Question

Find the largest positive integer n such that (n+2) divides (n^2+4) .

Answer 1

I remember a same question as this.

The expression (n^2)(n^2–1)(n^2-n-2), can be written as

(n^2)*(n+1)*(n-1)*(n-2)*(n+1)

= (n^2)*(n+1)^2*(n-1)*(n-2)

if we substitue k for n-2, we get (k+2)^2*(k+3)^2*(k+1)*k

= k*(k+1)*(k+2)^2 * (k+3)^2 — (Since n goes from 1 to … k goes from -1 to …)

When k = -1 or k = 0, we have 0, so we don’t have to worry, any number divides zero (0).

When k = 1, we have 1*2*(9)*(16) = 288

When k = 2, we have 2*3*(16)*25 = 2400

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Answer 2

Step-by-step explanation:

By division we find that n^3+100=(n+10)(n2−10n+100)−900.

Therefore, if n+10 divides n3+100, then it must also divide 900. Since we are looking for largest n, n is maximized whenever n+10 is, and since the largest divisor of 900 is 900, we must have n+10=900⇒n=890

The largest n is therefore 890 .

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