There was a star mark question. 8,9,7,6.... Series is such that if you add first 2 numbers of the series the third one will be the unit digit of the added sum. We have to fine the minimum number of terms n such that S(n)>2000. Only this much i remember. I calculated it and the answer was 520.

• the series is 897639213471... after 12 digits the series is repeating. sum of the 12 digits is 60. now 60*33=1980. so 20 is needed. so adding first 3 digits (8+9+7)=24. (1980+24)>2000.
  now in 33 cycles no of digits is 12*33=396. we have to add 3 with this. so total 396+3=399
• 11 years agoHelpfull: Yes(61) No(0)
• series is:8,9,7,6,3,9,2,1,3,4,7,1,8,9,7,6,3...
  we see after every 12 terms the digits are repeated.
  sum of the 1st cycle of digits=60
  =>sum of n cycle of digits=60n
  a/q, 60n>2000
  => n> 100/3=33.33333
  now in 1 cycle,no.of terms=12
  in 33.33333 cycle,no of terms=12*33.33333= approx 399.(ans)
• 11 years agoHelpfull: Yes(10) No(7)
• since,here we can see that 8,9,7,6,3,9,2,1,3,4,7,1,8,9,7,6,3,9,2,1,3,4,7,1...cycles repeat after every 12 terms..so every cycle sum is 60 therefore if therefore if there is 34 cycles then sum will be 60*34=2040>2000...so it is clear that cycle repeats 34 times..and in 1 cycle no of terms n=12 so total no 0f terms=12*34=408...but for no of terms to be minimum if we subtract sum of last 8 digits of last cycle(ie 2,1,3,4,7,1,8,9..sum=35)then also we can get sum =2040-35=2005>2000...therefore minimu no of terms=408-8=400...so n=400
• 11 years agoHelpfull: Yes(9) No(14)
• THE SERIES IS 8,9,7,6,3,9,2,1,3,4,7,1.8.7,.... THE SERIES REPEATS AFTER 12 TERMS. SUM OF THE 1ST 12 TERMS S 60 SO THE SUM OF THE SERIES WILL BE 60+60+...N TERMS.
  SO THE TOTAL SUM SHOULD BE >2000. SO A MULTIPLE OF 60 >2000 IS 2040(60*34). NO OF TERMS=34*12=408. N SHOULD BE MINIMUM TO GET SUM>2000. SO REMOVE THE LAST 9 TERMS OF THE LAST CYCLE. SUM=2004 AND N=399
• 11 years agoHelpfull: Yes(6) No(3)