Well done Daniel and Mariana from
St Julian's School; Eduardo from The British School, Manila; Wesley
from Milpitas High School; Chuyi Yang from Loughborough High School
and Andrei from Tudor Vianu National College, Bucharest.
Trial and improvement was the
chosen method for some of you, Mariana found all the solutions
using inequalities and Andrei used a similar method carefully
working through all the possibilities.
Removing 3 consecutive numbers
from n numbers the solutions are: n=11 (removing 1,2 and 3); n=15
(removing 9, 10 and 11) and n=17 (removing 15, 16 and 17). If the
numbers removed are not consecutive then the solutions are n=13
(where any three numbers adding up to 16 can be removed); n=15
(where any three numbers adding up to 30 can be removed and n= 17
(where any three numbers adding up to 48 can be removed).
Several of you noted that n is not
even because if it was then (n3) would be odd and 7.5 multiplied
by (n3) would not be a whole number, but as 7.5(n3) is equal to
the sum of whole numbers it has to be a whole number.
Chuyi, who is much younger than
any of the other contributors, beautifully explained a trial and
improvement method as follows showing that logical thinking can
sometimes go a long way without needing a lot of mathematical
knowledge.
Trial for n 
Trial 
Error 
21  This is just trying a high odd number 
If n is 21 then x/18=7.5 so x will be 135. The sum of all the
numbers between 1 and21 is 231 so the three removed numbers must
total 96. There are such numberswhich are 31, 32 and 33 but since
the numbersonly go up to 21 this is not possible.

This does not work because the numbers are out of range, so it
means that this number is too high.

19 '?? This is going down by one odd number. 
If n is 19 then x/16=7.5 so x will be 120. The sum of all the
numbers between 1 and 19 is 190 so the three removed numbers must
total 70. But there are no three consecutive numbers that total 70
so 19 cannot be a possible value for n. 
This does not work, the three removed consecutive numbers will
be around 23 and even that is out of range and too high. 
17 '?? Once more going down by an
odd number, is this low enough? 
If n is 17 then x/14=7.5 so x will be 105. The sum of all the
numbers between 1 and 17 is 153 so the three removed numbers must
total 48. There are such numbers which are 15, 16 and 17. 
This works well and therefore is a right answer. So this is
just low enough as one of the removed numbers is the highest in the
range. 
15 '?? Even though a correct answer
has been found already it always good to check if there is more
than one answer. This was also the first estimate made so this is
also to check the original estimate. 
If n is 15 then x/12=7.5 so x will be 90. The sum of all the
numbers between 1 and 15 is 120 so the three removed numbers must
total 30. There are such numbers which are 9, 10 and 11. 
This works well also and therefore is a right answer, so the
first estimate made was correct 
13 '?? This is going down now and I have realised that the
total of the three consecutive numbers are becoming smaller and
smaller and this is just experimenting on how low it can go 
If n is 13 then x/10=7.5 so x will be 75. The sum of all the
numbers between 1 and 13 is 91 so the three removed numbers must
total 16. There are no possible numbers so 13 is an impossible
value for n. 
This is an impossible possibility for n. The rough number for
the three consecutive numbers would be roughly five so it is still
possible to go down further. 
11 '?? This is going down further to
see if there might be any possible values for n which are quite
small. 
If n is 11 then x/8=7.5 so x will be 60. The sum of all the
numbers between 1 and 11 is 66 so the three removed numbers must
total 6. There are such numbers and they are 1, 2 and 3. 
This works well also and therefore is a right answer. Because
this goes down to 1 as a removed number then it is impossible to go
any further as 1 is the smallest value. Therefore 11 is the
smallest possible answer for n. 
Conclusion: n=17, 15 and 11
are the only possible
solutions.



Here is Mariana's solution
1.Solution with
consecutive numbers .
The sum of integers 1 to n is: $1 + 2 + 3 + \cdots + n =
{n(n+1)\over 2}.$ Let the three numbers removed be $x$, $x+1$ and
$x+2$ which add up to $3x+3$. Then $${{n(n+1)\over 2} (3x+3)\over
n3}=7.5 \quad (1)$$ This simplifies to: $${n^214n+39\over 6} = x.
\quad (2)$$ The smallest value of the first of the three
consecutive numbers is 1 and the largest value is $(n2)$.
Therefore $1\leq x \leq n2.$ Combining equation (2) with this
inequality: $$1 \leq {n^2  14n +39\over 6} \leq n2.$$ This
simplifies to $6 \leq n^214n+39 \leq 6n12.$ So $$0\leq n^2 14n +
33 =(n  11)(n  3)$$ and $$n^220n + 51 =(n3)(n17)\leq 0.$$ From
the first inequality $n\leq 3$ or $n\geq 11$ and from the second
inequality $3\leq n \leq 17.$ However, $n=3$ is not a valid
solution because, after removing three consecutive numbers, there
would not be any numbers left to yield an average of 7.5.
Therefore, $11\leq n \leq 17.$
Using equation (2) again, the following table contains the values
of $x$ for all integer values of $n$ between 11 and 17.
n 
11 
12 
13 
14 
15 
16 
17 
x 
1 
2.5 
4.3 
6.5 
9 
11.83 
15 
Since $x$ has to be a whole number, then the only values of $n$
that are valid are $n=11$ (removing 1,2 and 3); $n=15$ (removing 9,
10 and 11) and $n=17$ (removing 15, 16 and 17).
2.
Solution with nonconsecutive numbers
Let the sum of the three numbers removed be $S$, then from equation
(1): $${n(n+1)\over 2}  S = 7.5(n3)$$ which simplifies to $${n^2
 14n + 45 \over 2} = S.\quad (3)$$ The smallest value of $S$ is 6,
when the numbers removed are 1, 2 and 3. The largest value of $S$
is $n + (n1) + (n2) = (3n  3)$. Therefore $6\leq S \ leq 3n3$.
Combining this inequality with equation (3): $$6 \leq {n^2  14n +
45 \over 2} \leq 3n3.$$ This reduces to $0\leq n^2  14n + 33 =
(n11)(n3)$ and $n^220n+51 = (n3)n17)\leq 0.$
So, as in the first part $11\leq n \leq 17.$ The following table
shows the values of S for all integer values of $n$ from 11 to
17.
n 
11 
12 
13 
14 
15 
16 
17 
S 
6 
10.5 
16 
22.5 
30 
38.5 
48 
As $S$ has to be a whole number, then $n$ can only take the values
11, 13, 15 and 17 giving the only possible solutions.
For $n=11$ the only three numbers adding up to 6 are 1, 2 and 3
which are consecutive and this solution came up in the first
part.
For $n=13$ any three numbers adding up to 16 can be removed; for
n=15 any three numbers adding up to 30 can be removed and for n= 17
any three numbers adding up to 48 can be removed.