令 \( f(x) = a_4x^4 + a_3x^3 + a_2 x^2 +a_1x+a_0 \)
由 \( f(1) =39 \) 得 \( a_0 +a_1+a_2+a_3+a_4 = 39 \)
從同餘 \( f(7) \equiv a_0 \) (mod 7) 可得 \( a_0 = 6 + 7n_0 \)
再考慮 mod \( 7^n \) 可得
\( a_{1}=5-n_{0}+7n_{1} \)
\( a_{2}=4-n_{1}+7n_{2} \)
\( a_{3}=0-n_{2}+7n_{3} \)
\( a_{4}=6-n_{3} \)
\( 21+6(n_{0}+n_{1}+n_{2}+n_{3})=39\Rightarrow n_{0}+n_{1}+n_{2}+n_{3}=3 \)
