## Number System

- April 1, 2020
- CODE OF GEEKS
- 0

A Number is denoted by a group of digits, which is referred to as Numeral.

**Types of Numbers **

1. Natural Numbers : All counting numbers like 1, 2, 3, 4, … are all Natural numbers.

2. Whole numbers : All counting numbers and 0 (zero) are referred to as Whole numbers.

3. Integers : All whole numbers as well as negatives of counting numbers are referred to as Integers. For instance, 2, 4, 5, 0, -3, 23, -25 are all integers.

4. Odd Numbers : All counting numbers that are not divisible by 2 are referred to as Odd numbers. For instance, 1, 3, 5, 7, 9 are all odd numbers.

5. Even Numbers : All counting numbers that are divisible by 2 are referred to as Even numbers. For instance, 2, 4, 6, 8 are all even numbers.

6. Prime Numbers : All counting numbers are prime numbers if they have only two factors – i.e itself and 1. For instance, 2, 3, 5, 7, are all prime numbers.

7. Composite Numbers : All natural numbers which are not prime, are called composite numbers.

8. Co Primes : Two natural numbers – a & b, are co-prime if their HCF is 1. For instance, (8, 11) is a co-prime pair.

Prime Check : Test to check whether a number is prime or not.

Let ‘n’ be the given number which we need to check whether it is prime or not.

m2 >= n

where, m is any smallest counting number. Test whether n is divisible with any of the prime numbers less than or equal to m. If yes, then n is not prime or else it is.

Let us consider one example :

Check whether 437 is a prime number or not ?

We know that, 21*21 = 441

So, 441 > 437 or (21)2

So, prime numbers less than 21 are 2, 3, 5, 7, 11, 13, 17, 19.

Clearly, 437 is divisible by 19. Hence, 437 is not a prime number.

**TEST OF DIVISIBILITY**

Divisibility by 2 :

A number is divisible by 2 if its unit digit is any of 0, 2, 4, 6, 8.

Example : 3344554 is divisible by 2.

Natural

Divisibility by 3 :

A number is divisible by 3 only when the sum of its digits is divisible by 3.

Example : In a number 4453452, sum of digits i.e 4+4+5+3+4+5+2 is 27, which is divisible by 3. Hence, 4453452 is divisible by 3.

Divisibility by 4 :

A number is divisible by 4 if and only if its last two digits is divisible by 4.

Example : In a number 4534560, 60 gets divided by 4, hence 4534560 is divisible by 4.

Divisibility by 5 :

A number is divisible by 5 if and only if its unit digit is 5 or 0.

Example : In a number 4563560, digit at unit place is 0, hence it is divisible by 5.

Divisibility by 6

A number is divisible by 6 if and only if it can be divided by both 2 and 3.

Example : A number 34523448, is divisible by both 2 and 3, hence it is divisible by 6.

Divisibility by 8 :

A number is divisible by 8, if and only if its last three digits are divided by 8.

Example : In a number 34563800, last three digits constitute 800, which is divisible by 8. Hence, 34563800 is divisible by 8.

Divisibility by 9 :

A number is divisible by 9 only when the sum of digits is divisible by 9.

Example : In a number 4453452, sum of digits i.e 4+4+5+3+4+5+2 is 27, which is divisible by 9. Hence, 4453452 is divisible by 9.

Divisibility by 10 :

A number is divisible by 10, if and only if it ends with 0, in other words, it has zero (0) in its unit place.

Example : A number 3456430 ends with 0 as a unit place, hence it is divisible by 10.

Divisibility of 11 :

A number is divisible by 11, if and only if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.

Example : In a number 3456332,

Sum of digits of a number of odd places = 2+3+5+3 => 13 // s1

Sum of digits of a number of even places = 3+6+4 => 13 // s2

So, s1-s2 = 0, hence 3456332 is divisible by 11.

Divisibility by 12 :

A number is divisible by 12, if and only if it can be divided by 3 and 4 both.

