N-3548-9-2537-31-Determinate-the-last-digit-of-N-

Question Number 124110 by mathocean1 last updated on 30/Nov/20

N = {(3548)}^{9} \times {(2537)}^{31}

D e t e r m i n a t e t h e l a s t d i g i t o f N .

Answered by floor(10²Eta[1]) last updated on 24/Dec/20

N = {(3548)}^{9} {(2537)}^{31} \equiv 8^{9} 7^{31} (\mod 10)

ϕ (10) = 4 .

8^{9} 7^{31} = {(8^{2})}^{4 + 1} {(7^{7})}^{4 + 3} \equiv 8 . 7^{3} \equiv 2.7 \equiv 4 (\mod 10)

last digit of N is 4

Answered by MJS_new last updated on 30/Nov/20

the last digit of N is the same as that of

8^{9} \times 7^{31}

for n ⩾ 1

the last digit of 8^{n} is {\begin{cases} 8; n = 4 k + 1 \\ 4; n = 4 k + 2 \\ 2; n = 4 k + 3 \\ 6; n = 4 k + 4 \end{cases}; k ⩾ 0

9 = 4 k + 1 \Rightarrow last digit of 8^{9} is 8

the last digit of 7^{n} is {\begin{cases} 7; n = 4 k + 1 \\ 9; n = 4 k + 2 \\ 3; n = 4 k + 3 \\ 1; n = 4 k + 4 \end{cases}; k ⩾ 0

31 = 4 k + 3 \Rightarrow last digit of 7^{31} is 3

8 \times 3 = 24 \Rightarrow last digit of N is 4

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