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Integral solutions of a binomial-coefficient inequality and the divisors of their sum | JEE Advanced

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

Question
If SS is the sum of all the possible integral values of xx satisfying the equation
20[x3−5x2+6x4−x−1C5]−4[x−1C4−4(x−1C3)]>0,20\left[\frac{x^3-5x^2+6x}4-{}^{x-1}C_5\right]-4\left[{}^{x-1}C_4-4\left({}^{x-1}C_3\right)\right] > 0,
 then the number of divisors of SS is
Solution
Answer: 8
Step 1: x−1C5{}^{x-1}C_5 needs x−1x-1 to be an integer with x−1≥5x-1 \ge 5, so xx is an integer and x≥6x \ge 6. Putting n=x−1n = x-1,
n≥5n \ge 5
 Step 2: x3−5x2+6x=x(x−2)(x−3)=(n+1)(n−1)(n−2)x^3-5x^2+6x = x\left(x-2\right)\left(x-3\right) = \left(n+1\right)\left(n-1\right)\left(n-2\right), and
20 nC5=n(n−1)(n−2)(n−3)(n−4)620\,{}^nC_5 = \frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)}6
4 nC4=n(n−1)(n−2)(n−3)6,16 nC3=16n(n−1)(n−2)64\,{}^nC_4 = \frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}6 ,\qquad 16\,{}^nC_3 = \frac{16n\left(n-1\right)\left(n-2\right)}6
 Every term carries (n−1)(n−2)\left(n-1\right)\left(n-2\right). Step 3: Taking it out, the left side is
(n−1)(n−2)6[30(n+1)−n(n−3)(n−4)−n(n−3)+16n]\frac{\left(n-1\right)\left(n-2\right)}6\Big[30\left(n+1\right)-n\left(n-3\right)\left(n-4\right)-n\left(n-3\right)+16n\Big]
=(n−1)(n−2)6(−n3+6n2+37n+30)= \frac{\left(n-1\right)\left(n-2\right)}6\left(-n^3+6n^2+37n+30\right)
 Step 4: n≥5n \ge 5, so (n−1)(n−2)>0\left(n-1\right)\left(n-2\right)>0 and the inequality becomes
n3−6n2−37n−30<0n^3-6n^2-37n-30<0
 n=10n = 10 is a root, so
(n−10)(n2+4n+3)=(n−10)(n+1)(n+3)<0\left(n-10\right)\left(n^2+4n+3\right) = \left(n-10\right)\left(n+1\right)\left(n+3\right)<0
 Step 5: n≥5n \ge 5 makes n+1>0n+1>0 and n+3>0n+3>0, so
n−10<0  ⟹  5≤n≤9  ⟹  x=6,7,8,9,10n-10<0 \implies 5 \le n \le 9 \implies x = 6,7,8,9,10
 Step 6:
S=6+7+8+9+10=40=23×5S = 6+7+8+9+10 = 40 = 2^3\times5
number of divisors=(3+1)(1+1)=8\text{number of divisors} = \left(3+1\right)\left(1+1\right) = 8
 Check: the left side is 480,840,1200,1386,1120480, 840, 1200, 1386, 1120 at x=6,7,8,9,10x = 6,7,8,9,10, and 00 at x=11x = 11. Answer: 88.
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