There is nothing to confirm a pattern here.... Are you sure about the questioneaakbari wrote:1122334444...
In above sequence, what is the 100th digit
A 1
B 2
C 3
D 4
E 5
11223334444
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Source: Beat The GMAT — Problem Solving |
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ajith
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eaakbari
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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
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eaakbari
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Yeahneoreaves wrote:is the question in this pattern
1223334444...
There are two 2's ...three 3's ...four 4's .....
122333444455555666666...
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eaakbari
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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
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
Whether you think you can or can't, you're right.
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kstv
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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.
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.
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neoreaves
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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
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
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thephoenix
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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
45th digit is 9(last of 9 series)
since 46th digit is one and here onwards the even digit is one till 19 series
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pradeepkaushal9518
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