# Doubt in number theory

this can be interpreted as difference if tye numbers equals 1000x then have to use modulo

Thanks

This means 1000|2978^n-2978^m=2978^m(2978^{n-m}-1)

Since 8|1000 and highest power of 2 dividing 2978 is 1 we need m \ge 3

Also we have that 125|2978^{n-m}-1 or 2978^{n-m} \equiv 1 \mod{125}

The least value of r such that 2978^r \equiv 1 \mod{125} is equal to \phi(125)=100

Hence n-m \ge 100 or n \ge m+100 \ge 103