What is the reminder when 2^123456789 is divided by 7?

Question

find the middle term of the sequence formed by all three digits numbers which leave a reminder 3, when divided by 4. Also find the sum of all numbers on both sides of the middle term separately.

Answer 1

since
41152263*3=123456789 and (2^3) / 7 = 1 (remainder) , hence we have the remainder as 1

Answer 2

Ans is 1.
2^129456789 can be written as (2^3)^43152263.
= (7+1)^43152263
By expanding the above expression...we get the result as 7*k + 1^43152263.
So the remainder is 1.

Answer 3

Let's do it using Basic Level Division
Sooo......
Have a guess..There might be some pattern in this question...
So let me check:
(2^1) is not divisible by 7
(2^2) is not divisible by 7
(2^3) is divisible by 7 leaving 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...
.........so the pattern continues in the order 241 241 241....
So... let's check 123456789 is divisible by 3.
Yes....in a glance we can say that 123456789 is exactly divisible by 3.
So.... what's the remainder if (2^3n)/7?
From the pattern 241 241..... it's clearly understood that it's one(1).
So yeah...1 is the remainder

~Suryan Nair N