Let me see.
Is 46^78+89^67 dividable with 5?
If a number is divisable by 5 it must end in a 0 or a 5
6^1=6
6^2=36 which ends in a 6
663=36*6 which ends in a 6
6^(any positive integer) must end in 6
6+4=10 (ends in 0)
6+9=15 (ends in 5)
so the question becomes does 89^67 end in a 4 or a 9
what could it end in? 9,81,..9,..1.. Ok, it could end in a 1 or a 9 it cannot end in a 4
89^1 ends in a 9
89^2 ends in a 1
89^3 ends in a 9
89^(an odd number) ends in a 9
SO 89^(67) ends in a 9
9+6=15
So yes that number will be divisable by 5
Is 46^78+89^67 dividable with 5 ?
yes !
$$\boxed{46^{78} + 89^{67} \stackrel{?}{=} 0 \mod 5} \\\\
\small{
\underbrace{(46^{78}\mod 5)}_{ 1^{78} \mod 5} + \underbrace{(89^{67}\mod 5)}_{-1^{67}\mod 5} \stackrel{?}{=} 0 \quad | \quad (46 \mod 5 = 1) \ and \ (89 \mod 5 = 4 = -1)
} \\\\\\
\small{
1^{78} = 1 \ and \ (-1)^{67} = -1}\\\\
\small{
( 1 \mod 5) + (-1 \mod 5) = [ (1-1) \mod 5 ]=( 0 \mod 5) = 0 \quad q.e.d.
}$$
Let me see.
Is 46^78+89^67 dividable with 5?
If a number is divisable by 5 it must end in a 0 or a 5
6^1=6
6^2=36 which ends in a 6
663=36*6 which ends in a 6
6^(any positive integer) must end in 6
6+4=10 (ends in 0)
6+9=15 (ends in 5)
so the question becomes does 89^67 end in a 4 or a 9
what could it end in? 9,81,..9,..1.. Ok, it could end in a 1 or a 9 it cannot end in a 4
89^1 ends in a 9
89^2 ends in a 1
89^3 ends in a 9
89^(an odd number) ends in a 9
SO 89^(67) ends in a 9
9+6=15
So yes that number will be divisable by 5