not as simple as it looks

tarika bata diya

ab ginn lo apne aap[4]

im glad dat atleast someone bothered to end dis one...thanx celestine!!! ;-))

gordo thats the best approach

edit of gordos post

this as we are looking for positive values of p,

for r=K, we have K values of q so that r-q is positive,
ie 1 +2+......899 = 899*900/2

now when r=q ,p=0 means x=r is still valid

so in that case x can be 1 to 899

so final ans is 899*451 which is one less

all the others abv have considered x=o also as a solution thats the mistake

post is ended now !!!!

had hai
maine to aakhri digit tak bata di
please
now this is closed[4]

the actual answer is 1 less than that,why????? any explanation?

Proof:

double a,c;int b;
int ctr=0;

for(b=1;b<=(900*901);b++)
{
a=(int) (b/900);
c=(int) (b/901);
if(a==c){
ctr++;
}
}
System.out.println("Ctr:"+ctr);

Output:
init:
deps-jar:
compile-single:
run-single:
Ctr:405449
BUILD SUCCESSFUL (total time: 0 seconds) --> lol

Did this to confirm as i had modified a past olympiad question

1+2+...+899=450*899

r u sure ur question is complete??
please mention the interval

yep, now do the counting

[339]

GOOD OBSERVATION ITH POWER
AMAZING
THEN

[1,899]
[902,1799]
[1803,2699]
[2704,3599]
[3605,4499].
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[45051,45989]
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[54062,54899]

[72081,72899]

[90101,90899]

[450501,450899]

[809999,809999]

last number to be multiplied by 900 is 899
in lhs we add the number with which we r multiplying
and in the rhs we a r multiplying with n-1 and subtractng 1

i hope this is the required solution[1]

obviously the list will not count to infinity.
as you see the solutions you listed earlier, the range reduces by 1 each time.

positive integral solutions of x