Example : A number 6845628 is divisible by 3 and 4 both, hence it is divisible by 12 as well.

Divisibility by 14 :

A number is divisible by 14, if and only if it can be divided by 2 and 7 both.

Example : A number 70 is divisible by 2 and 7 both, hence it is divisible by 14 as well.

Divisibility by 15 :

A number is divisible by 15, if and only if it can be divided by 3 and 5 both.

Example : A number 900 is divisible by 3 and 5 both, hence it is divisible by 15 as well.

Divisibility by 16 :

A number is divisible by 16, if and only if the number formed by the last 4 digits is divisible by 16.

Divisibility by 24 :

A number is divisible by 24, if and only if it is divisible by 3 and 8 both.

Insights : Any number of the form paqbrc will have (a + 1) (b + 1)(c + 1) factors, where p, q, r are prime.

Insights : For any perfect square, all the powers of the primes have to be even numbers. So, if the factor is of the form 2a * 3b * 5c.

Progression

Progression denotes the number of things that are represented in the form of series.

We’ll be looking at two series progressions :

Arithmetic Progression

Geometric Progression

**Arithmetic Progression**

An arithmetic progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means the second term minus the first term.

For instance, the sequence 1, 4, 7, 10 . . . is an arithmetic progression with common difference of 3.

An A.P having ‘a’ as the first term and ‘d’ as the common difference is generally given as :

a, (a+d), (a+2d), (a+3d), …. , (a+(n-1)d)

where ‘n’ is the number of elements in the series.

Hence, a general formula to find the n-th term (Tn) of any given AP series is given as :

[latex] \boxed { a \;+\;(n-1)\;*\;d } [/latex]

A general formula to find the sum of first ‘N’ of an AP series is given as :

[latex] \boxed { \frac{n}{2}\;*\;[2a \;+\;(n-1)\;*\;d] } [/latex]

However, in some questions, we are not provided with common difference (d), but only first term and last term, so in that case sum of ‘n’ numbers of an AP series is given as :

[latex] \boxed { \frac{n}{2}\;*\;[first_{term}\;+\;last_{term}] } [/latex]

Some Important Series and their equivalent formula :

[latex] \boxed { 1\;+\;2\;+\;3\;+\;4\;+\;…\;+\;n \;=\; \frac{n\;*\;(n\;+\;1)}{2} } [/latex]

[latex] \boxed { 1^2\;+\;2^2\;+\;3^2\;+\;4^2\;+\;…\;+\;n^2 \;=\; \frac{n\;*\;(n\;+\;1)\;*\;(2n\;+\;1)}{6} } [/latex]

[latex] \boxed { 1^3\;+\;2^3\;+\;3^3\;+\;4^3\;+\;…\;+\;n^3 \;=\; \frac{n^2\;*\;(n\;+\;1)^2}{4} } [/latex]

Geometric Progression

Geometric Progression refers to the special series in which a new element is obtained by multiplying special constant to the the current element. This special constant is referred to as common ratio and is denoted by ‘r’.

For instance, 2, 4, 8, 16, … is a geometric progression with common ratio(r) = 2.

A GP having ‘a’ as the first term and ‘r’ as the common ratio for ‘n’ terms is generally given as

a, ar, ar^{2}, a^{;3}, … , ar^{n-1}

where ‘n’ is the number of elements in the series.

Hence, a general formula to find the n-th term (Tn) of any given GP series is given as :

[latex] \boxed {ar^{n\;-\;1}}[/latex]

A general formula to find the sum of first ‘N’ of an GP series is given as :

[latex]\boxed {S_n \;=\; \frac{a(1 \;-\;r^n)}{1\;-\;r}\; when, r\;>\;1}[/latex]

**Q1. Show that 52563744 is divisible by 24 ?**

A. To check the divisibility of 52563744 by 24, we need to ensure that our given number is divisible by 3 and 8 both.

Checking divisibility by 3:

Sum of digits of number : 36, divisible by 3.

