Question

Chứng minh rằng:3¹⁹⁹⁹ - 7¹⁹⁹⁷ chia hết cho 5.

Answer 1

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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Answer 2

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

=> \(3^{1999}-7^{1997}⋮5\left(\text{đpcm}\right)\)

Answer 3

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 )