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If k is the number formed by writing the integers from 1 to 999 sequentially (123456789101112...998999), what is the 341st digit in k?
A. 0
B. 1
C. 4
D. 5
E. 9
_________________
If k is the number formed by writing the integers from 1 to 999 sequentially (123456789101112...998999), what is the 341st digit in k?
A. 0
B. 1
C. 4
D. 5
E. 9
|
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| Range | Number of digits | Total |
| 1 - 9 | 1 | 9 |
| 10 - 99 | 2 | 90*2=180 |
| 100 - 149 | 3 | 50*2=150 |
Total so far = 9 + 180 + 150 =339
We are close to 341 digit which will likely be a digit in our next number - 150
340th digit - 1
341th digit - 5
Ans : D
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Bunuel
GMAT Club's Official Explanation:
We begin by determining the number of digits contributed by each group of numbers:
- 1-digit numbers (1 through 9) contribute 9 digits.
- 2-digit numbers (10 through 99) contribute 90 * 2 = 180 digits.
- 3-digit numbers contribute 3 * x digits, where x is the number of 3-digit numbers included.
Setting up the equation to locate the 341st digit:
9 + 180 + 3 * x = 341
x = 50 + 2/3
This means the 341st digit is in the 51st 3-digit number (since 50 + 2/3 indicates we need the next number after the first 50).
Starting from 100, the 51st 3-digit number is 100 + 51 - 1 = 150. Since we have 2/3 remaining, the required digit is the second digit of 150, which is 5.
Therefore, the 341st digit is 5.
Answer: D.
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1 -> 9 = 9 digits
10 -> 99 = 90*2 = 180 digits
100 -> 150 = 51*3 = 153 digits
Add them up
9 + 180 + 153 = 342
Till 150, 342 digits would be there. This means 0 in 150 is the 342nd digit and 5 in 150 is the 341st digit
Therefore (D) 5 is the answer
10 -> 99 = 90*2 = 180 digits
100 -> 150 = 51*3 = 153 digits
Add them up
9 + 180 + 153 = 342
Till 150, 342 digits would be there. This means 0 in 150 is the 342nd digit and 5 in 150 is the 341st digit
Therefore (D) 5 is the answer
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1. The 1-digit numbers (1 to 9) contribute 9 digits.
2. The 2-digit numbers (10 to 99) contribute 180 digits.
3. After subtracting 9 + 180 from 341, we have 341 - 189 = 152 digits remaining to be accounted for in the 3-digit numbers.
Each 3-digit number contributes 3 digits. So, dividing 152 by 3 gives 50 full 3-digit numbers, which account for 150 digits. This leaves us with 2 digits remaining.
The 50th 3-digit number is 149, and the next number is 150. The 152nd digit is the 2nd digit of 150, which is 5.
Therefore, the answer is D. 5
2. The 2-digit numbers (10 to 99) contribute 180 digits.
3. After subtracting 9 + 180 from 341, we have 341 - 189 = 152 digits remaining to be accounted for in the 3-digit numbers.
Each 3-digit number contributes 3 digits. So, dividing 152 by 3 gives 50 full 3-digit numbers, which account for 150 digits. This leaves us with 2 digits remaining.
The 50th 3-digit number is 149, and the next number is 150. The 152nd digit is the 2nd digit of 150, which is 5.
Therefore, the answer is D. 5
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Let's approach this step by step
1) First let's understand how k is formed: k = 12345678910........998999
2) we need to find the 341st digit in this sequence
3) let's count the digits:- 1 to 9: 9 digits
10 to 99 digits: 90 x 2 = 180 digits
100 to 341: (341-99) x 3 = 726 digits
4) Total digits up to 341: 9+180+726= 915
5) the 341st digit is within the three digit numbers specifically in the number 113
6) 341-(9+180)=152nd digit in the three digit number sequence
7) 152/3=50 remainder 2
8) this means its the 2nd digit of the 51st three digit numbers
9) the 51st three digit number is 150
10) the 2nd digit of 150 is 5
therefore the 341st digit in k is 5
1) First let's understand how k is formed: k = 12345678910........998999
2) we need to find the 341st digit in this sequence
3) let's count the digits:- 1 to 9: 9 digits
10 to 99 digits: 90 x 2 = 180 digits
100 to 341: (341-99) x 3 = 726 digits
4) Total digits up to 341: 9+180+726= 915
5) the 341st digit is within the three digit numbers specifically in the number 113
6) 341-(9+180)=152nd digit in the three digit number sequence
7) 152/3=50 remainder 2
8) this means its the 2nd digit of the 51st three digit numbers
9) the 51st three digit number is 150
10) the 2nd digit of 150 is 5
therefore the 341st digit in k is 5
If we write the numbers 1,2,3,...9 - 9 digits
10 - 99 -> 2*90 = 180 digits
From hundred each number has 3 digits.
