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Breaking the question:
1. Contribution by Single digits 1 to 9 = 9

2. Contribution by two digit numbers :
99-10+1=90 two digit numbers
So contribution is 180

So 189th digit is 9 (of the number 99)
We need 341st digit, so 341 - 189 = 152nd digit after 9

3. Contribution by three digit numbers:
Each 3 digit # contributes 3 to the count.
So 50 (3 digit) numbers would contribute 150 to the count.

What is the 50th number from 100 => 149, and next number is 150.
Since '9' {of 149} is 150th then '5'{of 150} is the 152nd (or overall 341st term).

Ans=5
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1-9= 9 digits
10-99=90*2 =180 digits
We have 189 digits now

100-149=50*3=150 digits

we have 339 digits
next numbers are 150, 151

5 is 341st digit
D)
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Bunuel
12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

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

 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

 

1-9 have 1 digit = 1*9 = 9
10-99 have 2 digits = 2*90 = 180

Remaining 341-189 = 152

Start from 100 have 3 digits so 152/3 = 50 ...2
So it represent 100 - 149

The next number would be 150 and we count the 2nd digit = 5
So D is answer
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From 1 to 9 - 1 digit * 9 numbers = 9;
From 10 to 99 - 2 digits * 90 numbers = 180;
From 100 to 149 - 3 digits * 50 numbers = 150;

So by adding all the three we get 339 digits. So the leftover is 2 more digits(From 1 to 149 in sequence).

The next number is 150, where 5 is the 341st digit.
Option D
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The answer is 5.

To find the 341st digit, 1 to 9 has 9 digits

10 to 99 has = 99-10+1 = 90 and each number has two digits = 90*2 = 180

100 to 999 has 999-100+1 = 900 and each number has three digits = 900*3 = 2700

Now, to find 341st digit.. first two groups has 189 digits in which we cannot determine the 341st digit, so 341-189 = 152

so now we find the 152nd digit in the 3 digit group (100 to 999) --> 152 /3 = 51st digit

100+51 -1 = 150 and 152nd digit means second digit so its 5.
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The digits are written first as 1 to 9
Total number of digits in 1 to 9 = 9
Then, 10 to 99
Total number of digits in 10 to 99 = 2 ( 99 − 10 + 1) = 180
So, from 1 to 99 number of digits = 9 + 180 = 189

[hr]
Locate the 341st digit
The 341st digit falls within the 3-digit numbers. To determine its exact position:
  1. Subtract the digits contributed by 1-digit and 2-digit numbers:
341−9−180=152. 341 - 9 - 180 = 152.341−9−180=152.
Thus, the 341st digit is the 152
The 152-nd digit corresponds to the 2nd digit of 150, which is 5.
[hr]
Final Answer:
The 341st digit in kkk is 5

Bunuel
12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

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

 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

 

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While writing 1 to 999 sequentially, to get 341st digit...

First remove total digits of 1-10 = 11 => 341-11 = 330

Now remove digits of 11-100 => 330 - 181 = 149

Now remove digits of 101-110 => 149 - 30 = 119
Similarly it will be removed till 131-140 => 119 - 90 = 29

Now, remove digits of 141-150 => 29-30 = -1

=> Second last digit of last number from 141-150 = 5

Answer: D. 5
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solution:
single digit numbers: 9
two digit numbers: 90
total digits of k with above two numbers: 9+180=189
to get 341st digit, we need, 341-189=152 more digits.
that is 50 three digit numbers plus 2 digits of 51st number after 99. 31st number is 150.
therefore the answer is 5

Bunuel
12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

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

 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

 

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9*1=9 one digits,,,, (99-10)+1=90 two digits,,90*2=180 digits,,,,341-9-180=152,,,,,3*50=150,,,, 2 nd digit of 51st 3digit integer.... 100+(51-1)1=150,,, so 5 ,,,
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There are 9 one digit no., so total no. of digits contributed by them= 9
There are 90 two digit no., so total no. of digits contributed by them = 90x2= 180
There are 900 three digit no., so total no. of digits contributed by them= 900x3= 2700.

Total no. of digits contributed by 1-99= 180+9= 189, so when we reach the 189th digit in the FORMED NO.(given in question), we would have completed upto 99 no.

