Question

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 ♥️❤️♥️❤️♥️❤️​

Answer 1

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 2

Answer:

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

Step-by-step explanation: