Answwer atleast one question please
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Answer:
I am answering the 4th question
Step-by-step explanation:
7^2 = 49, mod 100
7^4 = 49^2 = 2401 = 1 mod 100
Hence,
7^4×502 = 1^502 = 1 mod 100
that is ,
7^2008 = 1 mod 100
Therefore the last two digit number of 7^2008 is 01
Answered by
1
Step-by-step explanation:
Answer of your question is -
root 3 +2root 2. is exceed
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