N=(3548)^9 ×(2537)^(31) Determinate the last digit of N.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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
	$the last digit of N is the same as that of 8^9 ×7^(31) for n≥1 the last digit of 8^n is { ((8; n=4k+1)),((4; n=4k+2)),((2; n=4k+3)),((6; n=4k+4)) :}; k≥0 9=4k+1 ⇒ last digit of 8^9 is 8 the last digit of 7^n is { ((7; n=4k+1)),((9; n=4k+2)),((3; n=4k+3)),((1; n=4k+4)) :}; k≥0 31=4k+3 ⇒ last digit of 7^(31) is 3 8×3=24 ⇒ last digit of N is 4$
	$the last digit of N is the same as that of$ $8^{9} \times 7^{31}$ $for n ⩾ 1$ $the last digit of 8^{n} is {\begin{cases} 8; n = 4 k + 1 \\ 4; n = 4 k + 2 \\ 2; n = 4 k + 3 \\ 6; n = 4 k + 4 \end{cases}; k ⩾ 0$ $9 = 4 k + 1 \Rightarrow last digit of 8^{9} is 8$ $the last digit of 7^{n} is {\begin{cases} 7; n = 4 k + 1 \\ 9; n = 4 k + 2 \\ 3; n = 4 k + 3 \\ 1; n = 4 k + 4 \end{cases}; k ⩾ 0$ $31 = 4 k + 3 \Rightarrow last digit of 7^{31} is 3$ $8 \times 3 = 24 \Rightarrow last digit of N is 4$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum