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Grade: 12

                        

The last two digits of 7^(2008) are

6 years ago

Answers : (2)

bharat bajaj
IIT Delhi
askIITians Faculty
122 Points
							71= 07
72= 49
73= 343
74= 2401
75= 16807
76= 117649
77= 823543
78= 5764801
Here, note that the last twodigitsof 74and 78are 01. For every 4th power, the last digits are 01
Since 2008 is divisible by 4, the last digits of72008are 01.
THanks
Bharat Bajaj
IIT Delhi
askiitians faculty
6 years ago
Manas Satish Bedmutha
22 Points
							7       = 07 (mod 100)
7 ^ 2 = 49 (mod 100)
7 ^ 4 = 01 (mod 100)

So 7 ^ (4 x 502) = 7 ^ 2008 = 01 (mod 100)
Hence last 2 digits are 01
						
6 years ago
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