The last two digits of 7^(2008) are
Samir Madan , 11 Years ago
Grade 12
Algebra> The last two digits of 7^(2008) are...
2 Answers
bharat bajaj
Last Activity: 11 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
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
IIT Delhi
askiitians faculty
Manas Satish Bedmutha
Last Activity: 11 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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