- #1
rajatgl16
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Why Am I Off by One in My Mathematics Olympiad Answer?
- Thread starter rajatgl16
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SUMMARY
The discussion centers on a mathematical problem involving the sequence defined by \( a_n = \left\lfloor \frac{n}{\sqrt{n}} \right\rfloor \) and its behavior around perfect squares. Participants analyze the relationship between \( a_n \) and \( a_{n+1} \) for values of \( n \) ranging from 100 to 121 and up to 2010. The conclusion is that there are 43 instances where \( a_n > a_{n+1} \), specifically at perfect squares, confirming the correctness of the calculations presented.
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Mathematics enthusiasts, competitive programmers, and students preparing for mathematics Olympiads who seek to deepen their understanding of sequences and inequalities.
- #2
tiny-tim
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hi rajatgl16! 
(have a square-root: √
)
hint: try an for n = 100 up to 121
(have a square-root: √
)hint: try an for n = 100 up to 121
- #3
rajatgl16
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I;m not getting what you mean. Please elaborate
- #4
Borek
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Check for which n from 100..121 range an > an+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.
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.
- #5
naveeniitkgp
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#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
#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
- #6
Mandelbroth
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rajatgl16 said:
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 an > an+1.
- #7
SammyS
Staff Emeritus
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Do you realize that you've just responded to a pretty old thread ?Mandelbroth said:
- #8
Mandelbroth
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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.SammyS said:
Is that bad? If so, I apologize...
- #9
SammyS
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No. That's not necessarily bad.Mandelbroth said:
Your answer is correct.
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