Question
Question: Let us suppose that: \(n = 1234567891011......787980\). The integer \(n\) is formed by writing the p...
Let us suppose that: n=1234567891011......787980. The integer n is formed by writing the positive integers in a row, starting with 1 and ending with 80, as shown above. Counting from the left, what is the 90th digit of n?
A. 1
B. 2
C. 3
D. 4
E. 5
Solution
Hint: Here, in order to find out the 90th digit of n we will observe the pattern which is followed and through that pattern the 90th digit of n will be predicted easily. For analysing such patterns some of the initial digits of n are examined.
Complete step-by-step answer:
Given, n=1234567891011......787980
Here, the first nine digits will be 1 to 9 respectively. The 10th digit of n will be 1, 11th digit of n will be 0, 12th digit of n will be 1 and so on. By this pattern we can say that 28th digit of n will be 1 and 29th digit of n will be 9. Now after this, 30th digit of n will be 2 and 31st digit of n will be 0. Similarly, 32nd digit of n will be 2 and 33rd digit of n will be 1.
According to this pattern, 48th digit of n will be 2 and 49th digit of n will be 9. Now after this, 50th digit of n will be 3 and 51st digit of n will be 0. Similarly, 52nd digit of n will be 3 and 53rd digit of n will be 1. According to this pattern, 68th digit of n will be 3 and 69th digit of n will be 9.
Now after this, 70th digit of n will be 4 and 71st digit of n will be 0. Similarly, 72nd digit of n will be 4 and 73rd digit of n will be 1. According to this pattern, 88th digit of n will be 4 and 89th digit of n will be 9. Now after this, 90th digit of n will be 5.
So, the 90th digit of n will be 5 (90th digit and 91stdigit together makes the positive integer 50).
Therefore, option E is correct.
Note: In these types of problems, there always exists a pattern or sequence and we just have to analyse that pattern to reach the answer. In this case, patterns exist for 1 to 9 positive integers, 10 to 19, 20 to 29, 30 to 39 and so on.