We are interested in the 341th digit, So after removing the 1 and 2 digit numbers from 341 digits
-> 341-189 = 152
We are interested in 152th digit from 100, each number consists of 3 digits => 152/3 = 50,
From 100 to 50 numbers = 149, the next number is 150.
So, the 341st digit in k is 5.
10 - 99 -> 2*90 = 180 digits
From hundred each number has 3 digits.
We are interested in the 341th digit, So after removing the 1 and 2 digit numbers from 341 digits
-> 341-189 = 152
We are interested in 152th digit from 100, each number consists of 3 digits => 152/3 = 50,
From 100 to 50 numbers = 149, the next number is 150.
So, the 341st digit in k is 5.
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1 Digit = 9 Numbers = 9*1 = 9 Digits
2 Digit = 10-99 = 90 numbers = 180 digits
Total Digits till 2 digit Numbers complete = 189 Digits
Remaining = 341-189 = 152 digits
Now How many 3 Digit numbers we need = 152/3 = 51 Numbers will give me 153 Digits
so 51st number from 100
use AP formula
100+(50)*1 = 150 (so, 1 340th Digit, 5 341st Digit, and 0 would be 342nd Digit)
so answer = 5
2 Digit = 10-99 = 90 numbers = 180 digits
Total Digits till 2 digit Numbers complete = 189 Digits
Remaining = 341-189 = 152 digits
Now How many 3 Digit numbers we need = 152/3 = 51 Numbers will give me 153 Digits
so 51st number from 100
use AP formula
100+(50)*1 = 150 (so, 1 340th Digit, 5 341st Digit, and 0 would be 342nd Digit)
so answer = 5
Ans is D. The sequence begins with 1-digit numbers from 1 to 9, 2-digit numbers from 10 - 99, covering 9 + 180 = 189 positions of the sequence. So the remaining positions will be occupied by a 3-digit number. Counting in the similar manner we find that the 340th, 341st and 342nd position will be taken by the digits 1, 5, and 1 respectively.
There are 9 single digit numbers, 90 two digit numbers and 900 three digit numbers. To the final number i.e. 123456789101112...998999, the single digit nos. contribute 9 digits, two digit nos. contribute 90*2 = 180 digits, and three digit nos. contribute 900*3 = 2700 digits.
The 341st digit lies in three digit number section as, 341 - 9 - 180 = 152 and the 152nd digit in the three-digit section corresponds to the 152÷3 = 51st three digit numbers 2nd integer (as remainder is 2)
The 51st three-digit number is 100 + 51 - 1 = 150 and second digit is 5
Answer D.
_________________
The 341st digit lies in three digit number section as, 341 - 9 - 180 = 152 and the 152nd digit in the three-digit section corresponds to the 152÷3 = 51st three digit numbers 2nd integer (as remainder is 2)
The 51st three-digit number is 100 + 51 - 1 = 150 and second digit is 5
Answer D.
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#
#2 digits = 90.
So 2*90 digits in final umber.
So total 189 digits, From here it is 3 digits.
To get 341th digit we need to find the 341-189 = 152digit, Which is 3 digit number at the 50th position+...
1st - 100
2nd -101
.
.
.
.
50th - 149.
Till 50th 150 digits so we need 152 digit
So .............149150... 9 is 150th so 152 is 5
IMO D
Bunuel
#1 digit = 9,
#2 digits = 90.
So 2*90 digits in final umber.
So total 189 digits, From here it is 3 digits.
To get 341th digit we need to find the 341-189 = 152digit, Which is 3 digit number at the 50th position+...
1st - 100
2nd -101
.
.
.
.
50th - 149.
Till 50th 150 digits so we need 152 digit
So .............149150... 9 is 150th so 152 is 5
IMO D
So, digit 1-9 contribute only 1 digit, afterwards, 10-99 each number contributes two digits. thus number of digit after 99 is 9+(90*2)=189. Post that each number will contribute 3 digits. thus, we need another 152 digits out of which 150 will be contributed by 149. thus 150 will contribute to 151 and 152nd digit. thus answer is D=5.
Bunuel
Karanjotsingh
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Concentration: Finance, Entrepreneurship
Posts: 139
Explanation
To determine the 341st digit in the number k formed by concatenating the integers from 1 to 999 sequentially (i.e., k=123456789101112...998999), we can follow these steps:
Step 1: Categorize the Numbers by Digit Length
Step 2: Locate the 341st Digit
Since 341 is less than 2889, the 341st digit lies within the concatenation of the numbers from 1 to 999. Let's determine exactly where:
The 341st digit in the concatenated number k is 5.
