Numbers Basics


Number system is a very important chapter and you will get questions from this area in many competitive exams.  We start with classification of numbers.

Types of numbers :

1. Natural numbers (N) = 1, 2, 3, . . . .
2. Whole numbers (W) = 0, 1, 2, 3, . . . .
3. Intezers (Z) = −∞ . . . −2, −1, 0, 1, 2, 3, . . .∞
4. Rational numbers (Q) = The numbers of the form pq  where q ≠ 0.  Eg: 15 , 0.46, 0.333333
5. Irrational numbers ($\mathbb{R} - Q$) = The numbers of the form x1n  ≠ Intezer.  Also π and e also irrational numbers.
Rational and Irrational numbers together is called Real numbers. It is denoted by $\mathbb{R}$

Other types of numbers:
a. Even numbers : Numbers which are exactly divisible by 2.  These numbers are in the format of 2n.
b. Odd numbers: Numbers which gives remainder 1 when divided by 2. These numbers are in the format of 2n ± 1.
c. Prime numbers : The numbers which are divisible by 1 and the number itself are primes.  The least prime is 2.
Note: There are 15 primes below 50, 25 primes below 100, 168 primes below 1000.
d. Composite numbers : The numbers of which are divisible by more than 2 numbers.  First positive composite number is 4.

Important rules related to Even and Odd numbers:

odd ± odd = even;
even ± even = even;
even ± odd = odd

odd × odd = odd;
even × even = even;
even × odd = even.

odd(any number) = odd
even(any number) = even

Fundamental Theorem of Arithmetic: 

Any positive integer greater than 1 can be represented as a product of primes only in one way.  (Order may be different). Writing a number as a product of primes is called prime factorization format. For example, 100 can be written as 22 × 5in only one way.

Converting recurring decimals into p/q format:

Model 1: 
If all the 'n' digits after the decimal points are recurring, say,
\(x.\overline {\underbrace {abc.....}_{n - digits}} \) = \(x.\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\)
Then \(x.\overline {\underbrace {abc.....}_{n - digits}}  = x + \dfrac{{abc....}}{{\underbrace {999...}_{n - times}}}\)

Example:
 \(3.\overline {713} \) =
Solution:
 \(3.\overline {713} \) = \(3 + \dfrac{{713}}{{999}}\) = \(\dfrac{{2997 + 713}}{{999}} = \dfrac{{3710}}{{999}}\)

Model 2:
If certain 'n' digits after the decimal points are recurring, and 'k' digits are not recurring,
\(x.\underbrace {mnp..}_{k - digits}\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\) = \(x.\underbrace {mnp..}_{k - digits}\underbrace {\overline {abc.....} }_{n - digits}\)
In this case, subtract the digits which are not recurring from the whole number and put it in the numerator.  In the denominator, put 9's for the number of digits which recur, and put 0's for the number of digits which won't recur.
=\(x.\underbrace {mnp..}_{k - digits}\underbrace {\overline {abc.....} }_{n - digits}\) = \(x + \dfrac{{mnp...abc... - mnp...}}{{\underbrace {999....}_{n - digits}\underbrace {000....}_{k - digits}}}\)

Example:
 \(12.3\overline {45} \)=
Solution:
Here only 45 are recurring.
Therefore, \(12.3\overline {45}  = 12 + \dfrac{{345 - 3}}{{990}}\) = \(12 + \dfrac{{342}}{{990}}\) = \(12 + \dfrac{{38}}{{110}}\) = \(12 + \dfrac{{19}}{{55}}\) = \(\dfrac{{679}}{{55}}\)

Solved Examples

1.  Let \(x,y\) are even numbers and \(z\) is an odd number.  Then which of the following is false?
a. \({\left( {x - z} \right)^2}\) is odd number
b. \(\left( {x - y} \right)z\) is even number
c. \(\left( {z - x} \right){y^2}\) is even number
d. \({\left( {x - y} \right)^2} + z\) is even number
Correct option: d
Solution: Recap the rules related to odd and even numbers.
\(x - z\) = even - odd = odd and odd2 = odd. So option a is correct.
\(\left( {x - y} \right)z\) = (even - even) × odd = even × odd = even.  So option b is correct
\(\left( {z - x} \right){y^2}\).  We know that y is even. So \({y^2}\) is even. So anything multiplied  by even is even. So option c also correct.
\({\left( {x - y} \right)^2} + z\) = even2 + odd = even + odd = odd. So option d is false.

Shortcut:
To solve these type of questions, assume numbers and check options.
Take x = 2, y = 4, z = 3
Clearly, the fourth option = \({\left( {2 - 4} \right)^2} + 3 = 7\)  is odd.

