11223334444

eaakbari wrote:1122334444...
In above sequence, what is the 100th digit

A 1
B 2
C 3
D 4
E 5

There is nothing to confirm a pattern here.... Are you sure about the question

Its from a GRE Prep Material, but the question holds good. I have the OA and I did solve in some way but want to see better methods

is the question in this pattern

1223334444...


There are two 2's ...three 3's ...four 4's .....

neoreaves wrote:is the question in this pattern

1223334444...


There are two 2's ...three 3's ...four 4's .....

Yeah
122333444455555666666...

If we want to find the 100th digit,
We can observe this forms an A.P 1+2+3+4.....
Which is the number of digits

Hence for Sum = 100

100 = n(n+1)/2

Choose n to be a whole number for easier calculation which is closer to 100. We can take n = 14

So when n = 14, term will be 105. So 104 and 105 will be the last numbers with 14 . That means

1 4 1 4 1 4
100 101 102 103 104 105

Therefore Answer is 1

But to come to this conclusion I took a lot of time. Experts please comment on this
Hence answer = A

OA = A

I do not have OE

The 45th digit is the last single digit.
1+2+3...........9=45

After that all the even no digit till 190th digit are 1
cos after 9 (45th) it is 10
46th to digit 1 and 47th digit is 0
48th digit 1 49th digit 0

100th is an even no digit so it will be 1.

The pattern can be mapped to Arithmetic series sum which relates to the digits position

So,
Using Arithmetic series sum formula

for 1-9 the sum comes out to be 9/2(1+9) = 45

Thus 45th digit is 9 and 46th onwards are two-digit numbers starting from 10

Now we need to find 100th digit ...so 100 - 45 = 55 ...since after 45 we have double digits so we divided 55/2 = 27.5 ...thus we need to find the first digit of the two-digit number ...which in this case is 1


Thus A

IMO A
45th digit is 9(last of 9 series)
since 46th digit is one and here onwards the even digit is one till 19 series

1223334444....97 th will be 3 (13 ) then 141414

98th 1
99th 4
100th 1