# 123456789101112.... . 4465 . find the remainder when divided by 45?

asked Nov 8, 2013 in TCS
retagged Nov 8, 2013

Divisibility for 45 is, the given number should be divisible by 9 as well as with 5.
Let the given number is N.
N has unit digit 5 so it is divisible by 5. Now the divisibility for 9 is digit sum of N  should be divisible by 9.
Digit sum of N = (1+2+3+...9) + 1 x 10 + (1+2+3+.....9) + 2 x 10 + (1+2+3....9) + 3 x 10 + (1+2+3.....9) + 4 x 5 + (1 + 2 + 3 + 4) + 6 + 5
= 45 + 10 + 45 + 20 + 45 + 30 + 45 + 20 + 10 + 11 = 2 (or 1 + 2 + 3+....44 + 65 = $\displaystyle\frac{44\times 45}{2}+ 65 = 990 + 65 = 1055 = 2$
So N = 5K = 9L + 2
For L = 2 the given number gets satisfied. So the least number satisfies the condition = 9(2) + 2 = 20

Remainder = 20
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;