A=\(75.\left(4^{2004}+4^{2003}+4^2+4+1\right)+25\)

A=\(75.\left(4^{2005}-1\right):3+25\)

A=\(25.\left(4^{2005}-1+1\right)\)

A=\(25.4^{2005}⋮100\)

Nhớ tick cho mình nhé!

a, C= 75.( 4²⁰⁰¹+4²⁰⁰⁰+4¹⁹⁹⁹+ ... +4²+4¹+4⁰)+25

= \(75.\frac{4^{2002}-1}{3}+25\)

= 25.(4²⁰⁰²-1) +25

= 25.4²⁰⁰²

Vì 25.4²⁰⁰² chia hết cho 4²⁰⁰² nên C chia hết cho 4²⁰⁰²

b, Vì 25 chia cho 4 dư 1 nên 25.4²⁰⁰² chia cho 4.4²⁰⁰² dư 6

Vậy C chia 4²⁰⁰³ dư 6

câu b sai rồi đáng ra phải thế này

\(\frac{25.4^{2002}}{4^{2003}}=\frac{25}{4}=6,25\)

Do đó C chia cho 4²⁰⁰³ dư 25.4^{2002 _} 6.4²⁰⁰³=1

đặt biểu thức ban đầu là A, 4²⁰²⁰+4²⁰¹⁹+...+4+1=B

4B=4²⁰²¹ +4²⁰²⁰ +4²⁰¹⁹+...+4²+4

3B=4B-B=4²⁰²¹-1  => B= (4²⁰²¹-1)/3

A=75B+25=75(4²⁰²¹-1)/3 + 25= 25(4²⁰²¹-1)+25=25(4²⁰²¹-1+1)=25.4²⁰²¹=100.4²⁰²⁰

=> A chia hết cho cả 100 và 4²⁰²¹

mặt khác A=25.4²⁰²¹=4²⁰²¹.(24+1)=24.4²⁰²¹+4²⁰²¹=6.4²⁰²²+4²⁰²¹ 

vì 4^2021<4²⁰²² nên A chia 4²⁰²² dư 4²⁰²¹

tick cho mk nha!!!!!!!!

Theo bài ra, ta có: \(C=75\left(4^{2001}+4^{2000}+4^{1999}+...+4^2+4+1\right)+25\)

Đặt \(S=4^{2001}+4^{2000}+4^{1999}+...+4^2+4+1\)

\(\Rightarrow4S=4^{2002}+4^{2001}+4^{2000}+...+4^3+4^2+4\)

\(\Rightarrow4S-S=4^{2002}+4^{2001}+4^{2000}+...+4^3+4^2+4-4^{2001}-4^{2000}-4^{1999}-...4^2-4-1\)

\(\Rightarrow3S=4^{2002}-1\)

\(\Rightarrow S=\dfrac{4^{2002}-1}{3}\)

Khi đó \(C=75.\dfrac{4^{2002}-1}{3}+25=\dfrac{75}{3}.\left(4^{2002}-1\right)+25=25\left(4^{2002}-1\right)+25=25\left(4^{2002}-1+1\right)=25.4^{2002}⋮4^{2002}\)

Vậy \(C⋮4^{2002}\left(đpcm\right)\)

a) \(125a+25b-75c\)

\(=25\left(5a+b-3c\right)⋮25\)

\(\Rightarrow dpcm\)

b) \(39a+26b\)

\(=13\left(3a+2b\right)⋮13\)

\(\Rightarrow dpcm\)

a/

\(A=4^2.4^{37}+4^2.4^{38}+4^2.4^{39}=4^2\left(4^{37}+4^{38}+4^{39}\right)=\)

\(=2.8.\left(4^{37}+4^{38}+4^{39}\right)⋮8\)

b/

\(B=10^7\left(1+10+10^2\right)=10.10^6.111=\)

\(=5.10^6.222⋮222\)

c/

\(C=5^{2006}\left(1+5+5^2\right)=5^{2006}.31⋮31\)