Pătrate și cuburi perfecte
Tema 10
Lucian Maran, MateMaraton, 01-12-2024
Problema 1. Arătați că numărul A = 1 + 2 + 3 + … + 2018 + 2019 + 2018 + … + 3 + 2 + 1 A=1+2+3+ \ldots+2018+2019+2018+ \ldots + 3+2+1 A = 1 + 2 + 3 + … + 2018 + 2019 + 2018 + … + 3 + 2 + 1
Olimpiadă, etapa locală, Buzău, 2018, E.537
Soluție:
X = 1 + 2 + 3 + … + 2018 + 2019 = 2019 ⋅ 2020 : 2 = 2019 ⋅ 1010. X=1+2+3+ \ldots+2018+2019 = 2019 \cdot 2020 : 2 = 2019 \cdot 1010. X = 1 + 2 + 3 + … + 2018 + 2019 = 2019 ⋅ 2020 : 2 = 2019 ⋅ 1010. Y = 2018 + … + 3 + 2 + 1 = 2018 ⋅ 2019 : 2 = 2019 ⋅ 1009. Y=2018+ \ldots + 3+2+1 = 2018 \cdot 2019 : 2 = 2019 \cdot 1009. Y = 2018 + … + 3 + 2 + 1 = 2018 ⋅ 2019 : 2 = 2019 ⋅ 1009.
A = X + Y = 2019 ( 1010 + 1009 ) = 2019 2 , A=X+Y=2019(1010+1009) = 2019^2, A = X + Y = 2019 ( 1010 + 1009 ) = 201 9 2 , A A A
Problema 2. Se consideră numerele: A = ( 2 20 + 2 21 ) ⋅ ( 3 20 + 3 21 ) A=(2^{20}+2^{21})\cdot (3^{20}+3^{21}) A = ( 2 20 + 2 21 ) ⋅ ( 3 20 + 3 21 ) B = ( 2 21 − 2 20 ) ⋅ ( 3 23 − 3 21 ) . B=(2^{21}-2^{20})\cdot (3^{23}-3^{21}). B = ( 2 21 − 2 20 ) ⋅ ( 3 23 − 3 21 ) . A + B A+B A + B A ⋅ B A\cdot B A ⋅ B
Olimpiadă, etapa locală, Dâmbovița, 2018, E.538
Soluție:
A = 2 20 ( 1 + 2 ) ⋅ 3 20 ( 1 + 3 ) = 2 22 ⋅ 3 21 . A=2^{20}(1+2) \cdot 3^{20}(1+3) = 2^{22} \cdot 3^{21}. A = 2 20 ( 1 + 2 ) ⋅ 3 20 ( 1 + 3 ) = 2 22 ⋅ 3 21 . B = 2 20 ( 2 − 1 ) ⋅ 3 21 ( 3 2 − 1 ) = 2 23 ⋅ 3 21 . B=2^{20}(2-1) \cdot 3^{21}(3^2-1) = 2^{23} \cdot 3^{21}. B = 2 20 ( 2 − 1 ) ⋅ 3 21 ( 3 2 − 1 ) = 2 23 ⋅ 3 21 .
a) A + B = 2 22 ⋅ 3 21 + 2 23 ⋅ 3 21 = 2 22 ⋅ 3 21 ( 1 + 2 ) = 2 22 ⋅ 3 22 A+B = 2^{22} \cdot 3^{21} + 2^{23} \cdot 3^{21} = 2^{22} \cdot 3^{21}(1+2) = 2^{22} \cdot 3^{22} A + B = 2 22 ⋅ 3 21 + 2 23 ⋅ 3 21 = 2 22 ⋅ 3 21 ( 1 + 2 ) = 2 22 ⋅ 3 22 b) A ⋅ B = 2 22 ⋅ 3 21 ⋅ 2 23 ⋅ 3 21 = 2 45 ⋅ 3 42 = ( 2 15 ) 3 ⋅ ( 3 14 ) 3 A \cdot B = 2^{22} \cdot 3^{21} \cdot 2^{23} \cdot 3^{21} = 2^{45} \cdot 3^{42} = (2^{15})^3 \cdot (3^{14)^3} A ⋅ B = 2 22 ⋅ 3 21 ⋅ 2 23 ⋅ 3 21 = 2 45 ⋅ 3 42 = ( 2 15 ) 3 ⋅ ( 3 14 ) 3
Problema 3. a) Arătați că numărul A = 2 ⋅ 2019 2020 − 2020 ⋅ 2019 2019 − 2018 ⋅ 2019 2018 A=2 \cdot 2019^{2020} - 2020 \cdot 2019^{2019} - 2018 \cdot 2019^{2018} A = 2 ⋅ 201 9 2020 − 2020 ⋅ 201 9 2019 − 2018 ⋅ 201 9 2018 B = 27 10 ⋅ 32 11 − 16 7 ⋅ 6 27 ⋅ 23 B=27^{10} \cdot 32^{11} - 16^7 \cdot 6^{27} \cdot 23 B = 2 7 10 ⋅ 3 2 11 − 1 6 7 ⋅ 6 27 ⋅ 23
Olimpiadă, etapa locală, Botoșani, 2019, E.539
Soluție:
A = 2019 2018 ( 2 ⋅ 2019 2 − 2020 ⋅ 2019 − 2018 ) = A=2019^{2018}(2 \cdot 2019^2-2020 \cdot 2019 - 2018)= A = 201 9 2018 ( 2 ⋅ 201 9 2 − 2020 ⋅ 2019 − 2018 ) = = 2019 2018 [ 2019 ( 2 ⋅ 2019 − 2020 ⏟ 2018 ) − 2018 ] = =2019^{2018}[2019(\underbrace{2 \cdot 2019 - 2020}_{2018})-2018]= = 201 9 2018 [ 2019 ( 2018 2 ⋅ 2019 − 2020 ) − 2018 ] = = 2019 2018 ( 2019 ⋅ 2018 − 2018 ) = =2019^{2018}(2019 \cdot 2018 - 2018)= = 201 9 2018 ( 2019 ⋅ 2018 − 2018 ) = = ( 2019 1009 ) 2 ⋅ 2018 2 =(2019^{1009})^2 \cdot 2018^2 = ( 201 9 1009 ) 2 ⋅ 201 8 2
