Question

Q. Three natural numbers x, y , z leave remainders 18,21,20 respectively. When divided by 23. Find the remainder when (x + y + z) is divided by 23.

please tell me the answer with explanation ​

Answer 1

Hope this answer helps you

Answer 2

$\huge\mathtt\red{Answer}$

$\boxed{x+y+z\:=0}$

$\huge\mathtt\red{Solution}$

$\mathtt{By\:division\:algorithm,}$

$\mathtt\pink{→x\:= \:k(23)\:+\:18}$

$\mathtt\orange{→y\:= \:l(23)\:+\:21}$

$\mathtt\red{→m\:= \:k(23)\:+\:20}$

$\mathtt{Adding\:these\:3\:equation\:we\:get}$

$\mathtt\pink{→x+y+z\:=\:23(k+l+m)\:+\:69}$

$\mathtt\pink{→23\:=\:(k+l+m+3)}$

$\mathtt{x+y+z\:is\:exactly\:divided\:by\:23}$

$\mathtt\pink{Hence\:the\:remender\:when\:x+y+z\:is\:divided\:by\:23\:is\:0}$