Checking divisibility by 8:

Number formed by last 3 digits – 744, divisible by 8.

Hence, we see that given number is divisible by both 3 and 8, and hence, will be divisible by 24.

**Q2. Find the least number that should be subtracted from 1672 to obtain a number which is completely divisible by 17 ?**

A. On dividing 1672 by 17, you will get 6 as a remainder. That is our least number that must be subtracted.

**Q3. Find the least number that should be added to 2010 to obtain a number which is completely divisible by 19 ?**

A. On dividing 2010 by 19, you will get 15 as a remainder. So, required number is

19-15 = 4

**Q4. On dividing a number by 342, we get 47 as remainder. If same number is divided by 18, then find the remainder.**

A. Let the quotient be ‘x’. Then, on applying formula

Divisor = (Dividend – Remainder)/Quotient

Number = 342x + 47

=> (18 * 19x) + (18 * 2) + 11

=> 18 + (19x + 2) + 11

So, number when divided by 18 gives remainder 11.

**Q5. Find the least value of * for which 684*628 is divisible by 12 ?**

A. **Divisibility by 12 :**

A number is divisible by 12, if and only if it can be divided by 3 and 4 both.

**Checking for divisibility by 3 :**

A number is divisible by 3 only when the sum of its digits is divisible by 3.

So, 6 + 8 + 4 + * + 6 + 2 + 8 = 34 + *

Value of * may be 2, 5, 8 etc.

**Checking for divisibility by 4 :**

A number is divisible by 4 if and only if its last two digits is divisible by 4.

28 is divisible by 4.

So, least value is **2**.

**Q6. If we write down all the natural numbers from 259 to 492 side by side get a very large natural number 259260261….491492.
How many 8’s will be used to write this large natural number ?**

A. From 259 to 458, there are total 200 natural numbers so there will be 2×20=40 8’s

From 459 to 492 we have 13 more 8’s

So, answer, 40+13= 53

**Q6. What will be the remainder when 2 ^{89} is divided by 89 ?**

A. Using **Fermat’s little theorem** :

If p is a prime number, then for any integer a, the number a ^{p} – a is an integer multiple of p.

a^{p} ≡ a (mod p)

So, **2 ^{89}** = 2 mod 89 = 2.

**Q7. How many factors of 1080 are perfect squares ?**

A. 1080 = 2^{3} * 3^{3} * 5

As mentioned in our tutorial, For any perfect square, all the powers of the primes have to be even numbers. So, if the factor is of the form 2^{a} * 3^{b} * 5^{c}.

For ‘a’ values may be 0 and 2,For ‘b’ values may be 0 and 2, and for ‘c’ may have value 0.

Totally there are 4 possibilities : 1, 4, 9, and 36. So, our answer is 4.

**Q8. How many factors of 2 ^{4} * 5^{3} * 7^{4} are odd numbers?**

A. For 2^{4} * 5 ^{3} * 7^{4} :

Any factor of this number should be of the form

‘a’ should be zero since factors are to be odd**.**

**So,** b can take any value

e 0,1,2,3 i.e. 4

and c can take any value 0,1,2,3,4 i.e. 5

So, all odd factors = 20 (4*5)

**Q9. How many 3-digit numbers do not have an even digits or zero ?**

A. Non even digits are : 1, 3, 5, 7, 9

So, For a 3 digit number – We have 5 valid values to fill up in each of three places :

So, answer is 5 * 5 * 5 = 125

**Q10. Find the least value of * for which 4832*18 is divisible by 11?**

A. **Divisibility by 11 :**

A number is divisible by 11, if and only if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.

Now,

4+3+*+8=15+*

8+2+1=11

22-15=7

Hence, * = 7.

