The last two decimal places of a number is equal to its congruence modulo 100.
So basically, you're asked to find x in :
For that, use Euler's totient function :
If , then
I have to think how to deal with this one
Divide by 4 :Find the integers x and y that solve each of the following equations, or explain why no solution exists
Bézout's identity tells us :
there exist x and y such that ax+by=1 if and only if a and b are coprime.
Which is obviously not the case here, since 464 and 246 are both even.
Another way to see that x and y don't exist is that the LHS is even and the RHS, 1, is odd.
There are 4x2 digits in 11111111 and 4 in 3333 and 4444.(b)
so divide by 1111 :
subtract 3 on each side :
And this has a solution, since 10001 and 3 are coprime. Use the extended Euclidean algorithm to find x and y' (y'=y-1)
Divide by 4 :(c)
242 and 93 are coprime (because 93=3x31 and none of them divide 242)
Use the extended Euclidean algorithm too.