2026/055) show that 1994 devides $10^{900} - 2^{1000}$

Now Let us factor 1994

$1994 = 2 * 997$

We need to show that it is divisible by 2 and 997 

RHS is even so is is divisible by 2

We need to show that the RHS is divisible by 997 

We have as 997 is close to 1000 so let is brrin power of 10 and 2 to be as close as 997

$10^3 \equiv 3 \pmod {997}\cdots(1)$

and  $2^{10} \equiv 27 \pmod {997}\cdots(2)$

so $10^{900} - 2^{1000}$

 $=({10^9})^{100} - (2^{10})^{100}$ )

This is divisible by $10^9 - 2^{10}$

Now $10^9 - 2^{10}\pmod {997}$

$= (10^3)^3 - 2^{10}\pmod {997}$

 $= 3^3 - 27 \pmod {997}$

$= 27-27 \pmod {997}$

= 0

So the number is divisible by 997

As it is divisible by 997 and 2 so it is divisible by 1994