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Logical Reasoning
Logical Sequences
9+99+999+9999+............+999999.......23times. Find the last three digits
Read Solution (Total 9)
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- last three digit will be 087 ...
1. for unit place-> 9+9+..23 times = 23*9=207 (unit place is 7) (20 will be added to tenth place)
2. for tenth place-> 9+9+9..22 times =22*9=198(+20)=>218(tenth place is 8)(21 will be added to hundred place)
3. for hundred place-> 9+9+9..21 times=21*9=189(+21)=>210(hundred place is 0) - 11 years agoHelpfull: Yes(93) No(2)
- 9(1+11+111+.......+111...23times)
23times.....1111
+22times.....111
+21times......11
+ .
.
1
--------------------
23
+ 22
+ 21
-----------------
343
343*9=087
ans:=087 - 11 years agoHelpfull: Yes(12) No(1)
- 087
9(1 +11+111+................23 times)
9(last digit=3
2nd last digit=last digit of(22+2)=4
3rd last digit=21+2=3;)
9(343)=.............087
- 11 years agoHelpfull: Yes(6) No(0)
- it can be rewrite as (10-1)+(100-1)+......23 times
-->(10+100+1000....23terms)-23
sum of the GP above =1111-23 = 1088 so the last three digits are "088" - 11 years agoHelpfull: Yes(2) No(8)
- at the units place 9 appears 23 times so 9*23=227,now after addition 9 comes at the units place and 22 as carry,now at tens place 9 appears for 22 times so 9*22=198,now 198+22(previos carry)=220 so 0 comes at tens place nd 22 as carry,now 9 appears for 21 times s0 9*21=189+(22)=211 so finally last three digits are 109.....
- 11 years agoHelpfull: Yes(0) No(4)
- ans is 987
- 11 years agoHelpfull: Yes(0) No(7)
- there is 23 times then write 23 times one and then a zero.so,
111111111111111111111110-24=11111111111111111111186 - 11 years agoHelpfull: Yes(0) No(4)
- 087 will be the last three digits.
- 11 years agoHelpfull: Yes(0) No(0)
- 23*9=207=>unit place is 7 and 20 will be added to tenth place
22*9=198+20=218=>unit place is 8 and 21 will be added to hundredth place
21*9=189+21=210=>unit place is 0
so,answer is 087 - 8 years agoHelpfull: Yes(0) No(0)
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