# Remainder Theorem, Unit Digit, and Number of Zeros in expressions

## REMAINDER THEOREM, Unit Digit, and Number of Zeros

Find the number in Unit place of following:

1) (623)^49

2) (98)^43

Find the number of zeros in the following expressions:

1) 47!

2) 144! * 5 * 15 * 22 * 11 * 44 * 135

3) 1142! * 348 ! * 17!

4) 1^1 * 2^2 * 3^3 * 4^4 *......* 49^49

5) 1^1! * 2^2! * 3^3! * 4^4! *......* 10^10!

Find the remainder when

1) 17 * 23 is divided by 12

2) 37 * 43 * 51 is divided by 7

3) 43^197 is divided by 7

4) 21^875 is divided by 17

In this post, we are going to learn how to deal with these type of questions in depth.

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**How To Find UNIT DIGIT of a Mathematical expression:**

- If we raise any number to the power
, the unit digit of the new number comes as 1, 5 or 6.*4n* - Any number raised to the power
**4n**repeats its unit digit, - e.g. Unit digit of (3^4)^12 = (3^4)^22 = (3^4)^99 = (3^4)^234556
- Unit digit powered to 4n times:

*1,3,7, 9 - 1*

*5 - 5*

*2,4,6,8 - 6*Question: Find the Unit digit of (33)^123

Solution: for unit digit we can write 3^123 => (3^4)^30 * 3^3 => 1*3^3 => 7

Question: Find the Unit digit of (623)^49

Solution: for unit digit we can write 3^49 => (3^4)^12 * 3^1 => 3^1 =3

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How To Find **NUMBER OF ZEROS** of a Mathematical expression**:**

Zeros are formed by the combination of 2 * 5, the total number of pairs of 2 and 5 makes zeros in an expression. (Apart from multiples of 10)

To find the number of 2's and 5's we need to factorize the given mathematical expression.

*It is to be noted here that in factorial forms the number of 5's will always be lesser than 2's. Hence, we just need to count the number of 5's.*

Question: Find the number of zeros in 6!

solution: 6! = 6 * 5 * 4 * 3 * 2 * 1 = (3*2)* 5 * 2*2 * 3 * 2 * 1

This expression contains only one pair of 5 and 2, so only one zero.

Question: Find the number of zeros in 47!

*.*

Solution: 47/5 = 9 (quotient)

9/5 = 1

Hence total 10 zeros.

Question: Find the number of zeros in 1^1 * 2^2 * 3^3 * 4^4 *......* 49^49.

Solution: The 5's will be lesser than the 2's therefore we need to count a number of 5's only.

Thus: 5^5 * 10^10 * 15 ^15 * 20^20* 25^25 * 30^30 * 35^35 * 40 ^ 40 * 45 ^ 45

=> 5+10+15+20+50+ 30+35+40+45 => 250 zeros

*(25 means 5 * 5)#*

###
**REMAINDER THEOREM:**

Question: Find the remainder when 17 * 23 is divided by 12

Solution : (12+5)*(12+11) / 12

In this type of questions remainder only depends on the last term

=> 5*11 / 12 => 7

Question: Find the remainder when 37 * 43 * 51 is divided by 7

Solution : 2*1*2 / 7 => 4

**Negative remainder**

Consider the following case

14 * 15 /8 => 6*7 /8 => 2

However here we can use negative remainder

When 14 is normally divided by 8 remainders is seen as +6, however, there might be a time where negative remainder can be more useful.

Remainders, by concepts, are always positive, hence when we divide -27 by 5 we say that remainder is 3 and not (-2). however we can use the negative remainder to ease our calculations.

Thus above example may be solved as -2 * -1 => 2, which is same but less calculative.

If in this case answer comes out to be negative you need to subtract it with original number. e.g. (62 * 63 * 64)/66 => (-4*-3*-2 ) =>-24

Hence, required remainder will be 66-24 = 42

###
**How to Find Remainder while Dealing with Large Powers:**

1) If the expression can be expressed in the form of { (ax+1)^n } / a, the remainder will become 1 directly, the value of the power does not matter here.

e.g. 46^32624 / 9 => 1^32624 => 1

2) If the expression can be expressed in the form of { (ax-1)^n } / a , then

if n is even then, the remainder will be +1,

or n is odd then, the remainder will be -1.

e.g. 35^243 /9 => (-1)^243 => -1 => hence remainder = 8

Question: Find the remainder when 43^197 is divided by 7

Solution : 43^197 /7 => 1^197 => 1

Question: Find the remainder when 4^875 is divided by 17

Solution : 4^875 /17 => (4^4)^218 * 4^3 / 17 => 16^218 * 64/17 => 13

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Remainder Theorem, Unit Digit, and Number of Zeros in expressions
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