The principal at a school with N number of students asks them to clean up the school's large yard before the beginning of school.
To do this, the principal prepares a list with the name of every student in the school exactly once and divides the yard into various sized areas.
The students then start cleaning the yard one by one in the order listed. Each area of the yard is cleaned by only one student.
The principal, who is watching the students, takes note of the size of the area each student cleans.
After the entire courtyard has been cleaned, the principal notices a very interesting feature about the size of the areas cleaned by the students.
The total area cleaned by the first student on the list is exactly 3 times larger than the arithmetic average of the areas cleaned by the students following her on the list.
Similarly, the total area cleaned by every student on the list is exactly 3 times larger than the arithmetic average of the areas cleaned by the students who come after him on the list.
Accordingly, the total area cleared by the (N - 1)'st ranked student is exactly 3 times larger than the area cleared by the last student.
In addition to this property, the principal realizes that the total area cleaned by the first student on the list is exactly 595 times larger than the total area cleaned by the last student on the list.
How many students are there in the school?
Solution:
There are 34 students in the school.
Let the students be numbered 1, 2, 3, ... N
and their respective areas that they cleaned A1, A2, A3, ... AN
Then we know that A1 = 595 AN.
We also know that for any particular student k, the area cleaned by that student must be 3 times the average of the areas cleaned by the students following him, or
Ak = 3 ( (Ak+1 + Ak+2 + ... + AN ) / (N - k) )
Let AN = 1 and determine the pattern for the succeeding areas:
AN - 1 = 3 because it is equal to 3 times the average of AN, or 3 times 1 = 3.
AN - 2 = 6 because it is equal to 3 times the average of AN-1 and AN, or 3 times ((3 + 1) / 2) = 3 x 2 = 6.
AN - 3 = 10 because it is equal to 3 times the average of AN-2 and AN-1 and AN,
or 3 times ((6 + 3 + 1) / 3) = 3 x 5 = 10.
Here is the pattern: 1, 3, 6, 10, 15, 21, 28, coming from the last student back to the first.
The pattern for the areas cleaned by the students must be triangular numbers or multiples of them:
1, 3, 6, 10, 15, 21, ... 496, 528, 561, 595, ...
595 is the 34th triangular number, so one possible value of N is 34 and one posible value of A1 = 595 where A34 = 1.
It is not possible to determine the actual areas cleaned by the students. Here are some values that work for
A1, A2, A3, ... A32, A33, A34
595, 561, 528, ... 15, 10, 6, 3, 1
But all multiples of the above also work:
1190, 1122, ... 20, 12, 6, 2
Since we showed that the pattern for the areas is the sequence of triangular numbers, we can use the formula which generates the triangular numbers and set it equal to 595:
(N (N+1)) / 2 = 595
Then N2 + N = 2 (595)
N2 + N - 1190 = 0
(N +35) (N - 34) = 0
So, N = 34.
Correctly solved by:
| 1. Dr. Hari Kishan |
D.N. College, Meerut, Uttar Pradesh, India |
| 2. Kamal Lohia |
Holy Angel School, Hisar, Haryana, India |
| 3. Davit Banana | Istanbul, Turkey |
| 4. Colin (Yowie) Bowey | Beechworth, Victoria, Australia |
| 5. Kelly Stubblefield | Mobile, Alabama, USA |