From floor to ceiling

From the properties of floor and ceiling, $x-1 \leq \lfloor x \rfloor \leq x, y \leq \lceil y \rceil \leq y+1$ . So, we see that $x+y - 1 \leq \lfloor x \rfloor \leq + \lceil y \rceil \leq x + y + 1$ .

Thus $\frac{20n}{13} +\frac{13n}{20} -1 \leq \lfloor 20n/13 \rfloor + \lceil 13n/20 \rceil < \frac{20n}{13} +\frac{13n}{20} -1$

i.e. $\frac{569n}{260} -1 < 2013 \leq \leq {569n}{260} + 1$ . Simplifying gives $919 .3 < n < 920. 28$ . We then check that $n=920$ works.

[Latex edits]

We find n for the equation 20n/13 + 13n/20 = 2013. In this equation, n is 919.8242..., so we test n = 920, 919 for the problem above and find that n = 920 is correct.

We can approximate a value for the expression, and then use trial and error to find our answer. $\lfloor \frac{20n}{13} \rfloor + \lceil \frac{13n}{20} \rceil \approx \frac{20n}{13}+\frac{13n}{20}=\frac{569n}{260}$ Since we want the expression approximately 2013, $\frac{569n}{260}=2013 \Rightarrow n \approx 919.8$ . A quick check shows that the answer is $\boxed{920}$

We have $\left \lceil x \right \rceil = x$ or $\left \lceil x \right \rceil = \left \lfloor x \right \rfloor + 1$ ( $x$ is a real number).

Beside we have $\left \lfloor a \right \rfloor \leq a \leq \left \lfloor a \right \rfloor +1$ ( $a$ is a real number).

So, we have $2013 = \left \lfloor \frac{20n}{13} \right \rfloor + \left \lceil \frac{13n}{20} \right \rceil \leq \left \lfloor \frac{20n}{13} \right \rfloor + \left \lfloor \frac{13n}{20} \right \rfloor +1 \leq \frac{20n}{13} + (\frac{13n}{20} +1) \leq 2013 + 1 =2014$

So $\frac{2012.260}{569} \leq n \leq \frac{2013.260}{569}$ and n is an integer number, so $n=920$ .

From $\left\lfloor \frac { 20n }{ 13 } \right\rfloor +\left\lceil \frac { 13n }{ 20 } \right\rceil =2013$ , we can say

$\frac { 20n }{ 13 } +s+\frac { 13n }{ 20 } -t=2013$ , where

$0\le s< 1$ ,

$0\le t< 1$ ,

and therefore $-1< (s-t)< 1$ (we will remember this for later)

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$\large \frac { 20n }{ 13 } +\frac { 13n }{ 20 } =2013+t-s\\ \large \frac { 569 }{ 260 } n=2013-\left( s-t \right)$

$\large n=\frac { 260\times 2013 }{ 569 } -\frac { 260 }{ 569 } \left( s-t \right) \\ \large n=\frac { 523380 }{ 569 } -\frac { 260 }{ 569 } \left( s-t \right) \\ \large n=919+\frac { 469 }{ 569 } -\frac { 260 }{ 569 } \left( s-t \right)$

After calculating the bounds of n for when $(s-t)=1$ and $-1$ , we find that $\large 919.37< n<920.28$ (to 2 decimal places).

$\large \therefore n=\color{#20A900}{\boxed { 920 }}$

Let us consider 4 nos. x,y,r,R.

By Euclid's division lemma,

20n = 13x + r , r<13

13n = 20y - R , R<20

Examining carefully we see that x+y=2013

Now , $\frac{20n-r}{13}$ = x

$\frac{13n-R}{20}$ =y

Solving, we get,

569n-20r+13R=(2013)(260)

now for max value of n, r should be max i.e. 12 and R should be minimum i.e 1

max value of n comes out to be approx 920.

for min value of n, r should be min i.e. 1 and R should be max i.e 19

min value of n comes out to be approx 919

checking both 919 and 920 as n we get 920 as solution.

[a] + [b]=[a+b]. This statement is true in most cases. [20n/13 + 13n/20]=2013 [569n/260]=2013 now , this implies that L.H.S is always greater than or equal to R.H.S. n>=2013*260/569 n>=919.8 a positive integer greater than 919.8 is 920.

20n/13+13n/20=2013

400n+169n=2013 13 20

569n=523380

n=919.824

rounding off n we get n=920

First factorize the given equation :

\frac {20n}{13} + \frac {13n}{20} = 2013 n( \frac {20}{13}+\frac {13}{20}) = 2013 n(\frac {400+169}{260}) =2013 n(\frac {569}{260} = 2013 569n = 2013 x 260 n = 523380/569 n=919.82

Now we ceil n Therefore n = 920