Question

Cho a,b \(\in N\) thỏa mãn a+20, b+13 cùng chia hết cho 21. CMR \(4^a+9^b+a+b-10⋮21\)

Answer 1

Từ giả thiết => $a\equiv 1\left(mod3\right)$, a=3k+1 ($k\in N$); b$\equiv 2\left(mod3\right)$, b=3q+2 $\left(q\in N\right)$

=> $A=4^a+9^b+a+b=1=1+0+1+2\left(mod3\right)$hay $A\equiv 4\left(mod3\right)$(1)

Lại có $4^a=4^{3k+1}=4\cdot {64}^k\equiv 4\left(mod7\right)$

$9^b=9^{\mathrm{...}}$