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

• 123456789101112....4465
  =123456789101112...4400+65
  sum of digits of 123456789101112...4400=1+2+3+...+44=(44*45)/2=22*45
  clearly 123456789101112...4400 is divisible by 9 & 5, so divisible by 45
  hence when 65 is divided by 45,20 will be remainder
  ANS=20
  
• 11 years agoHelpfull: Yes(23) No(0)
• in the given question first calculate the sum upto 44 terms by the formula n(n+1)/2
  which comes:44(45)/2=(990+65)%9*5=Ans:2
• 11 years agoHelpfull: Yes(8) No(15)
• ans=0
  given num is multiply 0f 45
• 11 years agoHelpfull: Yes(7) No(11)
• First calculate sum upto 465 terms as the last term is 465 using n(n+1)/2.
  465(465+1)/2=108345.
  Divide 108345 by 45 which gives remainder 30.
• 11 years agoHelpfull: Yes(6) No(5)
• ANS WILL BE 20
  
• 11 years agoHelpfull: Yes(5) No(5)
• last term isn't clear
  4465 or 464 465 or 65
• 11 years agoHelpfull: Yes(4) No(0)
• 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 = 44×452+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
• 11 years agoHelpfull: Yes(3) No(1)
• 20 wl b remainder. confrm ans
  
• 11 years agoHelpfull: Yes(2) No(1)
• 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 = 44×452+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
• 11 years agoHelpfull: Yes(2) No(1)
• given nmbr is a multiple of 3...not of 9 @vankayala....
• 11 years agoHelpfull: Yes(1) No(1)
• 465*466/2 =
  remainder 2
• 11 years agoHelpfull: Yes(0) No(1)
• 24C7*2*17!*18
• 11 years agoHelpfull: Yes(0) No(2)
• First calculate the sum of the series using n2[2a+(n-1)d]
  i.e 652[2+(65-1)] = 2145,now you can divide dis by 45 ul get 30 as remainder
• 11 years agoHelpfull: Yes(0) No(2)