what-is-the-largest-positive-integer-n-such-that-n-3-100-is-divisible-by-n-10-

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

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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