AoPS Academy
written on the first page and 
empty pages. Suppose in each of the next seconds, Alice can flip to the next page, and write down the sum or product of two numbers (possibly the same) which are already written in her grimoire.
be the largest possible number such that for any 
, Alice can write down the number 
on the last page of her grimoire. Prove that there exists a positive integer 
such that for all 
, we have that ![\[n^{0.99n}\leqslant F(n)\leqslant n^{1.01n}.\]](http://latex.artofproblemsolving.com/5/7/e/57e2b7e3d97564665d8de28a3afd74bdcf74ba5b.png)
+2 w
of positive integers for which there exist positive integers 
and such that
holds for all integers 
(Note that 
denotes the greatest common divisor of integers 
and 
)
such that every collection of consecutive positive integers contains an integer whose sum of digits is divisible by 
. for which he can do this, or prove that none exist.
on one side and a 
on the other. Harry has of these coins arranged in a line from left to right. He repeatedly performs the following operation: if there are exactly 
coins showing , then he turns over the th coin from the left; otherwise, all coins show and he stops. For example, if 
the process starting with the configuration 
would be 
, which stops after three operations.
, let 
be the number of operations before Harry stops. For example, 
and 
. Determine the average value of over all 
possible initial configurations .
.
.
inequality holds.
= 
$(\frac{\frac{a_1+a_n}{2}+a_2+a_3+....+a_{n-1}+\frac{a_1+a_n}}{2})^2 \geq 0 $ by Cauchy for numbers 
, determine the largest real constant such that for all positive real numbers 
we have![\[\frac{a_1^2+\ldots+a_n^2}{n}\geqslant\left(\frac{a_1+\ldots+a_n}{n}\right)^2+C_n\cdot(a_1-a_n)^2\mbox{.}\]](http://latex.artofproblemsolving.com/7/c/0/7c0664071c568371c5746c2a560a55aeb6945168.png)
then can be any reals. Let us assume 
, then we just need to find the min of
, then 
and by Cauchy Inequality, we have
equality holds, so 
.
, determine the largest real constant such that for all positive real numbers we have then can be any reals. Let us assume , then we just need to find the min of, then and by Cauchy Inequality, we have equality holds, so .
is maximal constant (or at least I can't see how), and kuing probably guessed and then proved it. I have systematical way to prove that constant.![$a_2,...,a_{n-1} \in [a_1,a_n]$](http://latex.artofproblemsolving.com/b/d/4/bd4bfc5549eb73b16a1a5f1ab1c3d8e4b4b16aaa.png)
and since inequality is symmetric in variables 
, we can also suppose 
.
. With this change, RHS stays untouched, so we need to prove:
(because of homogenity), and 
. Using mixing variable method, we just need to prove inequality for case 
, ![$t\in [1,a]$](http://latex.artofproblemsolving.com/5/9/0/590e49143a73129afacec660322d044b248915ed.png)
.
.
must be greater than zero
.
what Ovchinnikov Denis has done very well. 
. Then inequality becomes 
, but 
in this example, so it is not true 
)
.The left hand side can be written as
where 
is the arithmetic mean of 
.We substitute 
.Then the inequality is equivalent to 
,and we have 
.By setting 
we find that 
.Thus we have to prove that 
.It is clear that 
so it remains to show that 
which is clearly equivalent to 
.QED
to get that .![\[\frac{a_1^2+\ldots+a_n^2}{n}\geqslant\left(\frac{a_1+\ldots+a_n}{n}\right)^2+\frac{1}{2n}\cdot(a_1-a_n)^2\mbox{.}\]](http://latex.artofproblemsolving.com/c/4/1/c414d1262218dcde3b275bec4ed85f73be50b479.png)
Since inequality is homogenous we can assume that 
. We can rewrite inequality as follows:
From Cauchy inequality we have:
Therefore:
It remains to show that:
where 
. Hence the conclusion.
for which 
and so 
for 
.
![\[((n-2)(a_1+a_n)-2(a_2+\ldots+a_{n-1}))^2+2n((n-2)(a_2^2+\ldots+a_{n-1}^2)-(a_2+\ldots+a_{n-1})^2) \ge n(n-2)(2nC_n-1)(a_1-a_n)^2\]](http://latex.artofproblemsolving.com/b/7/9/b79d10053dd5e899baa1382e1848236f0b5e65af.png)
is always nonnegative. Setting 
zero, thus satisfying the inequality.
,