Problem of the day!!! Firstly,we Claim: $5^n-1$ has $n$ nonzero digits all equal to 4 in base 5. Proof: This is clear by induction. Since the base case $n=1$ work,we use the inductive hypothesis for $n=k+1$ Now $5^{k+1}=5(5^k-1)+4=(44444.........4)_5$ in base 5 so our claim is correct. So let $a=k(5^{1994}-1)$,and if $k>4$ it would be obvious,so if $k=1,2,3,4$ then: $(5^{1994}-1)_{10}=(444.........4444)_5$ where there are 1994 4's all of which are non zero,so this satisfies. $2(5^{1994}-1)=(14444.......4443)_5$ again satisfying the condition. $3(5^{1994}-1)=(24444........42)_5$ again satisfying the condition. $4(5^{1994}-1)=(344444........41)_5$ again satisfying the condition. Hence proved.