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Numerical Ability
Number System
What is the last two digits of the given number having 200 digits?
123412344123444123444412.....
options-a)12
b)23
c)34
d)41
e)44
Read Solution (Total 11)
-
- 1st group 1234 number of digit 4
2nd group 12344 number of digit 5
3rd group 1234444 number of digit 6
so at the end of each group there will be series of 4s.
So the answer is 44 (option e)
- 9 years agoHelpfull: Yes(7) No(5)
- e) 44
iz the ryt answer ! - 9 years agoHelpfull: Yes(2) No(1)
- answer will be e)44
- 9 years agoHelpfull: Yes(1) No(1)
- series is n+3.
first term will have =4
so n+3=200
n=194
so it will end with 44 - 9 years agoHelpfull: Yes(1) No(4)
- 44 is correct ans
- 9 years agoHelpfull: Yes(1) No(0)
- correct answer is 44
- 8 years agoHelpfull: Yes(1) No(0)
- d) 41 is correct
- 8 years agoHelpfull: Yes(1) No(0)
- the given series is 1234 12344 123444.......
So the number of digits in each term are 4, 5, 6, ... or (3 + 1), (3+ 2), (3 +3),......upto nth terms are=3n [n(n 1)/2]
so 3n [n(n 1)/2]>=200
For n = 16, We get 184 in the left hand side. So after 16 terms the number of digits equal to 184. And 16 them contains 16 3 = 19 digits.
Now 17 term contains 20 digits and 123444......4(17times)... last two digits are 55. - 7 years agoHelpfull: Yes(1) No(0)
- 44 is correct answer.
- 9 years agoHelpfull: Yes(0) No(0)
- 44 last one
- 9 years agoHelpfull: Yes(0) No(0)
- every time the value of 4 is incresing by 1 unit then
in last group 183+3=186
186+17 times 44 == so the answer is 44 - 7 years agoHelpfull: Yes(0) No(0)
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