Find all $a, b, c, d, e\in \{0, 1, 2, 4, 8\}$ such that $ a + b + c + d + e = 8.$

Question

Find all $a, b, c, d, e\in \{0, 1, 2, 4, 8\}$ such that $$a + b + c + d + e = 8.$$

Is there any method for this? I have no idea. I can just fix the limit.

Answer 1

This should be done through brute force or using some tricks based on the fact that almost every element in the set is a power of $2$. One could use a generating function and look at the coefficient of $x^8$ in $(1+x+x^2+x^4+x^8)^5$. A quick calculation in Maple gave me the coefficient of $x^8$ in the product is $125$.

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There is also a dynamic programming solution (in the case where the numbers are nonnegative).

To do this with dynamic programming, we'll build an $6\times 9$ matrix $M$ where $M_{i,j}$ contains the number of ways to add to $j-1$ using $i-1$ variables. We know that $M_{1,1}=1$ and $M_{1,j}=0$ for $j\not=1$ since the sum of no variables is $0$. Then, for $i>1$, we can calculate $$ M_{i,j}=M_{i-1,j}+M_{i-1,j-1}+M_{i-1,j-2}+M_{i-1,j-4}+M_{i-1,j-8}. $$ Here, we assume that nonpositive subscripts on a matrix correspond to the value $0$. The result that you're looking for is in position $M_{6,9}$.

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Alternatively, you can look at the different ways to get $8$ and count rearrangements. For example, you could have

• $8+0+0+0+0$. There are $\binom{5}{1}=5$ places to place the $8$.

• $4+4+0+0+0$. There are $\binom{5}{2}=10$ places to place the two $4$'s.

• $4+2+2+0+0$. There are $\binom{5}{1}\binom{4}{2}=30$ places to place the $4$ and then the two $2$'s.

• $4+2+1+1+0$. There are $\binom{5}{1}\binom{4}{1}\binom{3}{2}=60$ places to place the $4$ and then the $2$, and then the $2$ $1$'s.

• $4+1+1+1+1$. There are $\binom{5}{1}=5$ places to place the $4$ and the remaining values are all $1$.

• $2+2+2+2+0$. There are $\binom{5}{4}=5$ places to place the $2$'s.

• $2+2+2+1+1$. There are $\binom{5}{3}=10$ places to place the $2$'s and the remaining spots are for $1$'s. If you add this all up, you get $125$.

Answer 2

I could not resist the brute force and told my friend Ruby about it:

def bruteforce
  count = 0
  s = [0, 1, 2, 4, 8]
  s.each do |a|
    s.each do |b|
      s.each do |c|
        s.each do |d|
          s.each do |e|
            if a+b+c+d+e == 8
              count += 1
              puts "#{count}.: #{a}+#{b}+#{c}+#{d}+#{e}"
            end
          end
        end
      end
    end
  end
  count
end

and she found $125$ combinations:

