Number System is all about finding the face value & place value of a digit, basic rules andÂ different types of numbers, rules for divisibility, factors and multiples.

- Natural numbersÂ areÂ defined as the numbers that occur commonly in nature. A natural numberÂ is a whole, non negative number andÂ the set of natural numbers is denoted by letter ‘N’.

Set of natural numbers is N = {1,2,3…..} - Whole numbersÂ are defined as all the numbers without fractions and no decimals.

Set of whole numbers is (W)={0,1,2,3…….}

Note:Â Every natural number is a whole number except zero is a whole number which is not a natural number. - Integers are defined as all the numbers i.e. zero, positive and negative numbers.

Set of integers = {……-3,-2,-1,0,1,2,3,…….} - Even numbers are defined as the numbers divisible by 2. i.e When an even number is divided by 2, the remainder is 0.

0,2,4,6,….. are even numbers. - Odd numbers are defined as the numbers which are not divisible by 2.Â i.e When an even number is divided by 2, the remainder is 1.

1,3,5,7,…… are odd numbers. - Prime numbers are those numbers which have exactly two factors namely itself and 1.

Example: 2,3,5,7,11,13,17….etc. - Composite numbers are defined as the numbers which are not prime.

Example: 4,6,8,12,15,….. etc.

**Rules for divisibility:Â **

- Divisibility by 2: A number is divisible by 2 if the last digit in a given numerical value is 0,2,4,6,8.

Example: 526492 is divisible by 2. Â Â (as last digit is 2) - Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

Example: 562185 is divisible by 3. Â Â (as sum of digits =27 is divisible) - Divisibility by 4:Â If theÂ number formed by last two digits is divisible by 4 then the given numerical value is also divisible by 4.

Example: 1459872 is divisible by 4. Â (as last two digits 72 is divisible by 4) - Divisibility by 5: If the units digit is 0 or 5, then the given number is divisible by 5.

Example: 256480 is divisible by 5. Â (as units digit is 0) - Divisibility by 8:Â If theÂ number formed by last three digits is divisible by 8 then the given numerical value is also divisible by 8.

Example: 45566568 divisible by 8. Â Â (as last three digits is divisible by 8) - Divisibility by 9:Â If the sum of its digits is divisible by 9 then the given numerical value is also divisible by 9.

Example: 21421899 divisible by 9. Â (as sum of its digits = 36 is divisible by 9) - Divisibility by 10: If the units digit is 0, then the given number is divisible by 10.

Example: 65712590 is divisible by 10. Â Â (as units digit is 0) - Divisibility by 11: If the difference between sum of its digits at odd places and sum of its digits in even places is either zero or a number divisible by 11.

Example: 25784 is divisible by 11. Â Â (as (2+ 7 + 4) – (5+8) = 13 – 13 =0)

- (Divisor Ã— Quotient) + Remainder = Dividend
- \((a+b)^2 = a^2 + b^2 +2ab\)
- \((a-b)^2 = a^2 + b^2 -2ab\)
- \(a^2 – b^2 = (a+b)(a-b)\)
- \((1+2+3+….+n) = \frac{1}{2n(n+1)}\)
- \(1^2+2^2+3^2+….+n^2 = \frac{1}{6n(n+1)(2n+1)}\)
- \(1^3+2^3+3^2+….+n^3 = \frac{1}{4n^2(n+1)^2}\)
- For Arithmetic Progression

(i) nth term = \(a + (n – 1)d\)

(ii) sum of n terms = \(\frac{n}{2(2a+(n-1)d)}\)

(iii) sum of n terms = \(\frac{n}{2(a+l)}\), where l is the last term - For Geometric Progression

(i) nth term = \(ar^{(n-1)}\)

(ii) sum of n terms = \(\frac{a(1-r^n)}{(1-r)}\) when r<1 ; \(\frac{a(r^n-1)}{(r-1)}\) when r>1

Problems

Take the unknown value as \(x\)

by substituting \(x\),

7429 – \(x\) = 4358 – 1587

7429 – \(x\) =Â 2771

\(x\) = 7429 – 2771

\(x\) = 4658

Therefore, the unknown value \(x\) = 4658

by substituting \(x\),

7429 – \(x\) = 4358 – 1587

7429 – \(x\) =Â 2771

\(x\) = 7429 – 2771

\(x\) = 4658

Therefore, the unknown value \(x\) = 4658

**Model 2:** Face value of 7 and Place value of 9 from the given digit 38745962 is?

**Solution:**

As, Face value is the value of digit itself and should relate place with ‘place value’

Given digit is 38745962

Therefore, face value of 7 is 7

(100 is multiplied since 9 is in hundred’s place)

Place value of 9 is (9 Ã— 100)=900

Given digit is 38745962

Therefore, face value of 7 is 7

(100 is multiplied since 9 is in hundred’s place)

Place value of 9 is (9 Ã— 100)=900

**Model 3:** On dividing 132 by a certain number, 12 as a quotient and 0 as a remainder is obtained. Find the divisor?

**Solution:**

Given that Dividend = 132, Quotient = 12, Remainder = 0

[Divisor Ã— Quotient] + Remainder=Dividend

by substituting the given values,

[DivisorÂ Ã— 12] + 0 = 132

Divisor = \(\frac{132}{12}\)

âˆ´ Divisor = 11

[Divisor Ã— Quotient] + Remainder=Dividend

by substituting the given values,

[DivisorÂ Ã— 12] + 0 = 132

Divisor = \(\frac{132}{12}\)

âˆ´ Divisor = 11

**Model 4:** What will be the units digit in the product (234 Ã— 256 Ã— 457 Ã— 952)?

**Solution:**Â

Given product is (234 Ã— 256 Ã— 457 Ã— 952)

Here units digits are 4,6,7,2

Therefore, product of Units digit in the given product = 4 Ã— 6 Ã— 7 Ã— 2 = 336.

Here units digits are 4,6,7,2

Therefore, product of Units digit in the given product = 4 Ã— 6 Ã— 7 Ã— 2 = 336.

**Model 5:** Test which of the following is not a prime numberÂ a)19Â b)11 c) 16 d) 13?

**Solution:**

As Prime number is the number which has exactly two factors namely itself

Factors of 19 = 1, 19

Factors of 11 = 1, 11

Factors of 16 = 1, 2, 4, 16

Factors of 13 = 1, 13

Therefore, 16 is not a prime as it is divisible by more than twice numbers.

Factors of 19 = 1, 19

Factors of 11 = 1, 11

Factors of 16 = 1, 2, 4, 16

Factors of 13 = 1, 13

Therefore, 16 is not a prime as it is divisible by more than twice numbers.

**Model 6:**Â Does the number 23679715 divisible by 11?

**Solution:**

Given digit is 23679715

(Sum of digits at odd places)-(Sum of digits at even places)

Digits at odd places = 5, 7, 7, 3

Digits at even places = 1, 9, 6, 2

â‡’ (5 + 7 + 7 + 3) – (1 + 9 + 6 + 2)

â‡’22 – 18 = 4

as 4 is not divisible by 11, 23679715 also not divisible by 11

(Sum of digits at odd places)-(Sum of digits at even places)

Digits at odd places = 5, 7, 7, 3

Digits at even places = 1, 9, 6, 2

â‡’ (5 + 7 + 7 + 3) – (1 + 9 + 6 + 2)

â‡’22 – 18 = 4

as 4 is not divisible by 11, 23679715 also not divisible by 11