fozzzy wrote:
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
1-9(9 digits)
now 10-19,20-29,.....,90-99....each set have 10 numbers with 2 digit....and there are 9 such sets=>therefore total digit in each set will be 10*2=20
and total digit in all 9 sets will be=20*9=180
uptil now we reached up to 180+9=189th digit.
we have been asked 333rd digit therefore remaining digits=333-189=144
now 100-109,110-119,120-129,130-139,140-149==>we have five sets having 10 numbers and each number has 3 digits==>therefore total no. of digit from 100-149 will be equal to 5*3*10=150
so 123456..........99100101....149==>1to 99 we came to 189 th digit therefore 149 will reach up to 339th digit
123456......100101....147148149=>9 is the 339th digit so by counting back 7 will be the 333rd digit.
i know thw explanation seems to be but when you will think its not so big.
press kudos if it helped.
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