If-a-n-6-n-8-n-find-a-1991-49-

Question Number 126400 by F_Nongue last updated on 20/Dec/20

I f a_{n} = 6^{n} + 8^{n} f i n d \frac{a_{1991}}{49} .

Commented by talminator2856791 last updated on 20/Dec/20

are you wanting remainder ?

Commented by MJS_new last updated on 20/Dec/20

a_{1991} > 10^{1798} \Rightarrow \frac{a_{1991}}{49} > 10^{1796}

Commented by MJS_new last updated on 20/Dec/20

the remainder is 42

Commented by F_Nongue last updated on 20/Dec/20

y e z I w a n t t h e r e m a i n d e r

Commented by MJS_new last updated on 20/Dec/20

the remainder you get by finding the periodical

remainders of \frac{a_{n}}{49} . start with n = 1, 2, 3, \dots

until you have one full period . then find the

position of 1991 within this period

Answered by floor(10²Eta[1]) last updated on 20/Dec/20

6^{1991} + 8^{1991} (\mod 49)

ϕ (49) = 7^{2} - 7 = 42

6^{1991} = {(6^{47})}^{42 + 17} \equiv 6^{17}

8^{1991} \equiv 8^{17}

a_{1991} \equiv 6^{17} + 8^{17} (\mod 49)

6^{2} = 36 \equiv - 13 (\mod 49)

6^{17} = {(6^{2})}^{8} . 6 \equiv 13^{8} . 6 = {(13^{2})}^{4} . 6 \equiv 22^{4} . 6

\equiv {(22^{2})}^{2} . 6 \equiv {(- 6)}^{2} . 6 \equiv 6^{2} . 6 \equiv - 13.6 \equiv - 78 \equiv 20 (\mod 49)

8^{2} = 64 \equiv 15 (\mod 49)

8^{17} \equiv {(8^{2})}^{8} . 8 \equiv 15^{8} . 8 = {(15^{2})}^{4} . 8 \equiv {(- 20)}^{4} . 8 = 20^{4} . 8

= {(20^{2})}^{2} . 8 \equiv 8^{2} . 8 \equiv 15.8 \equiv 22 (\mod 49)

a_{1991} \equiv 20 + 22 \equiv 42 (\mod 49)