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The last two digits of 7^(2008) are

The last two digits of 7^(2008) are

Grade:12

2 Answers

bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago
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
Manas Satish Bedmutha
22 Points
7 years ago
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

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