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2014-09-29T21:57:53+05:30

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Let N = (500+83)⁵¹²  Using binomial expansion,

N =  Sum of terms that have powers of 500 as a factor +  83⁵¹²

So on the R H S, only the last term contributes to the last two digits of N. Other terms have at least two zeros at the tenths and units positions. Last two digits of N are the last two digits of M = 83⁵¹².

M = 83⁵¹² = 

Now, the last two digits of the numbers which are powers of 3 are :

83⁴:  21           83⁸:  41            83¹⁶:  81               83³² : 61  
    83⁶⁴ :  21        83¹²⁸ :  41            83²⁵⁶ : 81              83⁵¹² : 61 

61 is last two digits of 83⁵¹² and 583⁵¹²

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(80+3)^512 :  the last two terms in the binomial expansion are:
  512 * 80 * 3^511 + 3^512 
These have the last two digits as 21 and 41 respectively.   So the total is 61.


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