7+77+777+............+7777777 23 times then what is the sum?

• rewrite above series as 7(1+11+111+1111......1111111)...observe that (1+11=12,1+11+111=123,1+11+111+1111=1234...this way 1+11+111+...1111111 leads to 1234567)...value of series=7(1234567)=8641969
• 10 years agoHelpfull: Yes(79) No(27)
• s=7+77+777+............+7777777 23 times
  s=7(1+11+111+.............+11111 23 times
  s=7/9(9+99+999+................)
  s=7/9(10+100+1000.......-23(1 fromk each term))
  s=7/9(10(10^23-1)/(10-1)-23)
  s=70/9((10^23-1)/9-23)
• 10 years agoHelpfull: Yes(37) No(14)
• Why this question is in blood relations category?
• 10 years agoHelpfull: Yes(18) No(0)
• i am sure this will be the right ans
  sn= 7/9[9+99+999+..........23 terms]
  =7/9[(10-1) (10^2-1) (10^3-1)...............]
  =7/9[(10+10^2+10^3+..)(1+1++.......23 terms)
  using gp series
  = 7/9[(10(10^23-1))/(10-9))-23]
  = 70/9[(10^23-1)-161/9]
• 10 years agoHelpfull: Yes(17) No(1)
• taking out 7 common we have
  7(1+11+111+...............+1111111 upto 23 times)
  now dividing it by 9 we have:-
  7/9(9+99+999+......+99999 23 times)
  which can be written as
  7/9((10-1)+(100-1)+(1000-1)...............(10^23-1))
  7/9[(10+100+1000+.............10^23)-23]
  7/9[(10(10^23-1)/9)-23]
• 10 years agoHelpfull: Yes(14) No(0)
• can any1 give crct ans plz
• 10 years agoHelpfull: Yes(5) No(1)
• 7+77+777+....+7777777 n times (n=23)
  =7(1+11+.......)
  =7/9(9+99+999+......)
  =7/9[ (10-1) + (100-1) + (1000-1) +.......]
  =7/9[ (10+100+1000+......) - (1+1+1+....)]
  =7/9[ (10*(10^n - 1/10-1 ) - n ]
  replace n by 23 as given in question
  =7/81 [ 10^24 - 10 -207 ]= 8.641975309 * 10^22
• 10 years agoHelpfull: Yes(5) No(0)
• this question took me 20 long minutes. i wasn't able to make out a minor error , due to which i kept on trying madly...i succeeded.
  a simpler solution is : 7(1+11+111+1111....+111...23 times)
  add the bracket .i....e 12345679012345679012343. multiply by 7 to get 86419753086419756086401.
  addition of bracket is a v.simple task:-
  last column is 23 in sum..secnd last is 22 in sum..third last is 21...fourth last is 20 and so on .
  thus satrting frm the extreme right. we get 23. so 2 carry forward and so on.
  this is an easier method than working it out with gp .
  
• 10 years agoHelpfull: Yes(5) No(1)
• take 7 common
  7(1+11+111+.......+1111..23times) {add the brackets} [1+11+111+1111=1234]
  = 7(12345678901234567890123) {multiply with 7}
  =86419752308641975230861
• 10 years agoHelpfull: Yes(4) No(0)
• 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7
  * 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
  = 86419753086419783086401 is the ans. time require 20 sec to solve
• 10 years agoHelpfull: Yes(3) No(3)
• 7(1+11+111+.....)
  a=1, r=11 by gp a((r^n)-1)/(r-1)
  7(1*(11^23)-1)/(11-1)
  7((11^23)-1)/10
• 10 years agoHelpfull: Yes(2) No(3)
• (7+77+777+...+7..)23 times
  7(1+11+111...+1..)23 times= 7(12345678909876543210321)= 86419752369135802472247
• 10 years agoHelpfull: Yes(1) No(0)
• 7(1+11+111+.......+11111 23 times)
  write like this
  1
  1 1
  1 1 1 like this 23 times
  sum last one's which gives 23 and keep 3 there and 2 is added to next position
  like wise do summation for 3 digits and multiply that number with 7 u will get 343*7=2401. so ans is 401.
  
  
• 10 years agoHelpfull: Yes(0) No(11)
• s=7+77+777+............+7777777 23 times
  s=7(1+11+111+.............+11111 23 times
  s=7/9(9+99+999+................)
  s=7/9(10+100+1000.......-23(1 fromk each term))
  s=7/9(1111....(23 times)0 -23) //after writting 1 (23 times) put 0 at the last/unit place
  s=7/9(1111....(21 times)087) //after subtracting 23 the last 3 digits of the series will always be 087, whether 1 is written 23 times/less/more, except 2 times
  s=(7777.....(21 times)609)/9 // on multiplying 7 inside, the last 3 digits will always be 609
• 10 years agoHelpfull: Yes(0) No(1)
• 7*23=161,so 16 is the carry and is the last term.
  7*22=154,154+16=170,so 17 carry and 0 is the last term and so on
• 10 years agoHelpfull: Yes(0) No(1)
• 7(1+11+11+.....+1111111)
  Multiply and divide by 9
  7/9(9+99+99+......+9999999)
  7/9((10-1)+(100-1)+..........+(10000000-1)) which can be written as
  7/9((10+100+.....+1000000)-n)
  now the part((10+100+.....+1000000)-n)comes under geometric series,therefore applying sum of series formula of GP ie a(r^n-1)/r-1.(a->first term,r->common ratio)
  (7/9(10*(10^23-1))/1(10-1)-23)(ie n->23 since the problem is 23 times)
  and finally the above equation is simplified to get the final answer.
  
• 10 years agoHelpfull: Yes(0) No(1)