N=123456789101112.....4465;

45=5*9;

unit digit of given no. is 5,so it is divisible by 5;

N=5K;

it will divisible by 9 if sum of the dgits is digits is divisible by 9;

1-9:sum=45;

100-199:sum=100+450+450;

200-299:sum=200+450+450;

.

.

900-999:sum=900+450+450;

so,100-999:sum=4500+4500+4500;

1000-1999:sum=1000+4500+4500+4500;

2000-1999:sum=2000+4500+4500+4500;

so,1000-3999:sum=1000*(1+2+3)+3*4500+3*4500;

4000-4099:sum=400+450+450;

4100-4199:sum=400+100+450+450;

.

.

4500-4599:sum=400+500+450+450;

so,4000-4599:sum=1600+600+8*450;

similarly,4400-4459:sum=240+240+10(1+2+3+4+5)+6.45;

4460-4465:sum:48+66+15;

finally,sum=900+12*4500+8*450+6*45+6000+1600+600+480+150+48+36+15;

sum=900+12*4500+8*450+6*45+8919;

this sum is divisible by 9;

so,N=9*L;

so,5*k=9*L;

K=(9*L)/5,so L=5;

least no. satisfies=45;

so,remainder=0;