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Question 764950: if 1st 44 positive integers form a no. N as N=12345678.........424344 and N is divided by 45, then what is the remainder?
Found 2 solutions by MathLover1, 9876:
Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!
for number to be divisible by 45 it should be divisible by both 5 and 9
as you can see, last digit is  and we know that number should be divisible by both 5 and 9
for number to be divisible by  last digit should be  or ; so, we can say that reminder should be because only if we deduct last digit will be
so, answer is: the remainder is
but to be sure, check the divisibility by 
in order to do that, we need to find the sum of the digits of N=12345678.........424344:
or, faster this way:
so sum of digits is 
 ; so, is divisible by
since and are co-prime numbers, the number given number should also be divisible by
dividing the number by we get  ;so, we got a remainder of
so, your answer is: the remainder is
Answer by 9876(1)  (Show Source):
You can put this solution on YOUR website! The answer is 9 not the 4 as solved by MathLover1(7369).I guanrentee This!
The number on dividing by 5 leaves remainder 4.
Digitsum of Numbers (1234......424344)-->
The number are from 0 to 9 (Sum of digits 45)
then 10 to 19 (Digit sum 45+1x10 = 55)
then 20 to 29 (Digit sum 45+2x10 = 65)
then 30 to 39 (Digit sum 45+3x10 = 75) and
then 40 to 44 (digit sum (10+4x5 = 30)
therefore total digit sum come out to be (45+55+65+75+30 = 270)
This is divisible by 9.
When the given number is divided by 5 it leaves a remainder 4. So the number is of the form 5A + 4.
When divided by 9 it leaves a remainder 0. Hence of the form 9B+0.
Equate both the equations 5A+4=9B.
Put A= 1 and B= 1 we get 9 which is the remainder.
Try Another Example:(Very Simple1):243/45 rem is not 3 isn't it?
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