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# Aptitude Question Discussion - NUMBER SYSTEM AND NUMBER THEORY

 Q. What is the last digit of the number 3579+ 1?
 A. 1 ✔ B. 4 C. 7 D. 3

Solution:

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 remainder of 1 when divided by 4.

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

Topic: Number System And Number Theory
4.5/5 (7)
Deepak says...
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
$\bo X^\bo Y$ (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 $X^Y$ is changing upto $Y=4$ when $Y=5$, the last digit is repeated. Hence the last digit is periodic (cyclicity) with 4.

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

METHOD: To find the last digit of number $X^Y$ Divide Y by 4, If the remainder is R, then the last digit of the number is $\text"(the last digit of given number)"^R$. Otherwise if remainder is zero, then the last digit of number is last digit of $X^4$ i.e $\text"(last digit of given number)"^4$

Example 1:  Last digit of the number $128^3409$
Last digit of $128$ is $8$ and when $3409$ is divided by $4$ , remainder is 1
Hence the last digit is $8^1=\bo8$ (by rule )

Example 2: Find the last digit of $4^2428$
Here when $2428$ is divided by 4 , Remainder is 0 , so the last digit of $4^2428$ is last digit of $4^4$ i.e $\bo6$

Example 3:
the last digit of the number
$N=237\bo 3^5203 +456\bo 7^105 - 65\bo 9^408$
$=237\bo 3^{4l+\bo 3} +456\bo 7^{4m+\bo 1} - 65\bo 9^{4n+\bo 4}$

So to get the last digit of the number, we take
$=\text"last digit of " \bo3^\bo3 +\text"last digit of " \bo7^\bo1 - \text"last digit of " \bo9^\bo4$
$=7+7-1$
$=14-1=13$,
the last digit is 3.

Now for the given question, start with the higher power i.e. $7^9=7^{4*2+\bo1}$ which gives last digit:
$=7^1=7$
(Don't forget we keep dividing the powers by 4 and use remainder to get the value of last digit.)
Using the result and moving forward we get: 3579 =3(5)--7=(3)--25

From the above relation last digit is $3^1=1$
So ast digit of the given number becomes $3+1=\bo 4$
13th May 2013 8:37pm
arshi says...
y we divide y by 4? is it due to cyclicity , or ny other reason
15th May 2013 5:48pm
Deepak says...
Yup its cyclicity.
Last digit is periodic with period 4 (can be confirmed from the table shown in the 1st comment).

So we always divide Y by 4 and use remainder to get the last digit.
15th May 2013 5:59pm
pragati says...
what is the last didgit of(3^11)^7
21st September 2015 10:51pm
sreekant singh says...
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
17th September 2014 10:24pm
vivek says...
didnt get that step 3^5^--7=3^-25
10th September 2014 5:57pm
Sahil says...
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 :-)
24th June 2014 12:13pm
dhruv says...

hellow,
as i was going thru 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 /ed by 4 gives remainder 0 not 1..............
jus needa clarification.
thankyou for all your time &support lofoya

-a loyal consumer
5th May 2013 11:43am
Gannu says...
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..................
14th August 2013 1:36pm

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5000
Type the numbers for four hundred seventy-two.