Chairs and Tables - Part 3
At a coffee shop that I visit, the table has 3 legs, and the chairs have 5 legs. If there are a total of 300 legs, and there are at most 4 chairs at each table, what is the most number of chairs in the coffee shop?
Note: Every chair must be at a table.
Image Credit: Wikimedia Chairs and Tables by wazo0p
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
Yes, most of the trick lay in figuring out "how many times you have to subtract 3 to obtain a multiple of 5".
Simple. Arranging the given options in aascending order, and by trial and error method, we first have a look at the smallest no., i.e., 51. Multiplying it by 5 (legs in one chair), we get 255. Subtracting from 300, we get 45, which is divisible by 3 (no. of legs per table). Also, 51 chairs and 15 tables go together, with 3-4 chairs per table. Logical
What is this 300? total legs are 500?
Since there can be at most 4 chairs at a table and every chair must be at a table, the number of legs at any table/chairs "set" can be any of 3 , 8 , 1 3 , 1 8 , 2 3 .
So we are looking for solutions to the equation 3 a + 8 b + 1 3 c + 1 8 d + 2 3 e = 3 0 0 that maximizes the number of chairs in the shop. Since the chair-to-table ratio is greatest when there are 4 chairs at a table, we will want to find solutions that maximize e .
Now 2 3 ∗ 1 3 = 2 9 9 , but as the least number of legs at a table/chairs set is 3 , (i.e., a table on it own), we cannot have 1 3 sets with 4 chairs.
Next, we have 2 3 ∗ 1 2 = 2 7 6 , which leaves 2 4 legs to distribute. After distributing 1 8 legs to a 3 -chair set we have 6 legs left, which can only be distributed to two lone tables. This gives us a total of 1 2 ∗ 4 + 1 ∗ 3 = 5 1 chairs. We could also have distributed 8 legs to each of three 1 -chair sets, which would also yield 1 2 ∗ 4 + 3 ∗ 1 = 5 1 chairs.
So as 5 2 chairs is not an achievable solution and 5 1 is, we can (reasonably) conclude that the maximum number of chairs in the shop is 5 1 .
What is this 300? Total legs are 500?
Log in to reply
The question started out with 300 legs, and 51 is the correct answer for that number of legs. Somehow the question got changed to 500 legs, so now none of the answer options are correct. I'll report the problem to get it changed back to 300 as Kevin had intended.
Log in to reply
Thank you. My solution is as under for 300 legs. Least number of tables is
2
3
3
0
0
=
1
3
.
.
.
If there are t tables than we have
3
0
0
−
3
t
≡
0
(
m
o
d
5
)
.
⟹
t=15,20,.....Trying out
1
5
,
3
0
0
−
3
∗
1
5
=
2
5
5
=
5
1
c
h
a
i
r
s
.
15 tables can accommodate 51 chairs. So the answer is 51.
Log in to reply
@Niranjan Khanderia – I like this approach. The first calculation shows that t > 1 3 and the second shows that t must be a multiple of 5 . The least such t value will then yield the greatest possible value for the number of chairs. So there is no need to determine particular configurations like I did in my solution.
For 5 0 0 legs we would have ⌊ 2 3 5 0 0 ⌋ = 2 1 so there must be at least 2 1 tables. The least multiple of 5 greater than 2 1 is 2 5 , and 5 0 0 − 3 ∗ 2 5 = 4 2 5 , which translates to 8 5 chairs. (By coincidence 8 5 = 3 5 ∗ 5 1 , but it won't always scale up this way in general.) One configuration with 8 5 chairs includes 2 1 four-chair tables, and 1 one-chair table. We could also have 2 0 four-chair tables, 1 three-chair tables and 1 two-chair tables. I suppose the next natural question to ask is how many distinct configurations involve 8 5 chairs. Maybe I'll work on this sometime. :)
Log in to reply
@Brian Charlesworth – I have always admire you. Your knowledge, intelligence and still unassuming nature, just like Ujjwal. You solve the problem from the fundamentals. So your methods can be applied to any similar problems. I greatly admire this approach. Some problems have some more special properties. Using them can make a short solution. If you can give such solution ALSO, it would be helpful. Say in the problem of cyclic quadrilateral, with sides 3, and 6 with angle 30, if we knew that 6 was in fact the diameter, we could have a short solution. ( I am unable to locate this problem now). Even in the present problem, the number of chair should be a multiple of 3, so also number of tables a multiple of 5, since 300 is a multiple of both 3, and 5. So the answer can be 51 or 54. But 54 gives 10 tables that can have only 40 chairs. Hence we try 51 and it fits. Greeting.
Log in to reply
@Niranjan Khanderia – Thank you; I appreciate the compliments. Ujjwal has an efficient and intuitive approach to mathematics that I admire greatly. My tendency is to go straight to work on the general case and then solve the specific case afterwards, but as you say, some problems have "special" properties that lend themselves to short solutions. Sometimes I see the short cut and will make mention of it, but with this problem the question I found more interesting was that of finding the number of possible configurations that involved the maximum number of chairs, hence my more general approach.
With the cyclic quadrilateral question you have in mind, the key was establishing that 6 was in fact the diameter, which made the problem quite quick to solve. I posted two versions of that problem: the first gave more information than I had intended, which made it rather easy to see that 6 was the diameter; and the second gave different information from which it was much more of a challenge to establish that 6 was the diameter. It was still the same cyclic quadrilateral, but the difficulty of the problem varied greatly depending on the information provided.
First think of taking 4 chairs per table. You get 23 legs per table.
Now multiply 23 with 12 to obtain 276 legs and 12 tables and 48 chairs.
Now for left over 24 legs you have to think of that how many times you have to subtract 3 to obtain multiple of 5 so 3 tables and 3 chairs.
So total of 5 1 chairs.