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Ta có: \(35^2\equiv375\)( mod 425)
\(35^3=35.35^2\equiv35.375\equiv375\)( mod 425)
\(35^4=35.35^3\equiv35.375\equiv375\)( mod 425)
\(35^8=35^4.35^4\equiv375.375\equiv375\)( mod 425)
\(35^{16}\equiv35^8.35^8\equiv375.375\equiv375\)( mod 425)
\(35^{32}\equiv35^{16}.35^{16}\equiv375.375\equiv375\)( mod 425)
=> \(35^2-35^3+35^4-35^8+35^{16}+35^{32}\equiv375-375+375-375+375+375\equiv325\)( mod 425)
Vậy số dư cần tìm là 325
a) \(S=1^5+3^5+....+75^5+99^5\)
\(\left(2a+1\right)^5-\left(2a+1\right)=2a\left(2a+1\right)\left(2a+2\right)\left[\left(2a+1\right)^2+1\right]\)
\(\left(2a+1\right)^5-\left(2a+1\right)=4a\left(2a+1\right)\left(a+1\right)\left[\left(2a+1\right)^2+1\right]⋮4\)
\(S=\left(1^5-1\right)+\left(3^5-3\right)+....+\left(75^5-75\right)+\left(99^5-99\right)+\left(1+3+5+...+75+99\right)\)
\(\Leftrightarrow\begin{matrix}1^5-1⋮4\\3^5-3⋮4\\5^5-5⋮4\\...........\\75^5-5⋮4\\99^5-99⋮4\end{matrix}\)
\(S_1=1+3+5+7+...+75+99=\frac{\left(1+75\right)\left[\frac{75-1}{2}+1\right]}{2}+99=38.38+96+3\)
\(\Rightarrow S_1:4\) dư 3
\(\Leftrightarrow S\) chia 4 dư 3
a/ 100 : { 2. [ 52 - ( 35 - 8 )]}
= 100 : { 2. [ 52 - 27 ]}
= 100 : { 2. 25 }
= 100 : 50
= 2
b/ 12 : { 390 : [ 500 - ( 125 + 35 . 7 )]}
= 12 : { 390 : [ 500 - ( 125 + 245 )]}
= 12 : { 390 : [ 500 - 370 ]}
= 12 : { 390 : 130 }
= 12 : 3
= 4
c/ 3. 52 - 16 : 22
= 3 . 25 - 16 : 4
= 75 - 4
= 71
Bài này dễ mà
a) \(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}+2\frac{14}{90}\)
= \(\left(74\frac{19}{35}+15\frac{16}{35}\right).\frac{7}{90}+2\frac{14}{90}\)
= 90 . 7/90 + 194/90
= 630/90 + 194/90
= 824/90 = 412/45
b) (-2/5 + 3/7) - (4/9 + 12/20 - 13/35) + 7/35
= -2/5 + 3/7 - 4/9 - 3/5 + 13/35 + 7/35
= (-2/5 - 3/5) + 3/7 - 4/9 + (13/35 + 7/35)
= -1 + 3/7 - 4/9 + 4/7
= -1 + (3/7 + 4/7) - 4/9
= -1 + 1 - 4/9 = -4/9
\(\frac{13}{21}\)+ \(\frac{8}{21}\) : \(\frac{16}{21}\)
= \(\frac{13}{21}\)+ \(\left(\frac{8}{21}:\frac{16}{21}\right)\)= \(\frac{13}{21}\)+ \(\frac{1}{2}\)
= \(\frac{47}{42}\)
\(\frac{36}{35}\): \(\frac{8}{3}\)- \(\frac{36}{35}\): \(\frac{6}{11}\)
= \(\frac{27}{70}\)- \(\frac{66}{35}\)= \(-\frac{3}{2}\)
\(A=\left(4^2\right)^{17}.\frac{64^{36}}{8^{35}.32^{34}}\)
\(A=16^{17}.\frac{64^{2.36}}{\left(2^3\right)^{35}.\left(2^5\right)^{34}}\)
\(A=...6.\frac{...6^{36}}{2^{105}.2^{170}}\)