What is the coefficient of x³³ in the expansion [(1+x)(1+3x³)(1+9x⁹)(1+27x²⁷)]²?

 

Solution 1: Bringing the square in gives

(1+x+x+x²)(1+3x³+3x³+3²x⁶)(1+9x⁹+9x⁹+9²x¹⁸)(1+27x²⁷+27x²⁷+27²x⁵⁴)

Notice that in each factor, the power of x is 0, 1 or 2 times a power of 3. Since the base 3 representation of 33 is 1020³, the only way to get x³³ from this expansion is to use the term x²⁷×x⁰×x⁶×x⁰. Hence, the coefficient of x³³ is (2×27)×1×3²×1=486.

Solution 2: Bring the square in as in Solution 1. Looking at the first three factors, the highest power we can achieve is 18+6+2=26<33, so we can't use the constant term in the fourth factor. We also can't use the x⁵⁴ term in the fourth factor, thus the only possibility of achieving x³³ is to use x²⁷ in the fourth, and get x⁶ from the first three. Now consider the third factor, we can't use x⁹ or x¹⁸, so we have to use the constant term. From the first 2 factors, it is clear that we can only use 1from the first factor and x⁶ from the second factor. Hence, the coefficient of x³³ is: 1×3²×1×(2×27)=486.

algebra

How many ordered triples of pairwise distinct, positive integers (a,b,c) are there such that abc=10⁶?

Clarification: Ordered triples - (1,10,10⁵) and (10⁵,10,1) count as different solutions.

Clarification: Pairwise distinct - a set of integers are pairwise distinct if no two of them are the same. For example, the set (1,1,10⁶) is NOT pairwise distinct, hence does not count as a solution.

Set a=2^a1⋅5^a2,b=2^b1⋅5^b2,c=2^c1⋅5^c2. We first count the number of triples without restriction. We seek the number of non-negative integer solutions to a₁+b₁+c₁=6,a₂+b₂+c₂=6. This is equivalent to arranging 6 1's with 2 0's (dividing blocks), thus there are (6+22) ways to solve either of the two equations. By the rule of product, there are (82)⋅(82)=784 solutions in total.

If a=b, we seek the number of non-negative integer solutions to 2a₁+c₁=6,2a₂+c₂=6. For each equation, there are 4 solutions, corresponding to ai=0,1,2,3. By the product rule, there are 4⋅4=16 solutions in all. The cases b=c and c=a are exactly the same. If a=b=c, there is exactly 1 solution.

Let A,B, and C be the set of triples where b=c,c=a, and a=b, respectively. By the Principle of Inclusion and Exclusion, |A∪B∪C|=|A|+|B|+|C|−|A∩B|−|B∩C|−|C∩A|+|A∩B∩C|=16+16+16−1−1−1+1=46. Hence, the number of triples with pairwise distinct values is 784−46=738.