AoPS Academy
![\[\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 4(xy+yz+zx),\]](http://latex.artofproblemsolving.com/0/4/7/047adb616a6be9f9ed52f2385fa592fb5464e420.png)
and 
.
and 
be elements of the set 
such that 
, 
is a divisor of 
, and 
is a divisor of 
. Determine the largest value that 
can take.
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
such that![\[\lim_{n \to \infty} \left( \frac{1 + 2^{1/3} + 3^{1/3} + \dots + n^{1/3}}{n^t} \right ) = c\]](http://latex.artofproblemsolving.com/e/1/a/e1a87bd7bee41a7d656e2ff067a87dff0c9929ec.png)
for some real 
. Also find the corresponding values of 
.
![$f\in C^1[a,b]$](http://latex.artofproblemsolving.com/8/4/8/848da12534d564560fe1cb016142159234be98b4.png)
, 
and suppose that 
, 
, is such that
for all ![$x\in [a,b]$](http://latex.artofproblemsolving.com/7/8/d/78d9dcd969a2c45f7cc9234aa1377b6ec1c2d5d3.png)
. Is it true that 
for all ?
is compact?
(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
(where are the digits and ) we define the \textit{permutation sum} as the value For example the permutation sum of 20253 is Let and be two fivedigit integers with the same permutation sum..
![\[
S(n) = 1111a_{n} + 10111b_{n} + 11011c_{n} + 11101d_{n} + 11110e_{n}
\]](http://latex.artofproblemsolving.com/d/c/6/dc68086c322208f7443e2192f623d3e1e06efe09.png)
![\[
S(m) = 1111a_{m} + 10111b_{m} + 11011c_{m} + 11101d_{m} + 11110e_{m}
\]](http://latex.artofproblemsolving.com/0/a/f/0af218635db948bbca114f174d600bb3945d9c76.png)
Also,![\[
S(n) + n = 11111(a_{n} + b_{n} + c_{n} + d_{n} + e_{n})
\]](http://latex.artofproblemsolving.com/1/0/3/10363f5e9a09ae68291dc24dee8347d9bae6685e.png)
![\[
S(m) + n = 11111(a_{m} + b_{m} + c_{m} + d_{m} + e_{m})
\]](http://latex.artofproblemsolving.com/8/8/c/88ca334b50162649bad05531dd2f89588f8b713f.png)
Since,![\[
S(n) = S(m)
\]](http://latex.artofproblemsolving.com/b/9/4/b94b10e0be06bdf17e09a93a9ae40e51ad0b459f.png)
![\[
\implies 11111(a_{n} + b_{n} + c_{n} + d_{n} + e_{n}) - n = 11111(a_{m} + b_{m} + c_{m} + d_{m} + e_{m}) - m
\]](http://latex.artofproblemsolving.com/f/e/d/fed32efe41634ce352403707b77dea0a3aa119e8.png)
![\[
m-n = 11111((a_{m} + b_{m} + c_{m} + d_{m} + e_{m})-(a_{n} + b_{n} + c_{n} + d_{n} + e_{n}))
\]](http://latex.artofproblemsolving.com/a/a/2/aa2f615fc83b016c8ee6dac955e0005dd048fcc6.png)
![\[
\implies m \equiv n \pmod{11111}
\]](http://latex.artofproblemsolving.com/7/f/9/7f95c981e2e2d4e4191b8c7ae8ea9f0711da5b06.png)
And,![\[
S(n) \equiv 4n \pmod{9}
\]](http://latex.artofproblemsolving.com/0/7/5/075a303409c5dbfaf66d9ac3afee9c632b43dc8a.png)
![\[
S(m) \equiv 4m \pmod{9}
\]](http://latex.artofproblemsolving.com/9/0/1/90109fa1fc52df75576af6150ad79937a27b01a7.png)
So,![\[
4m \equiv 4n \pmod{9}
\]](http://latex.artofproblemsolving.com/c/5/8/c5803068a7342d178221c049fde5f2a05550e7f1.png)
![\[
m \equiv n \pmod{9}
\]](http://latex.artofproblemsolving.com/4/b/4/4b4e83800ad64d3169f5a412d80a139ccf56d0d9.png)
Now, Using Chinese Remainder Theorem,![\[
m \equiv n \pmod{99999}
\]](http://latex.artofproblemsolving.com/d/c/b/dcb4eb91378d92060f4f3498194d27b659397ba1.png)
![\[
\boxed{m=n}
\]](http://latex.artofproblemsolving.com/5/8/4/5841b978a06e6e866677dc956130dcafcd7108d3.png)
:D :play_ball:
denote the digital sum. Then 
for two different 
and 
if they have the same permutation sum. But observe that 
must be 
.
,
,
.
is considered unique. Thus, only one value can be given for