Hello,

Note that a number with digits can be written :

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

Hence

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

(a)

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 :

10001x+3y=4

subtract 3 on each side :

10001x+3(y-1)=1

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)

242x+93y=1

242 and 93 are coprime (because 93=3x31 and none of them divide 242)

Use the extended Euclidean algorithm too.