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Q.

What is the last digit of the number 3579+ 1?

 A.

1

 B.

3

 C.

4

 D.

7

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Solution:
Option(C) is correct

Any power of 5 when divided by 4 gives a remainder 1.

Here, the power of 3 is itself a power of 5 and will give the remainder of 1 when divided by 4.

The last digit of the number will be 3.

And, hence, the last digit of the given number is $3+1 =$ 4.

Edit: For a detailed alternative solution, check comment by Deepak. 

Edit 2: For yet another alternative solution, check somment by Sravan Reddy.


(10) Comment(s)


Sravan Reddy
 ()

Key observation in this question -> Anything of power 5 is in the form of 4x+1. So the 3^5^7^9 simplifies to 3^(4x+1)

So units digit of 3^(4x+1) can be done in many ways and I like to use 'Modulus' function. Learning it helps in such cases (So,please try to learn if you find it good):

3^4 = 81 mod10 = 1 mod10

3^4x = 1 mod10

multiplying by 3 on both sides

3^(4x+1) = 3 mod10

so units digit of 3^5^7^9 = 3.. Answer is 3+1 = 4

Arithmetic way:

3^4= 81

3*4x = 81*81*81...x times. Still the units digit is 1

So answer is 3+1 = 4



Sreekant Singh
 ()

3^5^7^9 1st 9/4 R=1 , 7/4 R=3 , 5^3= 125 then 125/4 R=1 so 3^1 =3



Vivek
 ()

didnt get that step $3^5^--7=3^-25$



Sahil
 ()

3 raise to power will always give u a odd no.

and odd no. + 1 = even no.

and the only even no. in the options is 4 so the answer :-)



Deepak
 ()

To find the last digit of a number using a traditional method is not only cumbersome but difficult too. Here are few simple rules to make life easy:

LAST DIGIT of a number having form of XY (X-downwards, Y-Sideways)

X\Y

1

2

3

4

5

0

0

0

0

0

0

1

1

1

1

1

1

2

2

4

8

6

2

3

3

9

7

1

3

4

4

6

4

6

4

5

5

5

5

5

5

6

6

6

6

6

6

7

7

9

3

1

7

8

8

4

2

6

8

9

9

1

9

1

9


It is very clear from the above table that the last digit of XY is changing uptoY=4 when Y=5, the last digit is repeated. Hence, the last digit is periodic (cyclicity) with 4. 

So the last digit will always be periodic with period 4. We will make use of this property and follow the method given below, every time we need to find the last digit of a number.

METHOD: To find the last digit of number XY Divide Y by 4, If the remainder is R, then the last digit of the number is (the last digit of given number)R. Otherwise,, if remainder is zero, then the last digit of number is the last digit of X4 i.e (last digit of given number)4

Example 1:  Last digit of the number 1283409 
The last digit of 128 is 8 and when 3409 is divided by 4 , remainder is 1
Hence, the last digit is 81=8 (by rule )

Example 2: Find the last digit of 42428
Here when 2428 is divided by 4 , Remainder is 0 , so the last digit of 42428 is the last digit of 44 i.e 6 

Example 3:
 the last digit of the number
N=23735203+4567105−659408 
=23734l+3+45674m+1−6594n+4

So to get the last digit of the number, we take
=last digit of  33+last digit of  71−last digit of  94
=7+7−1
=14−1=13, 
the last digit is 3.

Now for the given question, start with the higher power i.e. 79=74*2+1which gives the last digit:
=7^1=7
(Don't forget we keep dividing the powers by 4 and use remainder to get the value of the last digit.)
Using the result and moving forward we get: 3579 =3(5)--7=(3)--25

From the above relation, last digit is 31=1
So the last digit of the given number becomes 3+1=4.

Arshi
 ()

y we divide y by 4? is it due to cyclicity , or ny other reason

Simi
 ()

Could have written where you writing exponents....readability is so poor. Good explanation but you have not even mentioned the exponents


Dhruv
 ()

Smile
Hello,

As I was going through the solution it occured to me that,

$2^5=32$

just as 3 in the above question has a power of 5 here too 2 has the power of ($5^1$)...&

32 when divided by 4 gives remainder 0 not 1..............

Just need a clarification.

Thank you for all your time and support lofoya

-a loyal consumer


Gannu
 ()

if you are trying to find out the last digit of $2^5$ then divide 5 by 4 then the remainder is 1. hence unit digit is 2 thus
$2^1$ is 2..................Smile