Now, remaing digits = 341- 189= 152

Now, from 100- 999, we have 3-digts no. only, so, 152/3= 50 (+2 remaining digits). We see there can be 50- 3 digits no. that can be formed with fully and still 2 digits will remain. So 149 will be the last 3- digit no. formed fully. So, the next no. is 150, but since we are left with only 2 digits, therefor k=5.

ans- k=5
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Ans D
When we write these in sequence
from 1 to 9 = single digit numbers to total digits written = 9
from 10 to 99 = 2 digit numbers = total digits written = (99-10+1) = 180 (10 an 99 both included in this)
till now total number of digits written = 9+180 = 189
from 100 to 999 = 3 digits numbers = total digits written = (999-100+1) = 900
this means that that number/digit lies in this range of sequence
now finding that 341st digit = 341-189 = 152 position
as 100 to 999 are 3 digit integers we can say that 152/3 = 150/3 + 2/3 => 50th + 2
the digit is the second digit of 51st after 100. =1(5)1
Hence 5 is the answer.
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IMO D

1. Single-Digit Numbers (1 to 9):
• There are 9 single-digit numbers.
• Total digits contributed: 9×1=9
2. Two-Digit Numbers (10 to 99):
• There are 99−10+1=90 two-digit numbers.
• Total digits contributed: 90×2=180
3. Three-Digit Numbers (100 to 999):
• There are 999−100+1=900three-digit numbers.
• Total digits contributed: 900×3=2700
Cumulative Digits:
• Digits from single-digit numbers: 9
• Digits from two-digit numbers: 180
• Digits from three-digit numbers: 2700
Finding the 341st Digit:
• Digits from single-digit numbers: 9
• Digits from two-digit numbers: 180
• Total digits from single and two-digit numbers: 9+180=189
Since 341 is greater than 189, the 341st digit is within the three-digit numbers.
Calculating the Position in Three-Digit Numbers:
• Digits needed beyond the first 189 digits: 341−189=152
• Each three-digit number contributes 3 digits.
• Number of complete three-digit numbers contributing to the 152 digits: ⌊152/3⌋=50complete three-digit numbers.
• Digits contributed by these 50 three-digit numbers: 50×3=150
• Remaining digits needed: 152−150=2
Identifying the Specific Number and Digit:
• The 50th three-digit number starting from 100 is 100+50−1=149
• The first digit of the next number (150) will be the 151st digit.
• The second digit of 150 will be the 152nd digit.
Thus, the 341st digit in the sequence is the second digit of 150, which is 5.
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This should be an easy question once we understand the pattern the digits will follow, as we can see this is an arithmetic sequence with a common difference of 1.

We want to find the 341st digit of this string (counting from the left)

If we see the sequence, first we have 9 one digit integers (from 1 to 9)

Each of the two-digit numbers will contribute a length of two to the string

Total length contributed by the two-digit numbers = Number of two-digit numbers * 2

Number of two-digit numbers =99 - 10 + 1 = 90

So their length = 90*2=180

Total Length = 180+9=189

Remaining length out of total = 341-189=152

Now all numbers will be three digit numbers i.e. each number will contribute a length of 3.

So for a length of 152, we need 50, 3 digit numbers and the 2nd digit of the 51st number.

50, 3 digit numbers will be from 100 to 149 and 51st number will be 150.

Hence the required value of the digit is 5.

Answer is D
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Bunuel
12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

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

 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

 

Total number of Single digit Integers = 9
Total number of double digit integers = 90
Total number of triple digit integers = 900

Number of digits for Single digit numbers = 9
Number of digits for double digit numbers = 90x2=180.
Remaining digits = 341 - 189 = 152.
MEANS 152nd digit will be in the three digit integers.
Which 3 digit integer? 152/3 = 50 + Remainder as 2.
2 as remainder means tens digit of 51st integer which will be tens digit of 150 which is 5.
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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

Single digit numbers=9
Sum of digits in 2 digit numbers= 90*2=180
Remaining digits= 341-189=152
Number of 3 digit numbers = 152/3= 50 + 1 numbers
50 the three digit number is 149 and second digit of the next number is 5

Ans D
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1-9 .. 9 digits
10-99...180 digits
100-149...150 digits

total 339 digits for 1 to 149

next is 150. so 340th will be 1 & 341th will be 5.

Answer D
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Let's understand what the question is asking -

if we write one big number by writing the integers from 1 to 999 in order, the number obtained will look like -

1234567891011.....998999

In this above number, the 7th digit is 7, 9th digit is 9, 10th digit is 1 and 11th digit is 0.
Similarly, we want to find what the 341st digit would be.

Starting from the number 10, till 99, we have 90 numbers and each number has 2 digits. Hence, there are 180 digits between 10 and 99. Adding the first 9 digits, the total makes it 189.
So, 189th digit is 9, of the number 99. Then, 190th digit is 1 of the number 100.
Now, from the numbers 100-150, we have 51 numbers, and each number has 3 digits. Therefore, 51 numbers will have 153 digits.

Adding previous digits 189 and new digits 153 -

189+153 = 342

that means, 342nd digit is 0 of the number 150.

Therefore, 341st digit would be 5.

Final Answer - Option D
Bunuel
12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

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

 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

 

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