Số dư trong phép chia $10^{10}+10^{10^2}+10^{10^3}+...+10^{10^{2015}}$ cho $7$ là?
Số dư trong phép chia $10^10+10^10^2+10^10^3+...+10^10^2015$ cho $7$ là?
Started By bovuotdaiduong, 04-04-2016 - 18:01
#1
Posted 04-04-2016 - 18:01
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#2
Posted 04-04-2016 - 18:35
Đặt $A=10^{10}+10^{10^2}+...+10^{10^{2015}}$
Dễ thấy $10^{10}\equiv 4\;(mod\;7)$
Nên $A \equiv 4+4^{10}+4^{10^2}+...+4^{10^{2014}}$
Dễ chứng minh được $4^{10}\equiv4\;(mod\;7)$
Nên $4^{10}\equiv4^{10^2}\equiv...\equiv4^{10^{2015}}\equiv4\;(mod\;7)$
Do đó $A\equiv4.2015\equiv3\;(mod\;7)$
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#3
Posted 15-04-2018 - 07:07
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