Mathematics olympiad question

[rajatgl16]

in the ques that i have attached as image with this thread..
I did it as a_n can be solved to $$\sqrt{}n$$ by rationalising it.

So as squre root of every natural no. 'n' is smaller than squre root of 'n+1' then in this ques. possible values comes out to be zero. Am i Right?

 

[tiny-tim]

hi rajatgl16!

(have a square-root: √ )

hint: try a_n for n = 100 up to 121

 

[rajatgl16]

I;m not getting what you mean. Please elaborate

 

[Borek]

Check for which n from 100..121 range a_n > a_n+1.

Do the same for any other range bounded by k² and (k+1)².

Look if there is some pattern.

If there is no pattern, I have no idea what tiny-tim aims at.

 

[naveeniitkgp]

#include <stdio.h>
#include <math.h>
#define MAX 2010

int gint(float x)
{
int n;
n=x;
return n;
}

int main()
{
int a[MAX], j,k, i, count1=0;
for(i=1; i<=MAX; i++){
k=gint(sqrt(i));
a=gint(i/k);
if(a[i-1]>a){
count1++;
printf("a(%d)=%d > a(%d)=%d\n", i-1,a[i-1], i, a);
}
else continue;
}
printf("%d", count1-1);}answer comes out to be 42

 

[Mandelbroth]

in the ques that i have attached as image with this thread..
I did it as a_n can be solved to $$\sqrt{n}$$ by rationalising it.

So as squre root of every natural no. 'n' is smaller than squre root of 'n+1' then in this ques. possible values comes out to be zero. Am i Right?

I'm thinking that $a_n > a_{n+1} \forall (n+1)^2 \in \mathbb{Z}$.

There are $\left\lfloor\sqrt{2010}\right\rfloor = 44$ perfect squares less than 2010, so I get 43 different values for n such that a_n > a_n+1.

 

[SammyS]

I'm thinking that $a_n > a_{n+1} \forall (n+1)^2 \in \mathbb{Z}$.

There are $\left\lfloor\sqrt{2010}\right\rfloor = 44$ perfect squares less than 2010, so I get 43 different values for n such that a_n > a_n+1.

Do you realize that you've just responded to a pretty old thread ?

 

[Mandelbroth]

Do you realize that you've just responded to a pretty old thread ?

Yes, I noticed. I'd like to know why I'm off by one from naveeniitkgp's answer, so I decided to respond here rather than make a new thread linking back to this one.

Is that bad? If so, I apologize...

 

[SammyS]

Yes, I noticed. I'd like to know why I'm off by one from naveeniitkgp's answer, so I decided to respond here rather than make a new thread linking back to this one.

Is that bad? If so, I apologize...

No. That's not necessarily bad.

Your answer is correct.