ai trả lời nhanh mình tích

ai tích mình mình tích lại cho

ta có:

\(2^{60}+5^{30}\)

=\(\left(2^4\right)^{15}+\left(5^2\right)^{15}\)

=\(16^{15}+15^{15}\) luôn chia hết cho 16 + 25 = 41

\(\Rightarrow2^{60}+5^{30}\) chia hết cho 41 ( đpcm )

chính sát

tra lời

link https://olm.vn/hoi-dap/detail/60197622644.html

hok tốt

Bài 2 thôi em dùng đồng dư cho chắc:v

a) \(21^2\equiv41\left(mod200\right)\Rightarrow21^{10}\equiv41^5\equiv1\left(mod200\right)\)

Suy ra đpcm.

b) \(39^2\equiv1\left(mod40\right)\Rightarrow39^{20}\equiv1\left(mod40\right)\)

Mặt khác \(39^2\equiv1\left(mod40\right)\Rightarrow39^{12}\equiv1\Rightarrow39^{13}\equiv39\left(mod40\right)\)

Suy ra \(39^{20}+39^{13}\equiv1+39\equiv40\equiv0\left(mod40\right)\)

Suy ra đpcm

c) Do 41 là số nguyên tố và (2;41) = 1 nên:

\(2^{20}\equiv1\left(mod41\right)\) suy ra \(2^{60}\equiv1\left(mod41\right)\)

Dễ dàng chứng minh \(5^{30}\equiv40\left(mod41\right)\)

Suy ra đpcm.

d) Tương tự

a) \(21^{10}-1=\left(21^5\right)^2-1^2=\left(21^5+1\right).\left(21^5-1\right)\)

\(21^5+1=\overline{...1}=2k+1+1=2n\)

\(21^5-1=\overline{...01}-1=\overline{...00}\)

\(\Rightarrow21^{10}-1=2n.\overline{...00}⋮200\left(đpcm\right).\)

b) \(39\equiv-1\left(mod40\right)\)

\(\Rightarrow39^{20}\equiv1\left(mod40\right)\)

\(\Rightarrow39^{19}\equiv-1\left(mod40\right)\)

\(\Rightarrow39^{20}+39^{19}\equiv1+\left(-1\right)\left(mod40\right)\)

\(\Leftrightarrow39^{20}+39^{19}\equiv0\left(mod40\right)\)

\(\Rightarrow39^{20}+39^{19}⋮40\left(đpcm\right).\)

d) \(2005\equiv-1\left(mod2006\right)\)

\(\Rightarrow2005^{2007}\equiv\left(-1\right)^{2007}=-1\left(mod2006\right)\)

\(2007\equiv1\left(mod2006\right)\)

\(\Rightarrow2007^{2005}\equiv1\left(mod2006\right)\)

\(\Rightarrow2005^{2007}+2007^{2005}\equiv-1+1=0\left(mod2006\right)\)

\(\Leftrightarrow2005^{2007}+2007^{2005}⋮2006\left(đpcm\right).\)

300

300

= 300