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WE have 9 numbers with 1 digit = 9
We have 89 numbers (99-10) with 2 digits = 178
Sum of both = 9+178=187
We are looking for the 341 digit - 187 = 154 digits
Next numbers are going to have 3 digits so 154/3 = 51 1/3. So the 51 number starting from 100 will be 150, as long as there was 1 remainder from the division. the 341digits will be 1


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

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A simple, computer scientist approach..
1-digit integers: 9 x 1 = 9 digits
2-digit integers: 90 x 2 = 180 digits (90 numbers, as 99-10+1=90)
3-digit integers: ?

To find the 341st digit, we take 341-189 (known digits up to number 99) = 152
152 / 3 = 50 remainder 2

As such, we are on our 50th number, 2nd digit.. As we are counting up from 100, which is zero-indexed, (e.g. 101 is the 2nd number with 3 digits), this represents 100+50-1 = 149. Thus, 149 is our final number, the second digit of which is 4.
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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-digit numbers (1 to 9):
  • There are 9 numbers.
  • Each contributes 1 digit.
  • Total digits contributed: 9
2-digit numbers (10 to 99):
  • There are 99−10+1=90 numbers.
  • Each contributes 2 digits.
  • Total digits contributed: 90×2=180

3-digit numbers (100 to 150):
  • There are 150−100+1=51 numbers.
  • Each contributes 3 digits.
  • Total digits contributed: 51×3 = 153

Since we have reached 189 (9 + 180) digits already, it is likely next 51 3-digit numbers will give us the desired answer therefore we have taken numbers till 150
Therefore, 189 + 153 = 342

The last 3-digit number added is 150, therefore 341st digit is 5
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Single digit numbers (1-9): 9
Two-digit numbers (10-99): 90 two-digits = 180
three-digit numbers: (100-xxx) ?
341-189= 152 -> so we need to know which 152nd number is.

152/3 = 50th 3-digit number+2 -> 150 is the 50th 3-digit number and the second digit of 150 is 5 -> correct answer D. 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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1- 9 : 9 digits
10-99: 90 numbers with 2 digits = 180
100-149: 50 numbers with 3 digits =150

Total digits till 149 will be 9 + 180+150 = 339
So after 149 next digits will be 150. hence 5 is 341th digit.
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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
for 1 - 9 we have 9 digits.
for 10 - 99 we have 2*90 = 180 digits
remaining digits = 341 - 189 = 152
rest are 3 digits starting with 100, 101, 102,....
152/3 = 50 + Remainder is 2
50th digit is 149 and
51st number is 150
Therefore which makes 341st digit is 5
Answer is D.
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Hi All,

According to me,

number 1-9 will have a total of 9 digits,

number 10-19 will have 20 digits, so 10-99 will have 20 x 9 = 180 digits

100-109 will have 30 digits, so 189+30n <=341

Calculating we can find that n=5 for the nearest value to 341, total count for digits - 339, for N =5 the number count will be for group 140-149.

Therefore the digit 9 in the number 149 will be the 339th digit, hence iterating further the digit 5 in number 150 will be the 341th digit in the sequence K.

Therefore 5 is our answer.
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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

123..9--->9 NUMBERS WITH 1 DIGIT
101112....99-->90 NUMBERS WITH 2 DIGITS
100101...999---> 900 NUMBERS WITH 3 DIGITS.
341-(90*2+9)=341-189=152. 152/3=50+2/3. SO 51ST NUMBER'S MIDDLE DIGIT. 100+51=151. IT IS 5 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

 

Series provided is 123...998999.

1 Digit number 1-9 (Both inclusive) = 9
2 Digit numbers 10-99 (both inclusive) = 90. Each contributes 2 digits so 90*2 = 180
3 Digit number 100 - 999 (Both inclusive) = 900. Each contributes 3 digits 900*3 = 2700

In total 9+180+2700 = 2889

341st digit is going to be from the 3 digit category. 189 digits have already been used up by the 1 and 2 digits. so 341-189 = 152. finding the 152nd digits from the 3 digit category.

