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What are the last two digits of $7^{2008}?$

  1. $21$
  2. $61$
  3. $01$
  4. $41$
  5. $81$
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Powers of 7 from 0 are as follows:
1 7 49 343 2401 ...
- so the cycle is repeating after every 4 powers. $7^{2008} = 7^{4 \times 502}$. So, the last two digits must be "01".
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answer c . That is 01. $7^{2008}$ equals can be written as $7^{4*502}$ now $7^4$ equals $49*49$ whose last 2 digits are 01 and then whenever a number has 01 as power something. take the  tens place and multiply by the last digit of power and that thats number unit digit and consider it as its tens digit.  ( case 1)

so here unit digit will be 1 and we have to find tens place. so take the tens place of the found last 2 digits 01 which is 0 . multiply with 2=0. so tens place will be also zero . so 01. here are all the case of such a problem . http://www.campusgate.co.in/2011/10/finding-unit-and-last-two-digits-of.html

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