2. Let \(x\) be prime and \(y\) be composite.  Then
a. \(xy\) is always even
b. \(y - x\) is never even
c. \(\dfrac{{x + y}}{x}\) is never even
d. None of the above
Correct option: d
Explanation: Interesting question.  We try to give counter examples to eliminate options one by one.
Take \(x = 3,y = 9\).  Then \(xy = 27\) is odd. So option 1 ruled out.
Take \(y = 4,x = 2\).  Then \(y - x = 2\) is even. So option 2 ruled out.
Take \(x = 2,y = 6\).  Then \(\dfrac{{x + y}}{x} = \dfrac{{2 + 6}}{2} = 4\) is even. So option 3 also ruled out.
So we choose option d.

3.  Prime factorization of 12600 is
a. \({2^3} \times {3^2} \times {5^2} \times 7\)
b. \({2^2} \times {3^2} \times {5^2} \times 7\)
c. \({2^2} \times {3^2} \times {5^3} \times 7\)
d. \({2^1} \times {3^1} \times {5^2} \times 7\)
Correct option: a
Explanation:
To find the prime factorization of a number, we continuously divide the given number by prime numbers until the we get prime number.  So \(12600 = {2^3} \times {3^2} \times {5^2} \times 7\)
prime factorization
Important Tip: The clue in solving this question is, when you observe the number, it is an even number ending with '0'.  So it must be divisible by both 2 and 5. So start with 2.  After 3 divisions we get 1575 which is clearly divisible by 5.  So after 2 divisions we got 63 which is divisible by 7 and 9.

4. What least number must be subtracted from 12702 to get number exactly 99 ?

a. 49
b. 30
c. 29
d. 31
Correct Option: B
Explanation:
Divide the given number by 99 and find the remainder.  If you subtract the remainder from the given number then it is exactly divisible by 99.
99)  12702 (128
         99 
         280
         198
           822
           792
             30
Required number is 30.


5. The largest number of four digits exactly divisible by 77 is 

a. 9768
b. 9933
c. 9988
d. 9944
Correct Option: B
Explanation:  Find the remainder when 10000 is divided by 77.  Then subtract that remainder from 10000.
77) 10000 (12
        77
         230
         154
           760
           693
             67
So 10000 – 67 = 9933 is exactly divisible by 77.

6. Sum of the numbers from 1 to 20 is 

a. 210
b. 110
c. 220
d. 105
Correct Option: A
Explanation: Sum of first n natural numbers = 1 + 2 + 3 + ..... n = $\displaystyle\frac{{n(n + 1)}}{2}$
Substitute n = 20.
So \({S_{20}} = \dfrac{{20 \times 21}}{2} = 210\)

7. Sum of even numbers between 15 and 25 is

a. 70
b. 80
c. 130
d. 100
Correct Option: D
Explanation: 16 + 18 + .......24.
Taking 2 common we get = 2 ( 8 + 9 + 10 +.....+ 12)
Sum of n natural numbers upto 12
8 + 9 + 10 +.....+ 12 = (1 + 2 + 3 + ......+  12) - ( 1 + 2 + 3 + ....+ 7)
By applying the formula for the first n natural numbers sum $\displaystyle\frac{{n(n + 1)}}{2}$ we get, $\displaystyle\frac{{12(12 + 1)}}{2} - \displaystyle\frac{{7(7 + 1)}}{2} = 50$
So 16 + 18 + .......24 = 2 × 50 = 100

Alternative Method:
16 + 18 + .......24
This is an Arithmetic progression with $a = 16, d = 2, l = 24$
Total terms in the sequence is given by = \(n = \dfrac{{l - a}}{d} + 1\)
So total terms = \(n = \dfrac{{24 - 16}}{2} + 1 = 5\)
Sum of the terms when first term and last term in known = \(\dfrac{n}{2}\left( {a + l} \right)\) = \(\dfrac{5}{2}\left( {16 + 24} \right) = 100\)

8. How many numbers between 1000 and 5000 are exactly divisible by 225?

a. 16
b. 18
c. 19
d. 12
Correct Option: B
Explanation: First multiple of 225 after 1000 is 1125 (225 × 5 ) and last multiple of 225 before 5000 is 4950 (225 × 22)
Total number are $\displaystyle\frac{{l - a}}{d} + 1 = \displaystyle\frac{{4950 - 1125}}{{225}} + 1 = 18$

9. If the first 200 numbers are written down and those divisible by 2 are deleted and again those divisible by 5 are deleted, how many numbers are left out ?

a. 80
b. 150
c. 200
d. 160
Correct Option: A
Total numbers divisible by 2 = 100
Total numbers divisible by 5 = 40
But there is double counting. So we have to subtract Total numbers which are divisible by both 2 and 5 i.e, 200 / 10 = 20
So Total numbers which are divisible by either 2 or 5 is 100 + 40 - 20 = 120
Number of numbers which are not divisible by any of those = 200 - 120 = 80

10. How many digits are required to write numbers between 1 to 100.

a. 196
b. 158
c. 192
d. 200
Correct Option: C
Explanation:
Single digits are from 1 to 9 = 9 digits
Doubt digits are from 10 to 99 = 90 x 2 = 180 digits
100 needs 3 digits.  Total 192 digits