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Number System Questions with Answers

Number System Basic Concepts, Formulas, Short tricks

Number System is an important topic for campus placement and TCS NQT Exam. It is also asked in government exams like CSAT, SSC, Banking, Railway and Defense. Full detailed study material, formulas, short tricks pdf and questions with answers from previous exams are given in this article. This article elaborates basic concepts of Number system. We regularly add new questions for competitive exams.

Number System Concepts

Types of Numbers :

Natural Numbers
Even Numbers
Odd Numbers
Prime Numbers and Composite Numbers
Whole Numbers
Integers and Real Numbers
Rational Numbers and Irrational Numbers

Divisibility Rules

Check the divisibility of any number by the rules given below:

Divisibility by 2: If the last digit is 0 and even number then the number can be divided by 2.
Divisibility by 3: A number is divisible by 3 if the sum of it’s digits are divisible by 3.
Divisibility by 4: A number is divisible by 4 if the last 2 digits are zero or divisible by 4.
Divisibility by 5: A number is divisible by 5 if it’s unit digit is 5 or 0.
Divisibility by 6: A number is divisible by 6 if it is divisible by 2 and 3 both.

Divisibility by 7: We use osculator (-2) as given below:
Example: 12642 : 1264 – 2 x 2 = 1260
126 – 0 x 2 = 126
126 is divisible by 7 so 12642 is divisible by 7

Divisibility by 8: A number is divisible by 8 if the last three digits are divisible by 8.
Divisibility by 9: A number is divisible by 9 if the sum of digits is divisible by 9.
Divisibility by 10: A number is divisible by 10 if the number is divisible by 5 and 2 both.

Divisibility by 11: In a number, if the difference between the sum of digits at even places and the sum of digits at odd places is either 0 or a multiple of 11, then no. is divisible by 11.
Example, 12342 : (1+3+2) – (2 +4) = 0 hence 12342 is divisible by 11.

Divisibility by 12: A number is divisible by 12 if it is divisible by 3 and 2 both.

Divisible by 13: Use (+4) as osculator.
Example: 876538 ÷ 13
876538: 8 × 4 + 3 = 35
5 × 4 + 3 + 5 = 28
8 × 4 + 2 + 6 = 40
0 × 4 + 4 + 7 = 11
1 × 4 + 1 + 8 = 13
13 is divisible by 13.
∴ 876538 is also divisible by 13.

Divisibility by 14: A number is divisible by 14 if the number is divisible by 7 and 2 both.
Divisibility by 15: To be divisible by 15, the number should be divisible by 5 and 3 both.
Divisibility by 16: To be divisible by 16, the number should be divisible by 4.

Divisible by 17: Use (-5) as osculator.
Example: 4318÷ 17
431-8×5=391
39-1×5=34
34 is divisible by 17 therefore 4318 is divisible by 17.

Divisible by 19: Use (+2) as osculator.
e.g., 149264: 4 × 2 + 6 = 14
4 × 2 +1 + 2 = 11
1 × 2 + 1 + 9 =12
2 × 2 + 1 + 4 = 9
9 × 2 +1 = 19
19 is divisible by 19

Divisibility by 25: A number is divisible by 25 if the last 2 digits are zero or divisible by 25.

Divisibility by a Composite number:
A number is divisible by a given composite number if it is divisible by all factors of that composite number. Example: To be divisible by 18, the number must be divisible by 9 and 2.

Division Theorem

Dividend = (Divisor × Quotient) + Remainder

REMAINDER Rule

A×B×C / n

Remainder = A_r×B_r×C_r / n ( A_r is remainder of A/n , B_rremainder of B/n, C_r remainder of C/n)

Example : Find the remainder of 16 × 17 × 19 when divided by 7.

16×17×19 / 7 remainder = 2

2×3×5/7 ( A_r=2 rem. of 16/7 , B_r =3 rem. of 17/7 , C_r =5 rem of 19/7)

= 30 / 7 remainder is 2.

Unit Digit of a Number

How to find unit digit of a number of power N

Rules 1 : Number ending with (0,1,5,6) will end with same unit digit

Example : unit digit of (1458141)¹²⁸⁷⁶⁴ = 1 , unit digit of (97936)²³⁴ is 6

Rule-2 : Number ending with other than (0,1,5,6)

first divide power by 4 and get the remainder

then unit digit of Number = unit digit of (last digit )^rem+1

Example: find unit digit of (23447)²⁵

25/4 remainder =1 (7)¹⁺¹ = (7)² =49 unit digit is 9

unit digit of (23447)²⁵ is 9

Number of factors or Number of Divisors

How to find number of factors of a number

first find prime factors = P₁^a. P₂^b P₃^c ……..

total number of factors = (a+1).(b+1).(c+1)….

Example: How many number of factors of 36 including 1 and number itself?

36 = 2² × 3²

number of factors = (2+1)×(2+1) = 9

total number of factors = 9

number of factors is also equal to the number of divisors so the total number of divisors of 36 is 9

if the factor of A is divisible by B then multiples of A is also divisible by B

Number System Questions and Answers

Q.1 What is the value of M and N respectively if M39048458N is divisible by 8 and 11, where M and N are single digit integers? (TCS-NQT)

A. 2 and 6
B. 6 and 4
C. 2 and 4
D. 3 and 6

Answer

B) A number is divisible by 8 if the last three digits of the number are divisible by 8. 58N is divisible by 8 if N=4, therefore N = 4

a number is divisible by 11 if the difference between the sum of digits at even places and sum of digits at the odd places is either 0 or is divisible by 11.
i.e. (M+9+4+4+8) – (3+0+8+5+N)
= M + 25 – (16 + N) put N = 4
= M + 5 must be zero or it must be divisible by 11
it is divisible by 11 if M = 6
⇒
Hence, M = 6; N = 4

