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Ta có :
\(5^{13}+5^{11}-5^{10}-40=5^{10}.\left(5^3+5-1\right)-40=5^{10}.129-40=5^{10}.43.3-40\)
Vì 510 . 43 . 3 chia hết cho 43 nên 510 . 43 . 3 - 40 chia cho 43 dư 43 - 40 = 3
\(A=\left(2^1+2^2+2^3+2^4\right)+....+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)+1\)
\(=2.15+2^5.15+...+2^{97}.15+1=15.\left(2+2^5+...+2^{97}\right)+1\)
A 15 dư 1
\(S=2^0+2^1+2^2+...+2^{99}+2^{100}\)
\(=1+2+\left(2^2+2^3+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=3+2^2.\left(1+2+4\right)+...+2^{98}.\left(1+2+4\right)\)
\(=3+7.\left(2^2+2^5+...+2^{98}\right)\)chia 7 dư 3
\(S=2^0+2^1+2^2+...+2^{99}+2^{100}\)
\(S=\left(2^0+2^1+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{98}+2^{99}+2^{100}\right)\)
\(S=\left(1+2+4\right)+2^3\left(1+2+4\right)+.....+2^{98}\left(1+2+4\right)\)
\(S=7+2^3\cdot7+....+2^{98}\cdot7\)
\(S=7\left(1+2^3+...+2^{98}\right)\)
=> S chia 7 dư 0 hay S chia hết cho 7
1b) Ta có: C= 2+22+23+24+...+299+2100
=> 2C= 22+23+24+...+2100+2101
=> 2C-C=(22+23+24+25+...+2101)-(2+22+23+24+...+2100)
=> C=2101-2
+) Ta có: 22x - 2 =C
<=> 22x=C+2
<=>22x=2101
<=>2x=101
<=>x=101:2
<=>x=50.5
a)\(C=2+2^2+2^3+...+2^{99}\)
\(C=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(C=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(C=2.31+...+2^{96}.31\)
\(C=31\left(2+...+2^{96}\right)\) chia hết cho 31
7.7.7....7=707
707:10=70(dư 7)
7101=74.74...74.7
=(..1)(...1)...(...1).7=...7 chia 10 dư 7