Count me III

Probability Level 4

Find the number of whole number solutions to the equation a + b + c + d = 30 a + b + c + d = 30 such that a ≤ b < c ≤ d a\leq b < c \leq d .


The answer is 249.

Jeffrey Robles by Jeffrey Robles , 26, Ireland

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1 solution

Jeffrey Robles
Aug 30, 2018

Redefine the variables as follows a + ( a + k 3 ) ⏟ b + ( a + k 3 + k 2 + 1 ) ⏟ c + ( a + k 3 + k 2 + k 1 + 1 ) ⏟ d = 30 a + \underbrace{\left(a + k_3\right)}_b +\underbrace{\left(a + k_3+k_2+1\right)}_c + \underbrace{\left(a + k_3+k_2 +k_1+1\right)}_d =30

where a , k i ∈ W a,k_i \in \mathbb{W} (whole numbers). Note that by this definition, the condition a ≤ b < c ≤ d a\leq b < c \leq d is always satisfied. Further note that c c is defined as b + 1 + k 2 b + 1+k_2 so that even when k 2 = 0 k_2 = 0 , c c is not equal to b b .

Simplifying this equation yields 4 a + 3 k 3 + 2 k 2 + k 1 = 28 3 k 3 + 2 k 2 + k 1 = 28 − 4 a \begin{aligned} 4a + 3k_3 + 2k_2 + k_1 &= 28 \\ 3k_3 + 2k_2 + k_1 &= 28 - 4a \end{aligned}

It follows that 0 ≤ a ≤ 7 0\leq a \leq 7 . Also, since the RHS of the last equation and the term 2 k 2 2k_2 are always even, k 3 k_3 and k 1 k_1 has the same parity, i.e k 3 ≡ k 1 m o d 2 k_3 \equiv k_1 \mod 2 .

Case: k 3 k_3 and k 1 k_1 are both odd 3 ( 2 k 3 ′ + 1 ) + 2 k 2 + ( 2 k 1 ′ + 1 ) = 28 − 4 a 6 k 3 ′ + 2 k 2 + 2 k 1 ′ = 24 − 4 a 3 k 3 ′ + k 2 + k 1 ′ = 12 − 2 a k 2 + k 1 ′ = 12 − 2 a − 3 k 3 ′ \begin{aligned} 3(2k_3' +1) + 2k_2 + (2k_1' + 1) & = 28 - 4a\\ 6k_3' + 2k_2 + 2k_1' &= 24 - 4a \\ 3k_3' + k_2 + k_1' &= 12 - 2a \\ k_2 + k_1' &= 12 - 2a - 3k_3' \end{aligned}

Note that since k 2 + k 1 ′ ≥ 0 k_2 + k_1'\geq 0 and 12 − 2 a − 3 k 3 ′ ≥ 0 12 - 2a - 3k_3' \geq 0 , the following should hold: 0 ≤ a ≤ 6 0\leq a \leq 6 and k 3 ′ ≤ ⌊ 12 − 2 a 3 ⌋ k_3' \leq \left \lfloor \frac{12-2a}{3} \right \rfloor .

Further note that for each pair ( a , k 3 ′ ) \left(a,k_3'\right) , there are a total of 13 − 2 a − 3 k 3 ′ 13 - 2a - 3k_3' non-negative solutions ( k 2 , k 1 ′ ) \left(k_2,k_1'\right) . Thus, the total number of solutions is ∑ a = 0 5 ∑ k 3 ′ = 0 ⌊ 12 − 2 a 3 ⌋ ( 13 − 2 a − 3 k 3 ′ ) = 102 \sum_{a=0}^5 \sum_{k_3' = 0}^{\left \lfloor \frac{12-2a}{3} \right \rfloor} \left(13-2a-3k_3'\right) = 102

Case: k 3 k_3 and k 1 k_1 are both even

Solution is analogous to the previous case

3 ( 2 k 3 ′ ) + 2 k 2 + ( 2 k 1 ′ ) = 28 − 4 a 6 k 3 ′ + 2 k 2 + 2 k 1 ′ = 28 − 4 a 3 k 3 ′ + k 2 + k 1 ′ = 14 − 2 a k 2 + k 1 ′ = 14 − 2 a − 3 k 3 ′ \begin{aligned} 3(2k_3') + 2k_2 + (2k_1') & = 28 - 4a\\ 6k_3' + 2k_2 + 2k_1' &= 28 - 4a \\ 3k_3' + k_2 + k_1' &= 14 - 2a \\ k_2 + k_1' &= 14 - 2a - 3k_3' \end{aligned}

Note that since k 2 + k 1 ′ ≥ 0 k_2 + k_1'\geq 0 and 14 − 2 a − 3 k 3 ′ ≥ 0 14 - 2a - 3k_3' \geq 0 , the following should hold: 0 ≤ a ≤ 7 0\leq a \leq 7 and k 3 ′ ≤ ⌊ 14 − 2 a 3 ⌋ k_3' \leq \left \lfloor \frac{14-2a}{3} \right \rfloor .

Further note that for each pair ( a , k 3 ′ ) \left(a,k_3'\right) , there are a total of 15 − 2 a − 3 k 3 ′ 15 - 2a - 3k_3' non-negative solutions ( k 2 , k 1 ′ ) \left(k_2,k_1'\right) . Thus, the total number of solutions is ∑ a = 0 5 ∑ k 3 ′ = 0 ⌊ 14 − 2 a 3 ⌋ ( 15 − 2 a − 3 k 3 ′ ) = 147 \sum_{a=0}^5 \sum_{k_3' = 0}^{\left \lfloor \frac{14-2a}{3} \right \rfloor} \left(15-2a-3k_3'\right) = 147

Therefore, there are a total of 102 + 147 = 249 102 + 147 = \boxed{249} solutions.

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