Quantitative Aptitude :: Numbers - Discussion
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2345 23455 234555 234555........... what was last 2 numbers at 200th digit?
A44
B55
C54
D66
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the given series is 2345 23455 234555......
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.
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the given series is 2345 23455 234555......
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.
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Now 17 term contains 20 digits and 2345555......4(17times)... last two digits are 55.
sorry its mistake in solution..:-)
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Solution:
1st group 2345 number of digit 4
2nd group 23455 number of digit 5
3rd group 234555 number of digit 6
so at the end of each group, there will be series of 5s.
So the answer is 55
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