Question

[I am➻Minh]

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

[mai đức anh]

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

$9^b=9^{3q+2}\equiv 2^{3q+2}\left(mod7\right)\Rightarrow 9^b\equiv 4\cdot 8^q\equiv 4\left(mod7\right)$

Từ gt => $a\equiv 1\left(mod7\right),b\equiv 1\left(mod7\right)$

Dẫn đến $A=4^a+9^b+a+b\equiv 4+4+1+1\left(mod7\right)$hay $A\equiv 10\left(mod7\right)$

Từ (1) => $A\equiv 10\left(mod3\right)$mà 3,7 nguyên tố cùng nhau nên $A\equiv 10\left(mod21\right)$

=> A chia 21 dư 10

Answer 2

[mai đức anh]

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

$9^b=9^{3q+2}\equiv 2^{3q+2}\left(mod7\right)\Rightarrow 9^b\equiv 4\cdot 8^q\equiv 4\left(mod7\right)$

Từ gt => $a\equiv 1\left(mod7\right),b\equiv 1\left(mod7\right)$

Dẫn đến $A=4^a+9^b+a+b\equiv 4+4+1+1\left(mod7\right)$hay $A\equiv 10\left(mod7\right)$

Từ (1) => $A\equiv 10\left(mod3\right)$mà 3,7 nguyên tố cùng nhau nên $A\equiv 10\left(mod21\right)$

=> A chia 21 dư 10