Q.2 By which of the following is 19541742 divisible?
I.11
II.9
III.12
A. Both I and III
B. Only III
C. Both II and III
D. Only I

Answer

D) check divisibility by 11: the difference of sum of digits at even and odd places is either 0 or divisible by 11 (1+5+1+4) – (9+4+7+2) = 11-22 = -11 divisible by 11

check divisibility by 9 : sum of the digits = 33 which is not divisible by 9

check divisibility by 12: number should be divisible by 4 and 3 both but it is not divisible by 3 so the answer is D

Q.3 If the 5-digit number 776xy is divisible by 3, 7 and 11, then the value of (5x+3y)?
A. 13
B. 21
C. 23
D. 26

Answer

C) to be divisibility by 11: the difference of the sum of digits at even and odd places is either 0 or divisible by 11 (7+6+y) – (7+x) = 6+y-x /11 (mod = 0) if y-x = 5 ………(1)

to be divisibility by 3 : sum of the digits is divisible by 3 = 20+x+y/ 3 ( mod=0) if x+y =1 or 7 ……(2)

from equation (1) and (2) y=6, x=1 possible values of x and y

check divisibility by 7: 77616/7 mod= 0 therefor x=1,y=6 5x+3y = 5×1+3×6=23

Q.4 A number 452p36 is such that it is divisible by 36. What can the value of p² be
A. 36
B. 49
C. 25
D. 16

Answer

B) Number divisible by 36, should be divisible by 9 and 4 also (9 and 4 are factors of 36 =9 x 4).

check divisibility by 4 :- last two digit (36) is divisible by 4.

to be divisible by 9 : Sum of all digits must be divisible by 9 = 4+5+2+p+3+6

20 +p is divisible by 9 if p = 7 (20+7 = 27)

P² = 49

Q.5 What is the remainder when 13⁸³is divided by 11?
A. 6
B. 3
C. 8
D. 5

Answer

C) By remainder rule 13⁸³ /11 = (2)⁸³ / 11 (on dividing 13 by 11 we get 2)

= (2⁵)¹⁶ x 2³ / 11 = (32)¹⁶ x 8 / 11

= (11+10)¹⁶ x 8/ 11 = (10)¹⁶ x 8/11 = (10²)⁸ x 8 /11 = (1)⁸ x 8 / 11 = 8/ 11 remainder is 8

Q.6 A number is divided by 342 remainder is 47. if it is divided by 18 then what will be the remainder?
A. 11
B. 0
C. 19
D. 18

Answer

C) By Division theorem Divided= divisor x Q + rem.

number = Q x 342+47

when this number is divided by 18, multiple of 342 is divisible by 18 because 342 is divisible by 18 therefore the remainder is 47/18 e.g. 11

Q.1:If the five-digit number 672 xy is divisible by 3, 7 and 11, then what is the value of (6x + 5y) ??

A) 16

B) 23

C) 17

D) 24

Explanation: Divisibility by 3: Sum of digits must be divisible by 3. 6+7+2+x+y=15+x+y it is divisible by 3 if x+y= 0 or 3 Divisibility by 11: Alternate sum and difference of digits should be either 0 or divisible by 11: 6−7+2−x+y=1−x+y=0 only if x=2 , y=1 now check Divisibility by 7, 67221/7≡0 (rem=0) so the answer is x=2, y=1
6*2+5*1= 17 .

Q.2: Six digit number 5a4b6c is divisible by 7 , 11 and 13 Find the value of (3a−5b+7c).?

A) 10

B) 12

C) 13

D) 21

Explanation: A six digit number in the form of xyzxyz is always divisible by 1001 i.e. 7*11*13
5a4b6c a=6, b=5, c=4
3a-5b+7c
3*6-5*5+7*4=21

Q.3: The number A4531B is divisible by 72. Where A and B single digit number then the number of factors of A+B is

A) 5

B) 9

C) 2

D) Can’t determine

Explanation: A number is also divisible by divisor’s co factors. A4531B is divisible by 72 (9*8) so it should be divisible by 9 and 8 also.
To be divisible by 8, the last 3 digits (31B) should be divisible by 8 therefore B = 2
A number is divisible by 9 when sum of all digits is divisible by 9
A+4+5+3+1+2 = A + 15
so A =3 A+B = 5 factors of 5 = 5*1 Ans is 2 .

Q.4: When (2²⁴ – 1) is divided by 7 , the remainder is:?

A) 4

B) 2

C) 1

D) 0

Explanation: (2³)⁸ -1/7 = 8⁸ – 1/7 By remainder rule 8/7 rem. Is 1 so 1-1 is 0 .

Q.5: A number is divided by 406 leaves remainder 115. What will be the remainder when it is divided by 29??

A) 31

B) 28

C) 27

D) can’t determine

Explanation: The required number N= 406*Q+115 by division theorem
when 406 /29 remainder=0 and 115/29 remainder= 28 so ans is 28 .

Q.6: If the 9-digit number 807x6y9z8 is divisible by 99, then the value of (x + y + z)^½ is

A) 4

B) 3√3

C) 6

D) 3√5

Explanation:given number is divisible by 99 if it is divisible by 9 and 11 both (99=11*9). check divisibility by 9 if the sum of its digits is divisible by 9. sum of the digits: 8+0+7+x+6+y+9+z+8 = x+y+z +38 is divisible by 9 if x+y+z = 7, 16
Now check Check divisibility by 11 : A number is divisible by 11 if the alternating sum of its digits is divisible by 11. alternating sum of the digits: 8−0+7−x+6−y+9−z+8 = 38 – x–y–z is divisible by 11 if x+y+z = 16 So x + y + z = 16 (x + y + z)^½ = 4

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