The large number (123456...424344) has been formed by concatenating the integers 1 to 44. What is the remainder when it is divided by 45?
Note: solve without calculation...
9 is the remainder 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.
My house number is such that, when divided by 2, 3, 4, 5 or 6 it will always leave a remainder of 1.
However, when divided by 11 there is no remainder. What is my house number?