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1616pragati's Comment
http://www.lofoya.com/aptitude/number-system/what-is-the-last-digit-of-the-number-3-5-7-9-1.php
Mon, 21 Sep 2015 22:51:18 +0530873sreekant singh's Comment
http://www.lofoya.com/aptitude/number-system/what-is-the-last-digit-of-the-number-3-5-7-9-1.php
Wed, 17 Sep 2014 22:24:21 +0530624vivek's Comment
http://www.lofoya.com/aptitude/number-system/what-is-the-last-digit-of-the-number-3-5-7-9-1.php
Wed, 10 Sep 2014 17:57:08 +0530618Sahil's Comment
http://www.lofoya.com/aptitude/number-system/what-is-the-last-digit-of-the-number-3-5-7-9-1.php
and odd no. + 1 = even no. and the only even no. in the options is 4 so the answer :-)]]>Tue, 24 Jun 2014 12:13:52 +0530523Gannu's Comment
http://www.lofoya.com/aptitude/number-system/what-is-the-last-digit-of-the-number-3-5-7-9-1.php
$2^1$ is 2..................]]>Wed, 14 Aug 2013 13:36:20 +0530183Deepak's Comment
http://www.lofoya.com/aptitude/number-system/what-is-the-last-digit-of-the-number-3-5-7-9-1.php
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.]]>Wed, 15 May 2013 17:59:47 +053036arshi's Comment
http://www.lofoya.com/aptitude/number-system/what-is-the-last-digit-of-the-number-3-5-7-9-1.php
Wed, 15 May 2013 17:48:48 +053035Deepak's Comment
http://www.lofoya.com/aptitude/number-system/what-is-the-last-digit-of-the-number-3-5-7-9-1.php
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: 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 $3+1=\bo 4$]]>Mon, 13 May 2013 20:37:32 +053034dhruv's Comment
http://www.lofoya.com/aptitude/number-system/what-is-the-last-digit-of-the-number-3-5-7-9-1.php
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]]>Sun, 05 May 2013 11:43:31 +053019