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What are the last two digits of the number 7245  ?

(A) 07 (B) 23 (C) 49 (D) 43

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For this we have to see the cyclicity of the numbers according to unit digit of base and format of exponent..

So in general :

a) 74n   will contain  last two digits as  01

b) 74n+1  will contain last two digits as 07

c) 74n+2  will contain last two digits as 49

d) 74n+3  will contain last two digits as 43

So here the exponent  =  245 which is of the form b) as mentioned above..

Hence last 2 digits of 7245   =   07

Hence A) is the correct answer. 

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Another approach to use Reminder Theorem

We have to find last two digits so when we divide a number by 100 we got reminder which are last two digits.

Using this concept we can find as

$\Phi (100) = 40$ (Euler totient function)

$245 mod 40 = 5$

Now we can easily handle given number

$7^5 mod 100 = 07$

Hence last two digits are 07

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