# Puzzle

The letters $$a, b, c, d, e$$ and $$f$$ represent single digits and each letter represents a different digit. They satisfy the following equations: $$a+b=d, \quad b+c=e \quad \text { and } \quad d+e=f \text {. }$$ Find all possible solutions for the values of $$a, b, c, d, e$$ and $$f$$.

This is clearly easily solved by computer, but what is an elegant and hopefully quick way of doing it by hand?

## 3 Answers

Solutions:

Eight possible permutations

Explanation:

By manupulating the equations, we get a+2b+c=f=d+e. This means that the least possible value of f must be 7, since none of the numbers can be zero; however, this means (a,c) belongs to {2,3} and b=1, but it doesn't satisfy e and f. So the only two possibilities for f are 8 and 9.

For f=8, b must be less than 3. For a,c = {1,3}, b=2, we don't have a solution so it is discarded. On to the next solution b=1, a,c={2,4}, we get d,e={3,5} so two solutions from f=8.

For f=9, b must be lesser than 4. For b=3, a,c={1,2} is the only possibility and d,e={4,5} hence it satisfies all equations. For b=2, a,c = {1,4} is the only possibility once again and d,e = {6,3}. This also satisfies all equations. For b=1, we have a,c = {2,5} or {3,4}. But a,c = {3,4} does not satisfy as d,e = {4,5} which has 4 as a common digit. For a,c = {2,5}, we have d,e = {3,6} and it satisfies all the equations.

Hence there are 6 solutions for f=9 and 2 solutions for f=8. Therefore the total number of solutions is 8.

• Maybe I don't understand your notation but how do you get 2 solutions in the second spoiler box? Either a,c={2,4} means a=2 and c=4 and similarly for d,e={3,5} so only one solution or it means {a,c}={2,4} with the two solutions a=2,c=4 and a=4, c=2 and similarly for d and e which gives 4 solutions in total. May 19 at 6:32
• @quarague Its only exactly two solutions for the second spoiler box. Look at the original equations to see the direct correspondence between a and d. May 19 at 6:54
• So you do mean {a,c}={2,4} and {d,e}={3,5} which on its own would give 4 solutions but the original equations give an additional condition so you only have 2 solutions. Maybe extend your explanation a little bit to include that. May 19 at 6:59
• d,e are governed by a,b and c so for different value of a,b,c there will only one possibility of d and e. I should have elaborated but I tried to keep it concise. May 19 at 10:04

Overall, I've counted

Eight solutions

so there will inevitably be some degree of case bashing. This is how I would proceed.

Notice firstly that

None of the digits is zero since that would make some of the others equal or negative.

Suppose $$f < 9$$

Then either $$d$$ or $$e$$ is less than $$4$$ (since they are distinct). Since both are the sum of distinct digits, it must be that the smaller of the two is $$3$$ and the other is $$4$$ or $$5$$.
Firstly, suppose that this $$d=3$$. Then $$a$$ and $$b$$ must be $$1$$ and $$2$$ in some order. Since $$e$$ must be $$4$$ or $$5$$ its clear that $$b$$ cannot be $$2$$ (since $$c$$ would have to be $$2$$ or $$3$$) and if $$b=1$$ then $$c=4$$ and $$e=5$$ is the only possible solution. This gives us the first solution $$(a,b,c,d,e,f) = (2,1,4,3,5,8)$$ Since the equations remain the same if we swap $$d$$ with $$e$$ and $$a$$ with $$c$$, we quickly obtain a second solution which comes from assuming $$e=3$$ instead $$(a,b,c,d,e,f) = (4,1,2,5,3,8)$$ and these are all the solutions in this branch

Now suppose $$f=9$$

Then $$d$$ and $$e$$ are either $$3$$ and $$6$$ or $$4$$ and $$5$$, in some order (since we've established that both $$d$$ and $$e$$ must be at least $$3$$).
$$d=3$$ gives us that $$a$$ and $$b$$ are $$1$$ and $$2$$ in some order.
With $$b=2$$ we must have $$c=4$$ and this gives another solution $$(a,b,c,d,e,f) = (1,2,4,3,6,9)$$ With $$b=1$$ we must have $$c=5$$ and we have yet another solution $$(a,b,c,d,e,f) = (2,1,5,3,6,9)$$ Again, we can reverse the role of $$(d,e)$$ and $$(a,c)$$ to retrieve the solutions in the $$e=3$$ branch $$(a,b,c,d,e,f) = (4,2,1,6,3,9)$$ $$(a,b,c,d,e,f) = (5,1,2,6,3,9)$$ Finally, $$d=4$$ gives us $$a$$ and $$b$$ being $$1$$ and $$3$$ in some order.
$$b=1$$ doesn't work because it would mean $$c=4=d$$.
$$b=3$$ gives us another solution $$(a,b,c,d,e,f) = (1,3,2,4,5,9)$$ and swapping $$(d,e)$$ and $$(a,c)$$ gives us the last solution which comes from the $$e=4$$ branch $$(a,b,c,d,e,f) = (2,3,1,5,4,9)$$

I just wanted to 'see' what the solutions looked like, but I disqualified myself by using a computer to get the solutions. Anyway, here is a visualization for your amusement.

# Visualization

The nodes of this hypergraph are the possible digit values that occurred in at least one solution. The coloured set covers are the hyperedges. The hyperedges are labelled 0-to-7 for the eight solutions.

# Observation

It looks like all solutions cover the set $$\{1,2,3\}$$. Maybe a simple rule based on these three special choices can be constructed to get what the remaining digits of the solution have to be (i.e. how to pick the remaining $$\{4,5,6,8\}$$). The number 7 doesn't get to join the party.

# Code for Visualization

import hypernetx as hnx
import matplotlib.pyplot as plt
from itertools import permutations

values = set(range(10))

solutions = {}
sol_count = 0
for perm in permutations(values, r=6):
a,b,c,d,e,f = perm
if a + b == d:
if b + c == e:
if d + e == f:
print(a,b,c,d,e,f)
solutions[str(sol_count)] = perm
sol_count += 1
else:
continue
else:
continue
else:
continue

H = hnx.Hypergraph(solutions)

hnx.drawing.draw(H)
plt.show()

• rot13: Zl pbqr vf cerggl anvir. Bar guvat lbh pbhyq punatr vs lbh jnagrq vg eha zber rssvpvragyl vf gb ercynpr inyhrf = frg(enatr(10)) jvgu inyhrf = frg(enatr(1,10)) - {7} fvapr mreb naq frira ner arire va gur fbyhgvbaf sbe guvf ceboyrz. May 18 at 20:48