Ta có:
\(3^{1999}=3^{2000}:3=\left(3^2\right)^{1000}:3=9^{1000}:3=...1:3=...7\)
\(7^{1997}=7^{1996}.7=\left(7^2\right)^{998}.7=49^{998}.7=...1.7=...7\)
Do đó: \(3^{1999}-7^{1997}=...7-...7=...0\)
Vì \(...0\)chia hết cho 5 \(\Rightarrow3^{1999}-7^{1997}\)chia hết cho 5
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Ta có : 31999 = 31996 . 33 = (34)499 . (....7) = (....1)499 . (...7) = ...7
71997 = 71996 . 7 = (74)499 . 7 = (....1)499 . 7 = ...7
Khi đó 31999 - 71997 = ...7 - ...7 = ...0
=> \(3^{1999}-7^{1997}⋮5\left(\text{đpcm}\right)\)
Ta có:
31999=32000:331999=32000:3
=(32)1000:3=(32)1000:3
=91000:3=91000:3
=.....:3=.....7=.....:3=.....7
71997=71996.771997=71996.7
=(72)998.7=(72)998.7
=49998.7=49998.7
=.....1.7=.....7=.....1.7=.....7
Do đó: 31999−71997=.....7−.....7=.....031999−71997=.....7−.....7=.....0
Vì .....0.....0 chia hết cho 5.5.
⇒31999−71997⇒31999−71997 chia hết cho 5.5. ( đpcm )