B = ( 3 3 ) 10 ⋅ ( 2 5 ) 11 − ( 2 4 ) 7 ⋅ 2 27 ⋅ 3 27 ⋅ 23 = B=(3^3)^{10} \cdot (2^5)^{11} - (2^4)^7 \cdot 2^{27} \cdot 3^{27} \cdot 23 = B = ( 3 3 ) 10 ⋅ ( 2 5 ) 11 − ( 2 4 ) 7 ⋅ 2 27 ⋅ 3 27 ⋅ 23 = = 3 30 ⋅ 2 55 − 2 28 ⋅ 2 27 ⏟ 2 55 ⋅ 3 27 ⋅ 23 = =3^{30} \cdot 2^{55} - \underbrace{2^{28} \cdot 2^{27}}_{2^{55}} \cdot 3^{27} \cdot 23= = 3 30 ⋅ 2 55 − 2 55 2 28 ⋅ 2 27 ⋅ 3 27 ⋅ 23 = = 3 27 ⋅ 2 55 ( 3 3 − 23 ⏟ 4 ) = =3^{27} \cdot 2^{55} (\underbrace{3^3 - 23}_{4})= = 3 27 ⋅ 2 55 ( 4 3 3 − 23 ) = = 3 27 ⋅ 2 57 = ( 3 9 ) 3 ⋅ ( 2 19 ) 3 =3^{27} \cdot 2^{57} = (3^9)^3 \cdot (2^{19})^3 = 3 27 ⋅ 2 57 = ( 3 9 ) 3 ⋅ ( 2 19 ) 3
Problema 5. Se consideră numerele x = 2023 2 n + 1 + 2024 n + n x=2023^{2n+1} + 2024^n + n x = 202 3 2 n + 1 + 202 4 n + n y = n + 2024 n , y=n+2024^n, y = n + 202 4 n , n n n x x x 5 , y 5,~y 5 , y
Olimpiadă, etapa locală, Bihor, 2024, E.542
Soluție:
U c ( x ) = 5 ⇒ x U_c(x)=5 \Rightarrow x U c ( x ) = 5 ⇒ x x ⏟ impar = 2023 2 n + 1 ⏟ impar + 2024 n ⏟ par + n ⇒ n \underbrace{x}_{\text{impar}}= \underbrace{2023^{2n+1}}_{\text{impar}} + \underbrace{2024^n}_{\text{par}} + n \Rightarrow n impar x = impar 202 3 2 n + 1 + par 202 4 n + n ⇒ n n = 2 k . \boxed{n=2k}. n = 2 k .
⇒ U c ( x ) ⏟ U c = 5 = U c ( 3 4 k + 1 ⏟ U c = 3 + 2024 n + n ⏟ y ) ⇒ U c ( y ) = 2 ⇒ y \Rightarrow \underbrace{U_c(x)}_{U_c=5}=U_c(\underbrace{3^{4k+1}}_{U_c=3}+\underbrace{2024^n + n}_{y}) \Rightarrow \boxed{U_c(y)=2} \Rightarrow y ⇒ U c = 5 U c ( x ) = U c ( U c = 3 3 4 k + 1 + y 202 4 n + n ) ⇒ U c ( y ) = 2 ⇒ y
Problema 6. Să se calculeze pătratul numărului a b ‾ , \overline{ab}, ab , a a a a ‾ + 101 ( a + b ) \overline{aaaa} + 101(a+b) aaaa + 101 ( a + b )
Olimpiadă, etapa locală, Prahova, 2024, E.543
Soluție:
a a a a ‾ + 101 ( a + b ) = a ⋅ 1111 ‾ + 101 ⋅ a + 101 ⋅ b = 1212 a + 101 b = 101 ( 12 a + b ) . \overline{aaaa} + 101(a+b) = a \cdot \overline{1111} + 101\cdot a+101 \cdot b = 1212a+101b=101(12a+b). aaaa + 101 ( a + b ) = a ⋅ 1111 + 101 ⋅ a + 101 ⋅ b = 1212 a + 101 b = 101 ( 12 a + b ) .
101 ( 12 a + b ) 101(12a+b) 101 ( 12 a + b ) 12 a + b = 101 ⋅ k 2 , \boxed{12a+b=101 \cdot k^2}, 12 a + b = 101 ⋅ k 2 , k = 1 , 2 , 3 , … . k=1,2,3, \ldots. k = 1 , 2 , 3 , … . a a a b b b ⇒ 12 a + b ≤ 12 ⋅ 9 + 9 = 114 , \Rightarrow 12a+b \leq 12 \cdot 9 + 9=114, ⇒ 12 a + b ≤ 12 ⋅ 9 + 9 = 114 , k k k 1. 1. 1. 12 a + b = 101 ⇒ a = 8 , b = 5 ⇒ a b ‾ 2 = 85 2 = 7225 . 12a+b = 101 \Rightarrow a=8, b=5 \Rightarrow \boxed{\overline{ab}^2= 85^2=7225}. 12 a + b = 101 ⇒ a = 8 , b = 5 ⇒ ab 2 = 8 5 2 = 7225 .