Ultime 4 cifre

[FreddyKruger]

Sia $P(x)= x \cdot 3^x $.
Calcolare le ultime 4 cifre di $ P(2019) $.

[axpgn]

$1873$

Cordialmente, Alex

[FreddyKruger]

Non ho il risultato ma mi fido, procedimento?

[axpgn]

È lungo

$ P(x)= x \cdot 3^x $

$ P(2019) = 2019 * 3^2019$

$ P(2019) = 673*3 * 3^2019=673*3^2020=673*(3^20)^101$

$3^20-=4401 mod 10000$

$ 673*4401^101=673*4401*4401^100$

$ 673*4401-= 1873 mod 10000$

$1873*4401^100$

$1873*(4401^2)^50$

$ 4401^2-= -1199 mod 10000$

$1873*(-1199)^25$

$1873*(-1999)*(-1199^3)^8$

$ -1999*1873 -= -4127 mod 10000$

$ -1999^3 -= 4001 mod 10000$

$-4127*4001^8$

$-4217*4001^2*4001^3*4001^3$

$4001^2-=8001 mod 10000$

$4001^3-=2001 mod 10000$

$-4127*8001*2001*2001$

$2001*2001-=4001 mod 10000$

$-4127*8001-= -127 -= 9873 mod 10000$

$9873*4001=39501873$

Quindi le ultime quattro cifre di $2019*3^2019$ sono $1873$

Sicuramente ci sarà un metodo più veloce




Cordialmente, Alex

[Super Squirrel]

$3^2019=3*(3^2)^1009=3*9^1009=3(-1+10)^1009-=$

$-=3[((1009),(0))(-1)^1009(10)^0+((1009),(1))(-1)^1008(10)^1+((1009),(2))(-1)^1007(10)^2+((1009),(3))(-1)^1006(10)^3](mod10000)$

$-=511943221467(mod10000)$

$2019*3^2019-=2019*1467(mod10000)$

$P(2019)-=1873(mod10000)$