Question 764950


for number to be divisible by 45 it should be divisible by both 5 and 9 

as you can see, last digit is {{{4}}} and we know that number should be divisible by both 5 and 9
for number to be divisible by {{{5}}} last digit should be {{{0}}} or {{{5}}}; so, we can say that reminder should be {{{4}}} because only if we deduct {{{4}}} last digit will be {{{0}}}

so, answer is: the remainder is {{{4}}}
  
but to be sure, check the divisibility by {{{9}}} 

in order to do that, we need to find the sum of the digits of N=12345678.........424344:

{{{1+2+3+4+5+6+7+8+9+1+0+1+1+1+2+1+3+1+4+15+1+6+1+7+1+8+1+9+2+0+2+1+2+2+2+3+2+4+2+5+2+6+2+7+2+8+2+9+3+0+3+1+3+2+3+3+3+4+3+5+3+6+3+7+3+8+3+9+4+0+4+1+4+2+4+3+4+4=279}}}

or, faster this way:

{{{1-9=45+1(of 10)=46}}}
{{{11-20=9+45+2=56}}}
{{{21-30=18+45+3=66}}}
{{{31-40=27+45+4=76}}}
{{{41-44=16+10=26}}}
so sum of digits is {{{279}}}

{{{279/9=31}}}; so, {{{279}}} is divisible by {{{9}}}
 
since {{{5}}} and {{{9}}} are co-prime numbers, the number given number should also be divisible by {{{5}}}

 dividing the number by {{{5}}} we get  {{{279/5 =55+4/5}}};so, we got a remainder of {{{4}}}

so, your answer is: the remainder is {{{4}}}