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Wittaler [7]
2 years ago
8

The Fibonacci sequence is defined by $F_1 = F_2 = 1$ and $F_{n + 2} = F_{n + 1} + F_n$. Find the remainder when $F_{1999}$ is di

vided by 5.
Mathematics
1 answer:
zavuch27 [327]2 years ago
6 0

Answer:

The remainder is 1

Step-by-step explanation:

Given the Fibonacci sequence

F_1 = F_2 = 1, and

F_(n + 2) = F_(n + 1) + F_n

We want to find the remainder when F_(1999) is divided by 5.

Let us write the first 20 numbers of the sequence in (mod 5). They are

F_1 = 1,

F_2 = 1,

F_3 = 2,

F_4 = 3,

F_5 = 5 = 0 (mod 5),

F_6 = 3,

F_7 = 3,

F_8 = 1

F_(9) = 4

F_(10) = 0

F_(11) = 4

F_(12) = 4

F_(13) = 3

F_(14) = 2

F_(15) = 0

F_(16) = 2

F_(17) = 2

F_(18) = 4

F_(19) = 1

F_(20) = 0

We have: 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0

Now, 1999 = 19(mod 20)

The 19th number in the sequence is 1.

So, the remainder is 1.

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25 divided by 8 is 3.125, but you want to have enough so the least amount you can buy is 4.

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A video game that usually costs $30.65 is marked down 60%. Kelvin determined that the new price of the game would be $18.39. Loo
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Un guardia forestal divisa, desde un punto de observación A, un incendio en la dirección N27° 10’ E. Otro guardia, que se encuen
Inga [223]

Answer:

la distancia del punto de observación A al incendio = 5.5 millas

la distancia del punto de observación B al incendio = 3.8 millas

Step-by-step explanation:

La expresión esquemática de la pregunta se puede ver en la imagen adjunta a continuación :

Del siguiente diagrama;

Usando la sine regla:

\dfrac{sin \  F}{f} = \dfrac{sin  \ A }{a}

\dfrac{sIn \  79}{6} = \dfrac{sin \   63}{a}

a sin 79 = 6 sin 63

a =\dfrac{6 \times sin \ 63}{sin \ 79}

a =\dfrac{6 \times 0.8910}{0.9816}

a = 5.446 millas

la distancia del punto de observación A al incendio = 5.5 millas (a la décima más cercana)

\dfrac{sin \  F}{f} = \dfrac{sin  \ B}{b}

\dfrac{sin \  79}{6} = \dfrac{sin  \ 38}{b}

b sin 79 = 6 sin 38

b  = \dfrac{6 \times sin  \ 38}{sin \  79}

b  = \dfrac{6 \times0.6157 }{0.9816}

b = 3.7635 millas

la distancia del punto de observación B al incendio = 3.8 millas (a la décima más cercana)

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