Algebra (Olympiad)hardFree

Unique Integer Making a Rational Function an Integer | IOQM

JEE Maths question with a full step-by-step solution.

Question
Let
f(n)=12n3−5n2−251n+3896n2−37n+45.f(n) = \frac{12n^3 - 5n^2 - 251n + 389}{6n^2 - 37n + 45}.
 There is a unique positive integer nn for which f(n)f(n) is an integer. Denote this unique value of nn by pp. Find the value of p+f(p)p + f(p).
Solution
Answer: 55
Step 1: Reduce the expression. By algebraic division,
f(n)=12n3−5n2−251n+3896n2−37n+45=2n+9+15n2−8n−166n2−37n+45.f(n) = \frac{12n^3 - 5n^2 - 251n + 389}{6n^2 - 37n + 45} = 2n + 9 + \frac{15n^2 - 8n - 16}{6n^2 - 37n + 45}.
 This is verified by expanding (2n+9)(6n2−37n+45)=12n3−20n2−243n+405(2n + 9)(6n^2 - 37n + 45) = 12n^3 - 20n^2 - 243n + 405 and adding the remainder 15n2−8n−1615n^2 - 8n - 16, which recovers the original numerator. Step 2: Factor the residual fraction.
15n2−8n−16=(3n−4)(5n+4),6n2−37n+45=(3n−5)(2n−9),15n^2 - 8n - 16 = (3n - 4)(5n + 4), \qquad 6n^2 - 37n + 45 = (3n - 5)(2n - 9),
 so
f(n)=2n+9+(3n−4)(5n+4)(3n−5)(2n−9).f(n) = 2n + 9 + \frac{(3n - 4)(5n + 4)}{(3n - 5)(2n - 9)}.
 Step 3: Set up the divisibility condition. Since 2n+92n + 9 is an integer, f(n)f(n) is an integer precisely when
(3n−4)(5n+4)(3n−5)(2n−9)\frac{(3n - 4)(5n + 4)}{(3n - 5)(2n - 9)}
 is an integer. If this ratio is an integer, then in particular (3n−5)∣(3n−4)(5n+4)(3n - 5) \mid (3n - 4)(5n + 4). The integers 3n−43n - 4 and 3n−53n - 5 are consecutive, hence coprime, so 3n−5∣(5n+4)3n - 5 \mid (5n + 4). Step 4: Reduce to a divisor of 37.
5(3n−5)−3(5n+4)=15n−25−15n−12=−37,5(3n - 5) - 3(5n + 4) = 15n - 25 - 15n - 12 = -37,
 so 3n−5∣373n - 5 \mid 37. Since 3737 is prime, 3n−5∈{1, −1, 37, −37}3n - 5 \in \{1,\ -1,\ 37,\ -37\}, giving
n=2,n=43,n=14,n=−323.n = 2, \quad n = \tfrac{4}{3}, \quad n = 14, \quad n = -\tfrac{32}{3}.
 The only positive integers among these are n=2n = 2 and n=14n = 14. Step 5: Test the two candidates.
f(2)=2(2)+9+(2)(14)(1)(−5)=13−285=375,f(2) = 2(2) + 9 + \frac{(2)(14)}{(1)(-5)} = 13 - \frac{28}{5} = \frac{37}{5},
 which is not an integer, so n=2n = 2 is rejected.
f(14)=2(14)+9+(38)(74)(37)(19)=37+2812703=37+4=41,f(14) = 2(14) + 9 + \frac{(38)(74)}{(37)(19)} = 37 + \frac{2812}{703} = 37 + 4 = 41,
 which is an integer. Step 6: Conclude. The unique positive integer for which f(n)f(n) is an integer is n=14n = 14, so p=14p = 14 and f(p)=f(14)=41f(p) = f(14) = 41. Therefore
p+f(p)=14+41=55.p + f(p) = 14 + 41 = 55.
 Answer: 55
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