kangaroo maths competition How many three-digit positive integers $ABC$ exist, such that $(A + B)^c$ is a three-digit integer and an integer power of $2$?
Note: An integer power of $2$ is a number in the form $2^k$
, where $k$ is an integer.
(A) $15$ 
(B) $16$ 
(C) $18$ 
(D) $20$ 
(E) $21$

Here is complete list of $21$ solutions.


*

*$2^7=128:$


*

*$2^7=(1+1)^7\implies ABC=117$

*$2^7=(2+0)^7\implies ABC=207$


*$2^8=4^4=16^2=256:$


*

*$2^8=(1+1)^8\implies ABC=118$

*$2^8=(2+0)^8\implies ABC=208$

*$4^4=(1+3)^4\implies ABC=134$

*$4^4=(2+2)^4\implies ABC=224$

*$4^4=(3+1)^4\implies ABC=314$

*$4^4=(4+0)^4\implies ABC=404$

*$16^2=(7+9)^2\implies ABC=792$

*$16^2=(8+8)^2\implies ABC=882$

*$16^2=(9+7)^2\implies ABC=972$


*$2^9=8^3=512:$


*

*$2^9=(1+1)^9\implies ABC=119$

*$2^9=(2+0)^9\implies ABC=209$

*$8^3=(1+7)^3\implies ABC=173$

*$8^3=(2+6)^3\implies ABC=263$

*$8^3=(3+5)^3\implies ABC=353$

*$8^3=(4+4)^3\implies ABC=443$

*$8^3=(5+3)^3\implies ABC=533$

*$8^3=(6+2)^3\implies ABC=623$

*$8^3=(7+1)^3\implies ABC=713$

*$8^3=(8+0)^3\implies ABC=803$