Answer: D) 5
To determine the 341st digit in the number k formed by concatenating the integers from 1 to 999 sequentially (i.e., k=123456789101112...998999), we can follow these steps:
Step 1: Categorize the Numbers by Digit Length
- 1-digit numbers (1 to 9):
- Count: 9 numbers
- Total digits: 9 × 1 = 99 digits
- 2-digit numbers (10 to 99):
- Count: 99 − 10 + 1 =90 numbers
- Total digits: 90 × 2 = 180 digits
- 3-digit numbers (100 to 999):
- Count: 999 − 100 + 1 = 900 numbers
- Total digits: 900×3=2700900 \times 3 = 2700900×3=2700 digits
Step 2: Locate the 341st Digit
Since 341 is less than 2889, the 341st digit lies within the concatenation of the numbers from 1 to 999. Let's determine exactly where:
- Digits covered by 1-digit and 2-digit numbers:
- Total: 9 + 180 = 189 digits
- Digits remaining to reach the 341st digit:
341 − 189 = 152 digits - Determine which 3-digit number contains the 152nd digit:
- Each 3-digit number contributes 3 digits.
- Number of complete 3-digit numbers in 152 digits:
1523 = 50 complete numbers with a remainder of 2 - Thus, 50 complete 3-digit numbers account for: 50 × 3 = 150 digits
- Next digit position: 152 − 150 = 2 (i.e., the 2nd digit of the next number)
- Identify the specific 3-digit number:
- Starting number: 100
- 50th 3-digit number: 100 + 50 − 1 = 149
- Next number: 150
- Find the 2nd digit of 150:
- Digits of 150: '1', '5', '0'
- 2nd digit: '5'
The 341st digit in the concatenated number k is 5.
Answer: D) 5
1 2 3 ... 9 -> 9 digits
10 12 ... 19 -> 20 digits
20 ... -> 20 digits
.
.
.
90 ...... 99 -> 20 digits
9 + 9(20) = 189 digits
100 101 ... 109 -> 30 digits
110
..
Now
341 - 189 = 152 digits
152 mod 30 = 2
2nd digit in the 6th row
i.e. 150 151 152 ...
Option D
10 12 ... 19 -> 20 digits
20 ... -> 20 digits
.
.
.
90 ...... 99 -> 20 digits
9 + 9(20) = 189 digits
100 101 ... 109 -> 30 digits
110
..
Now
341 - 189 = 152 digits
152 mod 30 = 2
2nd digit in the 6th row
i.e. 150 151 152 ...
Option D
Bunuel
Step 1: Count digits from one-digit numbers.
- There are 9 one-digit numbers (1 through 9), each contributing 1 digit.
- Total digits = \(9 \times 1 = 9\) digits.
Step 2: Count digits from two-digit numbers.
- There are 90 two-digit numbers (from 10 to 99), each contributing 2 digits.
- Total digits = \(90 \times 2 = 180\) digits.
Step 3: Calculate total digits so far.
- Total digits from one-digit and two-digit numbers = \(9 + 180 = 189\) digits.
Step 4: Find the position of the 341st digit.
- The 341st digit is found by calculating the position beyond the first 189 digits.
- Position in the three-digit numbers = \(341 - 189 = 152\).
Step 5: Determine which three-digit number contains the 152nd digit.
- Since each three-digit number contributes 3 digits, divide the position by 3 to find the target number:
- Number of full three-digit numbers = \(\left\lfloor \frac{152}{3} \right\rfloor = 50\) full numbers.
- Position within the next number = \(152 - (50 \times 3) = 2\).
Step 6: Identify the exact three-digit number and the specific digit.
- The 50th three-digit number after 99 is 149 (starting from 100).
- The 51st number, which contains the 152nd digit, is \(149 + 1 = 150\).
- The 2nd digit of 150 is 5.
Conclusion:
- The 341st digit in the sequence is 5.
Answer: D. 5
- There are 9 one-digit numbers (1 through 9), each contributing 1 digit.
- Total digits = \(9 \times 1 = 9\) digits.
Step 2: Count digits from two-digit numbers.
- There are 90 two-digit numbers (from 10 to 99), each contributing 2 digits.
- Total digits = \(90 \times 2 = 180\) digits.
Step 3: Calculate total digits so far.
- Total digits from one-digit and two-digit numbers = \(9 + 180 = 189\) digits.
Step 4: Find the position of the 341st digit.
- The 341st digit is found by calculating the position beyond the first 189 digits.
- Position in the three-digit numbers = \(341 - 189 = 152\).
Step 5: Determine which three-digit number contains the 152nd digit.
- Since each three-digit number contributes 3 digits, divide the position by 3 to find the target number:
- Number of full three-digit numbers = \(\left\lfloor \frac{152}{3} \right\rfloor = 50\) full numbers.
