not as simple as it looks

great work, both of you

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 !!!!

say x=900p + r
and also x=901p + q
where 0<=r<900
and 0<=q<901

we have 900p +r= 901p +q
or p=(r-q)
now r belongs to [0,899]
and q belongs to [0,900]
as we are looking for positive values of p,
for r=K, we have K+1 values of q so that r-q is positive,
so the number of solutions can go up to
900+899+898+....+1=(900*901/2)
mite be completely wrong...

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

not infinity,but quite a large number( btwn 100000 and 1 million -)

CAN WE COUNT THEM ?????????

THE LIST EXTENDS UPTO INFINITY

SORRY IF AM ASKING A DUMB
BUT
THIS IS AN OBVIOUS QUESTION

positive integral solutions of x