1.: 0+0+0+0+8
2.: 0+0+0+4+4
3.: 0+0+0+8+0
4.: 0+0+2+2+4
5.: 0+0+2+4+2
6.: 0+0+4+0+4
7.: 0+0+4+2+2
8.: 0+0+4+4+0
9.: 0+0+8+0+0
10.: 0+1+1+2+4
11.: 0+1+1+4+2
12.: 0+1+2+1+4
13.: 0+1+2+4+1
14.: 0+1+4+1+2
15.: 0+1+4+2+1
16.: 0+2+0+2+4
17.: 0+2+0+4+2
18.: 0+2+1+1+4
19.: 0+2+1+4+1
20.: 0+2+2+0+4
21.: 0+2+2+2+2
22.: 0+2+2+4+0
23.: 0+2+4+0+2
24.: 0+2+4+1+1
25.: 0+2+4+2+0
26.: 0+4+0+0+4
27.: 0+4+0+2+2
28.: 0+4+0+4+0
29.: 0+4+1+1+2
30.: 0+4+1+2+1
31.: 0+4+2+0+2
32.: 0+4+2+1+1
33.: 0+4+2+2+0
34.: 0+4+4+0+0
35.: 0+8+0+0+0
36.: 1+0+1+2+4
37.: 1+0+1+4+2
38.: 1+0+2+1+4
39.: 1+0+2+4+1
40.: 1+0+4+1+2
41.: 1+0+4+2+1
42.: 1+1+0+2+4
43.: 1+1+0+4+2
44.: 1+1+1+1+4
45.: 1+1+1+4+1
46.: 1+1+2+0+4
47.: 1+1+2+2+2
48.: 1+1+2+4+0
49.: 1+1+4+0+2
50.: 1+1+4+1+1
51.: 1+1+4+2+0
52.: 1+2+0+1+4
53.: 1+2+0+4+1
54.: 1+2+1+0+4
55.: 1+2+1+2+2
56.: 1+2+1+4+0
57.: 1+2+2+1+2
58.: 1+2+2+2+1
59.: 1+2+4+0+1
60.: 1+2+4+1+0
61.: 1+4+0+1+2
62.: 1+4+0+2+1
63.: 1+4+1+0+2
64.: 1+4+1+1+1
65.: 1+4+1+2+0
66.: 1+4+2+0+1
67.: 1+4+2+1+0
68.: 2+0+0+2+4
69.: 2+0+0+4+2
70.: 2+0+1+1+4
71.: 2+0+1+4+1
72.: 2+0+2+0+4
73.: 2+0+2+2+2
74.: 2+0+2+4+0
75.: 2+0+4+0+2
76.: 2+0+4+1+1
77.: 2+0+4+2+0
78.: 2+1+0+1+4
79.: 2+1+0+4+1
80.: 2+1+1+0+4
81.: 2+1+1+2+2
82.: 2+1+1+4+0
83.: 2+1+2+1+2
84.: 2+1+2+2+1
85.: 2+1+4+0+1
86.: 2+1+4+1+0
87.: 2+2+0+0+4
88.: 2+2+0+2+2
89.: 2+2+0+4+0
90.: 2+2+1+1+2
91.: 2+2+1+2+1
92.: 2+2+2+0+2
93.: 2+2+2+1+1
94.: 2+2+2+2+0
95.: 2+2+4+0+0
96.: 2+4+0+0+2
97.: 2+4+0+1+1
98.: 2+4+0+2+0
99.: 2+4+1+0+1
100.: 2+4+1+1+0
101.: 2+4+2+0+0
102.: 4+0+0+0+4
103.: 4+0+0+2+2
104.: 4+0+0+4+0
105.: 4+0+1+1+2
106.: 4+0+1+2+1
107.: 4+0+2+0+2
108.: 4+0+2+1+1
109.: 4+0+2+2+0
110.: 4+0+4+0+0
111.: 4+1+0+1+2
112.: 4+1+0+2+1
113.: 4+1+1+0+2
114.: 4+1+1+1+1
115.: 4+1+1+2+0
116.: 4+1+2+0+1
117.: 4+1+2+1+0
118.: 4+2+0+0+2
119.: 4+2+0+1+1
120.: 4+2+0+2+0
121.: 4+2+1+0+1
122.: 4+2+1+1+0
123.: 4+2+2+0+0
124.: 4+4+0+0+0
125.: 8+0+0+0+0

Answer 3

Try approaching it as one would an R-SUM problem, with only a slight modification:

First consider the 25 element multiset with 5 copies each of 0, 1, 2, 4, 8.

Iterate over all three tuples $(a,b,c)$ and hash their sums. Then iterate over all two-tuples $(x,y)$ and check (each check takes constant time) if the value of $8-x-y$ has been hashed already. If so, then (a,b,c,x,y) satisfies $a+b+c+x+y = 8$. We don't overcount in the $(a,b,c)$s because we are hashing by tuples but we have to be slightly careful in the $(x,y)$s. In order to avoid counting $(a,b,c,x,y)$ twice, keep a boolean hash of all $(x,y)$ which led to good tuples $(a,b,c,x,y)$ and only count a tuple $(a,b,c,x,y)$ as good if $(x,y)$ hasn't been hashed yet.

The time complexity is $O(n^3)$, which is better than the $O(n^5)$ brute force approach, though in a problem this small it doesn't practically matter, this approach can be simply generalized to larger problems in which the time difference matters.