what is the largest positive integer n such that n^3 +100 is divisible by n+10 ?


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Question Number 104038 by bramlex last updated on 19/Jul/20

$w h a t i s t h e l a r g e s t p o s i t i v e$ $i n t e g e r n s u c h t h a t n^{3} + 100 i s$ $d i v i s i b l e b y n + 10 ?$
	Answered by john santu last updated on 19/Jul/20
	$Let n^3 +100 = (n+10)(n^2 +an+b)+c = n^3 +n^2 (10+a)+n(b+10a)+10b+c equating coefficients yields { ((100+a = 0)),((10a+b = 0)),((10b + c =100)) :} solving this system yields a=−10, b = 100 & c = −900 therefore by the Euclidean Algorithm we get n+10 = gcd(n^3 +100, n+10) = gcd(−900,n+10) = gcd(900,n+10) . The maximum value for n is hence n = 890. (JS ⊛)$
	$L e t n^{3} + 100 = (n + 10) (n^{2} + a n + b) + c$ $= n^{3} + n^{2} (10 + a) + n (b + 10 a) + 10 b + c$ $e q u a t i n g c o e f f i c i e n t s y i e l d s$ ${\begin{cases} 100 + a = 0 \\ 10 a + b = 0 \\ 10 b + c = 100 \end{cases}$ $s o l v i n g t h i s s y s t e m y i e l d s$ $a = - 10, b = 100 & c = - 900$ $t h e r e f o r e b y t h e E u c l i d e a n$ $A l g o r i t h m w e g e t n + 10 =$ $g c d (n^{3} + 100, n + 10) =$ $g c d (- 900, n + 10) =$ $g c d (900, n + 10) . T h e m a x i m u m$ $v a l u e f o r n i s h e n c e n = 890 .$ $(J S ⊛)$
	Commented by bramlex last updated on 19/Jul/20

	$n i c e! ∙ ⌣ ∙$
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