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Maths Puzzle
find the reminder when (2222^5555+5555^2222)/7?
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2222/7=3
5555/7=4
3^5555+4^2222/7
(3^5)^1111+(4^2)^1111/7
3^5/7=5
4^2/7=2
5^1111+2^1111/7
5^1=5
5^11=*******5
and same for 2^1111
so last digits after powering both of them must be...5 and 2
so 5+2/7
7/7=1
remainder=0
- 11 years agoHelpfull: Yes(1) No(2)
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