Algebra (Olympiad)hardFree

Positive Integral Solutions of a Partial-Sum Equation | IOQM

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

Question
Find the number of positive integral solutions of the equation
x1+2(x1+x2)+3(x1+x2+x3)+4(x1+x2+x3+x4)+⋯+15(x1+x2+⋯+x15)1255.x_1 + 2(x_1 + x_2) + 3(x_1 + x_2 + x_3) + 4(x_1 + x_2 + x_3 + x_4) + \cdots + 15(x_1 + x_2 + \cdots + x_{15}) 1255.
Solution
Answer: 1
Step 1: Express the left side as a single linear combination of the variables. The left side is ∑k=115k(x1+x2+⋯+xk)\sum_{k=1}^{15} k(x_1 + x_2 + \cdots + x_k). The variable xjx_j appears in the group (x1+⋯+xk)(x_1 + \cdots + x_k) for every k≥jk \ge j, so its total coefficient is ∑k=j15k\sum_{k=j}^{15} k. Hence
LHS=∑j=115cjxj,cj=∑k=j15k=120−j(j−1)2,\text{LHS} = \sum_{j=1}^{15} c_j x_j, \qquad c_j = \sum_{k=j}^{15} k = 120 - \frac{j(j-1)}{2},
 using the standard formula ∑k=115k=15⋅162=120\sum_{k=1}^{15} k = \dfrac{15 \cdot 16}{2} = 120. Step 2: Compute the minimum value. Each xjx_j is a positive integer, so xj≥1x_j \ge 1. The minimum of the left side occurs at x1=x2=⋯=x15=1x_1 = x_2 = \cdots = x_{15} = 1, where each partial sum satisfies x1+⋯+xk=kx_1 + \cdots + x_k = k. Then, using the standard formula ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6},
LHSmin⁡=∑k=115k⋅k=∑k=115k2=15⋅16⋅316=1240.\text{LHS}_{\min} = \sum_{k=1}^{15} k \cdot k = \sum_{k=1}^{15} k^2 = \frac{15 \cdot 16 \cdot 31}{6} = 1240.
 Step 3: Determine the excess to be distributed. Since the required value is 1255>12401255 > 1240, the variables must exceed the all-ones configuration. Write xj=1+djx_j = 1 + d_j with each djd_j a non-negative integer. Then
LHS−LHSmin⁡=∑j=115cjdj=1255−1240=15.\text{LHS} - \text{LHS}_{\min} = \sum_{j=1}^{15} c_j d_j = 1255 - 1240 = 15.
 Step 4: Examine the coefficients. Since cj=120−j(j−1)2c_j = 120 - \dfrac{j(j-1)}{2} is a decreasing function of jj, the two smallest coefficients are
c15=15,c14=14+15=29,c_{15} = 15, \qquad c_{14} = 14 + 15 = 29,
 and cj≥c14=29c_j \ge c_{14} = 29 for every j≤14j \le 14. Step 5: Locate the only admissible increase. If dj≥1d_j \ge 1 for any index j≤14j \le 14, then cjdj≥29>15c_j d_j \ge 29 > 15, which already exceeds the required total. Therefore d1=d2=⋯=d14=0d_1 = d_2 = \cdots = d_{14} = 0, and the equation reduces to
c15d15=15  ⟹  15d15=15  ⟹  d15=1.c_{15} d_{15} = 15 \implies 15 d_{15} = 15 \implies d_{15} = 1.
 Step 6: State and verify the unique solution.
x1=x2=⋯=x14=1,x15=2.x_1 = x_2 = \cdots = x_{14} = 1, \qquad x_{15} = 2.
 The partial sums are 1,2,…,14,161, 2, \ldots, 14, 16, so
LHS=∑k=114k2+15⋅16=1015+240=1255,\text{LHS} = \sum_{k=1}^{14} k^2 + 15 \cdot 16 = 1015 + 240 = 1255,
 which confirms the value. Exactly one positive integral solution exists. Answer: 1
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