More easier one:
What's the reminder when 2^(123456789) is divisible by 7?
No need of calculators.No rough work necessary... It's quite easy and can be solved mentally (If I can do, then definitely u can♥️✌️)
[100 Points]
Leaving Brainly ♥️
Answers
Answer:
Step-by-step explanation:
Easy method ✔✔✔
(2^1) is not divisible by 7
(2^2) is not divisible by 7
(2^3) is divisible by 7 which leaves remainder 1
(2^4) ÷ 7 will give u remainder 2
(2^5) ÷ 7 will give u remainder 4
(2^6) ÷ 7 will give u remainder 1
(2^7) ÷ 7 will give u remainder 2
pattern continues in the order 241 241 241(according to upper eq)
NOW
let's see wheather 123456789 is divisible by 3.
we can say that 123456789 is exactly divisible by 3.
if (2^3n)/7?
From the pattern 241 241
WE CONCLUDED THAT
THE ANS IS 1 .
1 is the remainder
NOTE:-
From given solution when we divide all the terms by 7 we doesn't get remainder 1 but when we divide 2^6÷7
Than we got the remainder 1
Answer:
Step-by-step explanation:
Your answer is in attachment
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