 99, the quotient is (a – c). Therefore, 634 – 436
when divided by 99, the quotient is 2.
(b) We
know that abc – cba when divided by 2, the quotient
is 99. Therefore, 634 – 436 when
divided by 6 – 4, that is 2, the quotient is 99.
(c) Take
note that when abc – cba is divided by 33, the quotient
is 3 x (a – c). Therefore,
634 – 436 when divided by 33, the quotient is 3 x (6 – 4), that is, 6.
(d) Take
note that when abc – cba is divided by 3, the quotient is
33 x (a – c). Therefore,
634 – 436 when divided by 3, the quotient is 33 x (6 – 4), that is, 66.
(e) Take
note that when abc – cba is divided by 22, the quotient
is 32. Therefore, 634 –
436 when divided by 22, the quotient is 32
or 9.
Exercise
Solve
each of the following.
1. Without
performing actual addition and division, find the quotient when the sum of 36
and 63 is
divided by (a)
9 (b) 11
2. Without
performing actual computation, find the quotient when 72 – 27 is divided by
(a) 9 (b) 5
3. Without
performing actual addition, find the quotient when 347 + 473 + 734 is divided
by
(a) 111 (b)
14 (c) 37 (d) 3 (e) 42 (f) 518
4. Find
the quotients when the difference of 519 – 915 is divided by
(a) 99
(b) 4 (c) 33 (d) 3 (e) 44
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