Hmm interesting. In order to find the last 3 digits of 101^{107}, then we have to take it to mod 1000.

101^{107} (mod 1000)

= 201^{53} * 101 (mod 1000)

= 401^{26} * 201 * 101 (mod 1000)

= 801^{13} * 201 * 101 (mod 1000)

= 601^6 * 801 * 201 * 101 (mod 1000)

= 210^3 * 801 * 201 * 101 (mod 1000)

=210^4 * 801 * 101 (mod 1000)

= 401^2 * 801 * 101 (mod 1000)

= 801 * 801 * 101 (mod 1000)

= 601 * 101 (mod 1000)

= 701 (mod 1000)

The last three digits are therefore 701.