Pre-RMO sample question

$7^{100}-3^{100}$ is divisible by 7 - 3 =4

Also $7^{100}-3^{100}$ = $16807^{20}-243^{20}$ which is divisible by 16807+243=17050 So $7^{100}-3^{100}$ is divisible by 100 So last 2 digits are 00.

There is an easy rule in mathematics that if a number has a power then we could subtract 1 from the power and multiply the number of power to the number eg.4^100=100×4^100-1=100×4^99. If we do this till the power 1 then we will get 100^100×4. 3^100=100^100×3=3×100^100 7^100=100^100×7=7×100^100

100^(7-3)=4×100^100. Sum of last 2 digit is 0 because if hundred 0s are placed after 4 the last digit is 0

By Euler's Theorem , we get

$7^{80} = 1 \mod100$ and $3^{80} = 1 \mod100$

Now $7^3=43 \mod100$ = $7^6 = 49 \mod100$ = $7^{12}= 1\mod100$

So $7^{18}= 49 \mod100$ And $7^2 = 49 \mod100$ So

$7^{20} = 1 \mod100$ or $7^{100} = 1\mod100$ .

Again with 3, we get $3^{20} = 1 \mod100$

So $3^{100} = 1 \mod100$

So $7^{100} - 3^{100} = 0\mod100$

So $\boxed{00}$ are the last two digits