Unit digit&last 2 digits of powers&mathematical expressions|Best&fast trick

Sohel Sahoo: Hello Guys,

In this article we will discuss the shortcut tricks on finding out  Unit digit (last digit) & last 2 digits, last 3 digits and so on… in a interesting way.

By reading this article heartly & focusing the highlighted points you would solve big complicated question within a matter of seconds. I have explained every such type of questions asked in competitive exams in detail with simple and easy solutions to save the priceless time.

CONCEPTS:-    

Any number can be expressed in its place value decimal representation ; where digits are written in unit place,10^th place,100^th place&so on… considering from right hand side of the number.

In order to get the unit digit of a number , we have to consider that digit after leaving 10^th place& higher places. It will be better for you to divide that number by ‘10’& get the remainder , since the remainder so obtained will itself the last digit.

 Likewise the last two digits  are obtained by leaving 100^th place &higher places Or better you can attain it by the remainder so obtained by dividing the no. by ‘100’.

In more general  you can  get the last n digits by dividing that no. /expression by 10ⁿ;where n=1,2,3,4,5 …. i.e. a natural number.

NOTE:-

Ø  If a no./expression is divided by 10, then the remainder obtained is the unit digit of that no./expression  and vice-versa.

Ø  If a no./expression is divided by 100, then the remainder obtained is the last 2 digits of that no./expression  and vice-versa.

 v  Find the last Digit of the expression : 102 x 1299 x 13704 x 1198.

Sol:-  (102 x 1299 x 13704 x 1198)/10

      = [(2)(-1)(4)(-2)] /10 [Any no. is divisible by 10 if last digit is 0]

    = 16/10

    = 6

Since,remainder is 6 on dividing by 10;we have the unit digit as ‘6’.(Ans)

  v  Find the last 2 Digits of the expression: 101 x 297 x 405 x 137798 x 4115297.

Sol:- (101 x 297 x 405 x 137798 x 4115297) / 100

     = 1(-3)( 5)(-2)(-3) /100     [Any no. is divisible by 100 if last 2 digits are 00]

    = -90

  = -90+100 = ‘10’ ; Remainder obtained on dividing by 100

So,clearly the last 2 digits of the expression is ‘10’(Ans)

  v  Find the last 2 Digits of the expression: 55 x 13978 x 372 x 497 x 331 x 27 .

Sol:-  (55 x 13978 x 372 x 497 x 331 x 27)/100

= (11 x 13978 x 93 x 497 x 331 x 27) /5   [Simplified by 5&4 i.e.5x4=20]

= 1(-2)(-2)2 x 1 x 2/5

= 16/5 =>R[16/5]=1

But,the actual remainder will be 1 x 20= 20

Hence,the last 2 digits are ‘20’(Ans)

    Unit Digit of a number raised to Powers:-

First, examine the unit digit carefully in the powers of decimal no.s from 0-9 as given below in each cases.

      Powers of  0,1,5&6 :-

Powers of ‘0’	Powersof ‘1’	Powers of ‘5’	Powers of ‘6’
01=0	11=1	51=5	61=6
02=0	12=1	52=25=…5	62=36=…6
NOTE:- Any power of 0,1,5&6 will give themselves i.e. 0,1,5&6 respectively as unit digit.

      Powers of 4 & 9 :-

Powers of ‘4’	Powers of ‘9’
41=4	91= 9
42=16=…6	92 = 81=…1
43=64=…4	93=….9
44=…6	94=…81=…1

NOTE:-       The unit digit of powers of 4&9 repeats after every 2 powers periodically ;which may be stated as : 

Powers	Unit digit of 4	Unit digit of 9
Odd	4	9
Even	6	1

           Powers of 2,3,7 & 8:-

Powers of ‘2’	Powers of ‘3’	Powers of ‘7’	Powers of ‘8’
21=2	31=3	71=7	81=8
22=4	32=9	72=49=…9	82=64=…4
23=8	33=27=…7	73=..63=…3	83=..32=…2
24=16=…6	34=..21=…1	74=…21=…1	84=…16=…6
25=..12=…2	35=….3	75=…7	85=..48=…8

NOTE:-   Observe that the unit digit of powers of 2,3,7&8  starts repeating after 4 powers cyclically ; So to find the unit digit 1^st divide the power given in question by 4 & obtain the reaminder. According to the remainder the unit digit will be as given above table.

Question: Find the unit digit of following numbers:

• 145563
  Answer= 5
• 2416387
  Answer= 1
• 135625369
  Answer= 6
• 190654729321
  Answer= 0

Question: Find the unit digit of following numbers:

• 239562679743
  Answer = 9 (since power is odd)
• 729698745832
  Answer = 1(since power is even)
• 454258741369
  Answer = 4 (since power is odd)
• 17465478932
  Answer = 6 (since power is even)

Question: Find the Unit digit of 287562581

Solution: we know that after 4 powers the unit digit of 7 repeats.
 Divide the power 562581 by 4.               [Any no. is divisible by 4 iff last 2 digits divisible by 4]                     
By doing that, we get a remainder=1.       
 1st power in the power cycle of 7 is 7.
Hence, the answer is 7.

Question:- Find the last digit of 2⁴⁰⁰³ .

^Sol:- we know that after 4 powers the unit digit of 2 repeats.

 Divide the power 4003 by 4.               [Any no. is divisible by 4 iff last 2 digits divisible by 4]                     
By doing that, we get a remainder=3.       
 3rd  power in the power cycle of 2 is 8.
Hence, the answer is 8.

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