find the remainder of 7^3000+1/6 and 6^3000003+1/7?

Question

brain developing

Answer 1

for the first part ((7^3000)+1)/6,
((7^3000)+1)/6 = ((7^3000)/6) + (1/6)
= (((6+1)^3000)/6) + (1/6)
=(A)/6 + (B)/6
when A is divided by 6, the remainder would be 1^3000 which is same as 1
when B is divided by 6, the remainder would be again 1

so for the complete expression the remainder comes out to be 2

similarly for the second part 6^3000003+1/7,
6^3000003+1/7 = (6^3000003)/7 + 1/7
= A/7 + B/7
A = 6^3000003 = (7-1)^3000003 which when divided by 7 would give remainder of : (-1)^3000003 which is same as -1
B when divided by 7 would give a remainder of 1

so just add both the remainders to get the result, which comes out to be 0

Answer 2

I think the questioner intended to say ...

Find the remainder of ((7 ^ 3000) + 1) / 6 and ((6 ^ 3000003) + 1) / 7.

... but needs a lesson in the standard order of operations and use of parentheses in mathematical notation.

p.s. 7 is 1 mod 6, so 7 ^ 3000 is 1 mod 6, and that plus 1 is 2 mod 6.

6 is -1 mod 7, as is any odd power such as 6 ^ 3000003, and that plus 1 is 0 mod 7.

Answer 3

r u suppose to divide the numbers?

there is no remainder without division.

Answer 4

I wonder if the question is complete!!!