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Aptitude Question Discussion  NUMBER SYSTEM AND NUMBER THEORY
Q.  What is the last digit of the number 3^{579}+ 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.

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Comments (9)
LAST DIGIT of a number having form of $\bo X^\bo Y$ (Xdownwards, YSideways)
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 $
$=7+71$
$=141=13$,
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: 3^{579} =3^{(5)7}=(3)^{25}From the above relation last digit is $3^1=1$
So ast digit of the given number becomes $3+1=\bo 4$
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.
and the only even no. in the options is 4 so the answer :)
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
$2^1$ is 2..................
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