# 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

**is very easy if we exercised on the various problems on the**

*Number system***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.

**Question 1):** 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

**Question 2):** 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

**Question 3):** 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.

**Question 4):** 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

**Question 5):** 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 =
11x127 => required three digit number is 127, whose digit sum = 10

**Question 6):** 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

**Question 7):** 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.

**Question 8):** 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

**Question 9):** 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

**Question 10):** (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.

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