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Under this post "

**Number System: Types of Questions Asked in Competitive Exams**" we will discuss various types of questions asked in competitive exams on number system.

Please go through all types of questions carefully, so that you can easily crack it in your exams.

**Type 1:**Find units digit of a number in form a

^{b}

^{}

^{To find the unit digit of number which is in the form ab}

^{}

^{1) If b is not divisible by 4}

^{Step 1: Divide b by 4, if it is not divisible then find the remainder of b when divided by 4. }

^{Step 2: Units digit = ar, r is the remainder. 2) If b is multiple of 4}

^{Units digit is 6: When even numbers 2, 4, 6, 8 are raised to multiple of 4. }

^{Units digit is 1: When odd numbers 3, 7 and 9 are raised to multiple of 4. }

^{}

**Q 1.**Find the unit digit of (4137)

^{754 }

**Solution:**Divide b by 4, if it is not divisible fully, write the remainder

The number is in the form a

^{b}i.e (4137)

^{754}

4 × 188 = 752, therefore we get remainder as 2

Units digit = (4137)

^{2}= 17114769

**Type 2:**Divisibility of Numbers

**Q 2.**Find the largest 4 digit number which is divisible by 88.

**Solution:**We know that the largest 4 digit number is 9999.

Simply divide 9999 by 88.

After dividing 9999 by 88 we get 55 as remainder.

The number is said to be completely divisible, only if the remainder is zero.

Hence, we can find the required answer by subtracting the remainder obtained from the 4 digit number.

Therefore, required number = 9999 – 55 = 9944

**Type 3:**Place Value and Face Value

**Q 3.**Find the difference between the place value and face value of digit 6 in numeral 2960543.

**Solution :**Place value of digit 6 in 2960543 = 60000

Face value of digit 6 in 2960543 = 6

Their difference = 60000 - 6 = 59994

**Type 4:**Find the result of different operations (additions, subtractions, multiplications, divisions, etc) on given integers.**Q 4.**Find the solution of (935421 × 625) =?

**Solution:**In this type of questions, try to simplify the numbers, to find the answer easily.

We are asked to multiply (935421 × 625)

625 is divisible by 5 and 5

This 5

625 is divisible by 5 and 5

^{4}= 625This 5

^{4}can be written as (10 /2)^{4}Hence, 935421 × | 10^{4} | = | 9354210000 | = 584638125 |

**Type 5:**Number a when divided by b gives remainder r, what will be the remainder when a is divided by c.

**Q 5.**The remainder is 29 when a number is divided 56. If the same number is divided by 8, then what is the remainder?

**Solution :**Dividend = [(Divisor × Quotient)] + Remainder

It is given that, the remainder is 29 when a number (dividend) is divided 56(divisor).

Dividend and quotient are unknown, hence assume dividend as X and quotient as Y.

Therefore,

X = 56 × Y + 29

56 is completely divisible by 8, but 29 is not completely divisible and we get remainder as 5, which is the required answer.

OR

X = 56 × Y + 29

= (8 × 7Y) + (8 × 3) + 5

5 is the required remainder.

**Type 6:**Arithmetic and Geometric Progression

**Q 6.**If 6 + 12 + 18 + 24 + --- = 1800, then find the number of terms in the series.

**Solution :**This is an Arithmetic Progression, in which x = 6, y = 6, sum of terms = 1800

Sum of n terms = | n | [2x + (n – 1)y] |

2 |

1800 = | n | [2 × 6 + (n – 1)6] |

2 |

1800 = 3n (n + 1)

n(n +1) = 600

n

^{2}+ n = 600

25 × 24 = 600

Therefore,

n

(n + 25) (n – 24) = 0

24 and -25 are the two solutions obtained. Only positive value can be considered. Hence, n = 24

The number of terms in the series = 24

n

^{2}+ 25n – 24n – 600 = 0(n + 25) (n – 24) = 0

24 and -25 are the two solutions obtained. Only positive value can be considered. Hence, n = 24

The number of terms in the series = 24

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**Number System: Types of Questions Asked in Competitive Exams**", Please comment in the comment box below.
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