**Q11 . In an AP 2, 5, 8, 11, 14, … . what will be the 20th term ?**

A. In the given AP series,

d = T_{2} – T_{1} => 3

a = 2

T_{20} = a + (20-1)*d

*T _{20} = a + 19d*

*T _{20} = 2 + 19*3

T_{20} = 2 + 57

T_{20} = 59

**Q12 . How many terms are there in AP : 7, 13, 19, … , 205 ?**

A. Here, T_{n} = 205 and a = 7, d = 6, we have to find ‘n’ :

T_{n} = a + (n-1)*d

*205 = 7 + (n-1)*6*

*205 = 7 + 6n + 6*

205 – 13 = 6n

n = 32

**Q13 . The fifth and thirteenth term of an AP are 5 and -3 respectively. Find its first term and common difference ?**

A. Here, T_{5} = 5 and T_{13} = -3, we have to find ‘a’ & ‘d’ :

T_{5} = a + 4d = 5 … *1*

T_{13} = a + 12d = -3 … *2*

Now, eq. 2 – eq. 1 :

8d = -8

So, d = -1

Putting the value of ‘d’ in eq. 1 :

a – 4 = 5

a = 9.

**Q14. Find the sum of 24 terms of the AP 5, 9, 13, 17, … ?**

A. In order to find AP series :

a = 5 and d = 4

Sn =n/2*[2*a + (n-1)*d]

=> 12 * [10 + 23*4]

=> 12 * [10 + 92]

=> 12 * [102]

=> 1224

**Q15. Find the sum of all odd integers from 1 to 1001 ?**

A. Odd integers from 1 to 1001 are 1, 3, 5, 7, … , 1001.

So, this is an AP with ‘a’ = 1 and ‘d’ = 2

Tn = 1001

Also,

T_n = a + (n-1) * d

T_n = 1 + (n-1)*2

1001 = 1 +2n -2

1002 = 2n

n = 501

So,

S{501} =501/2[1 + 1001]

=> 251001

**Q16. Find the 10-th terms of the GP 3, 6, 12, 24, … ?**

A. In the given GP,

a = 3

r = 2

**Q17. Find the sum of 8 terms of the GP 3, 6, 12, 24, … ?**

A. In the given GP,

**Q18. How many terms in the GP 1 + 4 + 16 + 64 + … will make the sum 5461 ?**

A. In the given GP,

**Q19. Find the sum of the series 2 + 6 + 18 + 54 + … + 4374 ?**

A. In the given GP,

**Q20. Find the sum of the series 1 + 1 + 1 + 1 + … ?**

A. On dividing 1672 by 17, you will get 6 as a remainder. That is our least number that must be subtracted.

Q3. Find the least number that should be added to 2010 to obtain a number which is completely divisible by 19 ?

A. On dividing 2010 by 19, you will get 15 as a remainder. So, required number is 19-15 = 4

Q4. On dividing a number by 342, we get 47 as remainder. If same number is divided by 18, then find the remainder.

A. Let the quotient be ‘x’. Then, on applying formula

Divisor = (Dividend – Remainder)/Quotient

Number = 342x + 47

=> (18 * 19x) + (18 * 2) + 11

=> 18 + (19x + 2) + 11

So, number when divided by 18 gives remainder 11.

Q5. Find the least value of * for which 684*628 is divisible by 12 ?

A. Divisibility by 12 :

A number is divisible by 12, if and only if it can be divided by 3 and 4 both.

Checking for divisibility by 3 :

A number is divisible by 3 only when the sum of its digits is divisible by 3.

So, 6 + 8 + 4 + * + 6 + 2 + 8 = 34 + *

Value of * may be 2, 5, 8 etc.

Checking for divisibility by 4 :

A number is divisible by 4 if and only if its last two digits is divisible by 4.

28 is divisible by 4.

So, least value is 2.

Q6. If we write down all the natural numbers from 259 to 492 side by side get a very large natural number 259260261….491492.

How many 8’s will be used to write this large natural number ?

A. From 259 to 458, there are total 200 natural numbers so there will be 2×20=40 8’s

From 459 to 492 we have 13 more 8’s

So, answer, 40+13= 53

Q6. What will be the remainder when 289 is divided by 89 ?

