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24 Jan 2022, 07:15
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D
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58% (02:22) correct
42%
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The integer n is formed by writing the integers 1 through 333 in order as shown below:
\(n = 123456789101112...331332333\).
Counting from the left, what is the 333rd digit of n?
A. 0
B. 1
C. 3
D. 7
E. 8
Are You Up For the Challenge: 700 Level Questions
\(n = 123456789101112...331332333\).
Counting from the left, what is the 333rd digit of n?
A. 0
B. 1
C. 3
D. 7
E. 8
Are You Up For the Challenge: 700 Level Questions
24 Jan 2022, 09:12
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Bunuel
First we have single digit 1 to 9, so total 9
Next we have 2-digit numbers. Total digits used 2*90 as there are 90 2-digit numbers.
So we are looking for 333-9-90 or 144th digit in 3-digit numbers. => 144 = 3*48, that is units digit of 48th number
48th 3-digit number is 147, so 333 digit is 7.
D
General Discussion
10 May 2024, 19:02
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1 to 9 ==> contributes 9 digits
10 to 99 ==> 90 numbers ==> contributes 90*2 = 180 digits
100 to 149 ==> 50 numbers => 150 digits
total 150+ 189 = 339 digits
149 => 9 here is the 339 th digit
count backwards..
In 147 , 7 is the 333rd digit.
D is the answer.
10 to 99 ==> 90 numbers ==> contributes 90*2 = 180 digits
100 to 149 ==> 50 numbers => 150 digits
total 150+ 189 = 339 digits
149 => 9 here is the 339 th digit
count backwards..
In 147 , 7 is the 333rd digit.
D is the answer.