152 / 3 = 50 + 2(remainder). This is second digit of the 51st 3 digit number
Hence the digit is 5.
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k=12345678910111213141516.......998999
Find 341st digit in k
1-9 = 9 numbers
10-99 = 90 numbers (2 digits each), so total numbers 90*2=180
100-149 = 50 numbers having 3 digits each, so total numbers 50*3=150
Total digits till now= 9+180+150=339
150= 340-342th digits
341st digit= 5

Answer: D
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There are 9 single digit numbers & 90 two digit numbers so the number total digits keeping those two in mind would be equal to 9+90*2=189 digits. For the three digit numbers, 341-189=152 digits. 152 mod 3 would be equal to 2 so this would be the middle digit of the 51st three digit number.

100-149 would comprise of 50 three digit numbers amounting to a total of 150 digits so the next number's middle digit would be the 341st digit. In this case the digit is 5 hence the answer to this question is (D) 5
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From 1 to 9: 9 digits
From 10 to 99: 180 digits
From 100 to 199: 300 digits
So that mean the 341st digit must be within the range of 100 and 199.

Total digits from 1 to 99 are 189. That means to reach the 341st digit, we need another 152 digits = 50 x 3 + 2.
So the 341st digit must be the 2nd digit of number 151 (=100 + 50 +1).

Answer: 5
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1 to 10 → 11 nos

11 to 20 → 20 nos . So till 90 →160 no’s and from 90 to 100 is 21 no’s
So total 11 + 160 + 21 =192 192 nos

Now 101 to 110 is 30 and we need additional 149 which is 120 + 29
Now with 30 we would have completed 150 .

But since it is 1 short , the digit will be 5 . D answer
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There are infinite digits in integer k. But our focus needs to be to find 341st digit.

Let's count...

single digit integers: (1-9) - 9 integers ----> 9 digits
two digit integers: (10-99) - 90 integers ---> 180 digits

So far we got 189 digits. Remaining digits = 341-189 = 152 digits.

From three-digit integers, we need to find 152nd digit. 50 three-digit integers fill the 150 digits, then middle of 51st three-digit integer's digit will be 341st digit of k.

Since we started counting from 100, 150 will be the 51st three-digit integer. So middle digit is 5.

Answer is D.
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To find the pattern, we can write the numbers as

1234567891
0111213141
5161718192
0212223242
5262728293
0...

From this we get, for every 11th, 31st, 51st, 71th, 91th, 111th, 131th, 151th.. terms will be 0
& every 21st, 41st, 61st, 81st, 101th, 121th, 141th.. term would be 5

Hence, 341st term would be 5

IMO 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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let us count the total number by their size
size 1: 1 to 9 so total 9
size 2: 10 to 99 so total 99-10+1 = 90 now each number is associated with 2 digits hence total digits would be 90*2=180
size 3: 100 to 200 {let us say 200 because we only have to go till 341st no}=200-100+1= 101 numbers so total numbers would be {101*3=303}
upto this point the total numbers we have is 9+180+303=492 but we need to go till 341st
so backtracking and going till 150 from 100
s0
150-100+1=51--->51*3=153 so total numbers till here {9+180+153=342} very nearby where we have to go

so numbers are 1234567891011.......................148149150 {now we we are at 150 and we know the position of Rightmost Number is 342 so the number left of it is 341st hence 5}
mark 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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Number of single digit integers =9
Number of double digit integers = 90
so total number of digits until now = 9+ (90*2) =189
To find 341 digit we can find what is the 3 digit number that occurs on 341st digit .i.e 341-189 =152 digits.
so 150/3 = 50 numbers starting from 100 (until 149) occupy places from 190 to so on.
Therefore 1 is 340 digit , 5 is 341 digit and 2 is 342 digit .
Hence 5 is the answer
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