Simple Integers

n 3 432 n + 2018 n + 12 = k \frac{n^3-432n+2018}{n+12}=k

For n N n\in\mathbb N find the number of values of n n such that the k k as defined above is an integer and also find the largest possible value of n n for which k k is an integer.

If the the number of values of n n is a a and the largest possible value of n n is b b , then enter your answer as a + b a+b .

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The answer is 5475.

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1 solution

X X
Apr 30, 2018

k = n 2 12 n 288 + 5474 n + 12 k=n^2-12n-288+\frac{5474}{n+12} ,5474 has 16 factors,but 1,2,7 is smaller than 12,so a = 16 3 = 13 , b = 5474 12 = 5462 , a + b = 5475 a=16-3=13,b=5474-12=5462,a+b=5475

same way :)

Jun Endo - 3 years, 1 month ago

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