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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