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Number System formulas and Tricks for Competitive Exams

## Number System formulas and Tricks for Competitive Exams

**What is Number System?
Definition:** The

**number system**or

the numerical system which are representing

**numbers**.

The **number system** is very important section in any type competitive exams under

**aptitude section**. This ** Number system** is very easy if we exercised on the

various problems on the

**Number system questions.**

Today we are sharing different type of **questions** are often

asked in various **competitive exams** like Bank Clerk, Bank PO, Railways, SSC,

UPSC. Aspirants suggested to exercise various types of questions and can help

them in the upcoming the **sarkari exams.** we are providing below some different

questions on ** number system **on this article, review it once.

**1 Question):** On dividing a certain

number by 342, we get 47 as a remainder. If the same number is divided by 18,

what will be the remainder?

Solution:

Let k be the

quotient,

then Number

= 342 k + 47

=

(18*19k) + (18*2) + 11

hence,

remainder will be 11

Note: First

check whether the first Divisor (342) is divisible by new divisor (18), if that

is true, then the new remainder will be the remainder from the first remainder

(47) and new divisor(18).

i.e.

remainder of 47/18 = 11

**2 Question):** What is the unit digit in

the product (3547)153 x (251)72 ?

Solution:

Required

digit = unit digit in 7153 x 172

= ( 74)38 x

7 x 1 = 7

**3 Question):** A number when successively

divided by 3, 5 and 8 leaves remainder 1, 4 and 7 respectively. Find the

respective remainders if the order of divisors is reversed.

Solution:

In this type

of questions, we need to find the Number and then divide it with given divisors

in reversed manner.

Dividend,

when divisor is 8 and remainder is 7 = 8 x 1 + 7 = 15

Dividend,

when divisor is 5 and remainder is 4 = 5 x 15 + 4 = 79

Dividend,

when divisor is 3 and remainder is 1 = 3 x 79 + 1 = 238

Hence,

Number or Dividend is 238, and we divide it with 8, 5 and 3 (normal division)

it will leave the remainders 6,4 and 2 respectively.

**4 Question ):** On dividing a number by 5,

we get 3 as a remainder. what will be the remainder when the square of this

number is divided by 5?

Solution:

remainder =

remainder of square of 3 divided by 5 = 4

**5 Question):** On dividing 2272 as well

as 875 by 3 digit number N, we get the same remainder. The sum of the digits of

N is:

Solution:

Here (2272 –

875)= 1397 is also exactly divisible by N,

1397 =

11×127 => required three digit number is 127, whose digit sum = 10

**6 Question):** What will be the remainder

when (6767 + 67) is divided by 68?

Solution:

(xn + 1)

will be divisible by (x+1) when x is odd.

Therefore, (6767 + 1) will be divisible by (67 +1)

or, (6767 + 1) + 66 when divided by 68 will give

remainder 66

**7 Question):** Which one of the following

numbers will completely divide (461 + 462 + 463 + 464)?

Solution:

461 ( 1 + 4 + 16 + 64 ) = > 461 x 85 =>

460 x 360 => which is divisible by 10.

**8 Question):** When a number is divided

by 13, the remainder is 11. When the same number is divided by 17, the

remainder is 9. What is the number?

Solution:

Number =

13p+11 = 17q+9 => q= (2+13p)/17

the least

value of p for which q is whole number is 26.

hence,

number = (13 x 26 + 11) = 349

**9 Question):** The sum of how many terms

of the series 6 + 12 + 18 + 24 + … 1800 ?

Solution:

here, a = 6; d=6,

we know, Sn

= n/2 [ 2a + (n-1)d ] => 1800 = n/2[

12 +(n-1)6 ] => n=24

**10 Question):** (22 + 42 + 62 +…+202)

Solution:

22(12 + 22 +

32 +…+102) => 4 (1/6 * 10 * 11 * 21) = 1540

**Question 11**): Find the sum of all the

odd numbers from 1 to 101.

Solution:

The required

Sum = Sum of all the odd numbers from 1 to 100 – Sum of all the odd numbers

from 1 to 20.

= Sum of

first 51 odd numbers – Sum of first 10

odd numbers

= 502 – 202

= 2501

Question

12): Find the greatest number of 5 digits which is exactly divisible by 137.

Solution:

The greatest

number of 5 digits is 99999, on dividing it by 137 we get 126 as a remainder.

therefore

the required number = 99999 – 137 = 99873

**Question 13):** The Sum of two numbers is

14 and their difference is 10. Find the product of the two numbers.

Solution:

(x+y)2 –

(x-y)2 = 4xy

142 -102 =

4xy

xy = 24

**Question 14):** The sum of two digit

numbers is 8. If the digits are reversed, the number is decreased by 54. Find

the number.

Solution:

x-y = dec.

in number /9 = 54/9 = 6

x+y = 8

solving

both, we get, x=7 and y=1.

**Question 15):** How many numbers between

100 and 300 are divisible by 7?

Solution:

On diving

the difference (300 – 200) by 7 we get 28 as quotient and 4 as a remainder,

Hence, there are 28 such numbers.

P.S. : This

Set of questions is prepared For Last Minutes Revision for ssc cgl, chsl, fci ,

ibps bank po and other exams.

**Number System:****Also Read:**

Number System formulas and Tricks for Competitive Exams

Reviewed by SSC NOTES

on

May 17, 2022

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