- Position within the next number = \(152 - (50 \times 3) = 2\).
Step 6: Identify the exact three-digit number and the specific digit.
- The 50th three-digit number after 99 is 149 (starting from 100).
- The 51st number, which contains the 152nd digit, is \(149 + 1 = 150\).
- The 2nd digit of 150 is 5.
Conclusion:
- The 341st digit in the sequence is 5.
Answer: D. 5
Bunuel
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If k is the number formed by writing the integers from 1 to 999 sequentially (123456789101112...998999), what is the 341st digit in k?
Digit number 1 = 1
Digit number 2 = 2
...
Digit number 9 = 9
Digit numbers 10-11 = 10
Digit numbers 12-13 = 11
...
Digit numbers 10+89*2=188, 189 = 99
Digit numbers 190-191-192 = 100
Digit numbers 193-194-195 = 101
Digit numbers 196-197-198 = 102
Digit numbers 199-200-201 = 103
Digit numbers 190+3*50=340-341-342 = 150
Digit number 341 = 5
IMO D
Digit number 1 = 1
Digit number 2 = 2
...
Digit number 9 = 9
Digit numbers 10-11 = 10
Digit numbers 12-13 = 11
...
Digit numbers 10+89*2=188, 189 = 99
Digit numbers 190-191-192 = 100
Digit numbers 193-194-195 = 101
Digit numbers 196-197-198 = 102
Digit numbers 199-200-201 = 103
Digit numbers 190+3*50=340-341-342 = 150
Digit number 341 = 5
IMO D
Bunuel
Single-digit numbers (1-9): These take up 9 digits.
Two-digit numbers (10-99): There are 90 of these, each taking up 2 digits, so 90 * 2 = 180 digits.
Three-digit numbers (100-999): There are 900 of these, each taking up 3 digits, so 900 * 3 = 2700 digits.
add up the digits:
9 digits (single-digit numbers)
180 digits (two-digit numbers)
Total so far: 189 digits
We need to find the 341st digit, which means we need to go further into the three-digit numbers.
Remaining digits needed: 341 - 189 = 152 digits.
Since each three-digit number contributes 3 digits, we can find out how many complete three-digit numbers we need to reach the 152nd digit:
153 / 3 = 51
The 50th three-digit number is 149, and the 51st is 150.
The 152nd digit will be the second digit of the number 150, which is 5.
IMO D
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Hello ,
first nine digit will be single and from 10 to 99 it will double digit so there will be total 9 single and 90 double so it will be 189 digit .
from 100 it will be 3 digit so we another 152 digit
from 100 to 149 we will have 150 more digit
so we need two more digit next digit will be 150
hence 5 will be 341 th digit
Hence answer is D
first nine digit will be single and from 10 to 99 it will double digit so there will be total 9 single and 90 double so it will be 189 digit .
from 100 it will be 3 digit so we another 152 digit
from 100 to 149 we will have 150 more digit
so we need two more digit next digit will be 150
hence 5 will be 341 th digit
Hence answer is D
Posts: 40
total digits from 1 to 9 = 9
total digits from 10 to 19 = 20
total digits from 20 to 29 = 20
.
.
total digits from 90 to 99 = 20
total digits from 100 to 109 = 30
total digits from 110 to 119 = 30
.
.
total digits from 140 to 149 = 30
adding all this gives 150
the 2nd digit in 150,151,.. is 5
Option D is correct
total digits from 10 to 19 = 20
total digits from 20 to 29 = 20
.
.
total digits from 90 to 99 = 20
total digits from 100 to 109 = 30
total digits from 110 to 119 = 30
.
.
total digits from 140 to 149 = 30
adding all this gives 150
the 2nd digit in 150,151,.. is 5
Option D is correct
Bunuel
Posts: 1
Digit are written first as 1 to 9. Therefore 9 digits are done.
Then we have number from 10 to 99 which are 90 numbers. Hence number of digits are 90 x 2 = 180.
Total as of now we have 189 digits done.
Now 341-189 = 152nd digit is required.
For nearest 3rd multiple is 150 which is 50 numbers from 100.
Hence starting from 100 next 50 numbers are till 149.
After ...........149150. in this digit 9 is at 339 position and hence at 341 we have digit 5.
Then we have number from 10 to 99 which are 90 numbers. Hence number of digits are 90 x 2 = 180.
Total as of now we have 189 digits done.
Now 341-189 = 152nd digit is required.
For nearest 3rd multiple is 150 which is 50 numbers from 100.
Hence starting from 100 next 50 numbers are till 149.
After ...........149150. in this digit 9 is at 339 position and hence at 341 we have digit 5.
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