A. Using Fermat’s little theorem :

If p is a prime number, then for any integer a, the number a p – a is an integer multiple of p.

ap ≡ a (mod p)

So, 289 = 2 mod 89 = 2.

Q7. How many factors of 1080 are perfect squares ?

A. 1080 = 23 * 33 * 5

As mentioned in our tutorial, For any perfect square, all the powers of the primes have to be even numbers. So, if the factor is of the form 2a * 3b * 5c.

For ‘a’ values may be 0 and 2,For ‘b’ values may be 0 and 2, and for ‘c’ may have value 0.

Totally there are 4 possibilities : 1, 4, 9, and 36. So, our answer is 4.

Q8. How many factors of 24 * 53 * 74 are odd numbers?

A. For 24 * 5 3 * 74 :

Any factor of this number should be of the form

‘a’ should be zero since factors are to be odd.

So, b can take any value 0,1,2,3 i.e. 4

and c can take any value 0,1,2,3,4 i.e. 5

So, all odd factors = 20 (4*5)

Q9. How many 3-digit numbers do not have an even digits or zero ?

A. Non even digits are : 1, 3, 5, 7, 9

So, For a 3 digit number – We have 5 valid values to fill up in each of three places :

So, answer is 5 * 5 * 5 = 125

Q10. Find the least value of * for which 4832*18 is divisible by 11 ?

A. Divisibility by 11 :

A number is divisible by 11, if and only if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.

Now,

4+3+*+8=15+*

8+2+1=11

22-15=7

Hence, * = 7.

Q11 . In an AP 2, 5, 8, 11, 14, … . what will be the 20th term ?

A. In the given AP series,

d = T2 – T1 => 3

a = 2

T20 = a + (20-1)*d

T20 = a + 19d

T20 = 2 + 193

T20 = 2 + 57

T20 = 59

Q12 . How many terms are there in AP : 7, 13, 19, … , 205 ?

A. Here, Tn = 205 and a = 7, d = 6, we have to find ‘n’ :

Tn = a + (n-1)*d

205 = 7 + (n-1)*6

205 = 7 + 6n + 6

205 – 13 = 6n

n = 32

Q13 . The fifth and thirteenth term of an AP are 5 and -3 respectively. Find its first term and common difference ?

A. Here, T5 = 5 and T13 = -3, we have to find ‘a’ & ‘d’ :

T5 = a + 4d = 5 … 1

T13 = a + 12d = -3 … 2

Now, eq. 2 – eq. 1 :

8d = -8

So, d = -1

Putting the value of ‘d’ in eq. 1 :

a – 4 = 5

a = 9.

Q14. Find the sum of 24 terms of the AP 5, 9, 13, 17, … ?

A. In order to find AP series :

a = 5 and d = 4

Sn =n/2[2a + (n-1)*d]

=> 12 * [10 + 23*4]

=> 12 * [10 + 92]

=> 12 * [102]

=> 1224

Q15. Find the sum of all odd integers from 1 to 1001 ?

A. Odd integers from 1 to 1001 are 1, 3, 5, 7, … , 1001.

So, this is an AP with ‘a’ = 1 and ‘d’ = 2

Tn = 1001

Also,

T_n = a + (n-1) * d

T_n = 1 + (n-1)*2

1001 = 1 +2n -2

1002 = 2n

n = 501

So,

S{501} =501/2[1 + 1001]

=> 251001

Q16. Find the 10-th terms of the GP 3, 6, 12, 24, … ?

A. In the given GP,

[latex] \boxed {ar^{n\;-\;1}}[/latex]

Q17. Find the sum of 8 terms of the GP 3, 6, 12, 24, … ?

A. In the given GP,

Q18. How many terms in the GP 1 + 4 + 16 + 64 + … will make the sum 5461 ?

A. In the given GP,

Q19. Find the sum of the series 2 + 6 + 18 + 54 + … + 4374 ?

A. In the given GP,

Q20. Find the sum of the series 1 + 1 + 1 + 1 + … ?