Question 18

We will count the number of digits written for different number ranges.

Step 1 — Finding the 1000th digit

First, let us count digits for single-digit numbers. Numbers from 1 to 9 are single-digit numbers. There are 9 such numbers. Each number uses 1 digit. Total digits used are $9 \times 1$.

$9 \times 1 = 9$

Next, let us count digits for two-digit numbers. Numbers from 10 to 99 are two-digit numbers. There are $99 - 10 + 1$ numbers. This is 90 numbers. Each number uses 2 digits. Total digits used are $90 \times 2$.

$90 \times 2 = 180$

Total digits written up to number 99 are $9 + 180$.

$9 + 180 = 189$

We need to find the 1000th digit. Since 189 is less than 1000, the 1000th digit must be in a three-digit number. Let us find how many digits are left to count. We subtract 189 from 1000.

$1000 - 189 = 811$

These 811 digits belong to three-digit numbers. Each three-digit number uses 3 digits. Let us find how many full three-digit numbers these 811 digits cover. We divide 811 by 3.

$811 \div 3 = 270 \text{ with a remainder of } 1$

This means we use 270 full three-digit numbers. Then we use 1 more digit from the next three-digit number. The first three-digit number is 100. The 270th three-digit number is $100 + (270 - 1)$.

$100 + 269 = 369$

So, we have written all digits up to the number 369. The next number is 370. The 1000th digit is the first digit of 370. The first digit of 370 is '3'. It occurs in the number 370.

$\boxed{\text{1000th digit is '3', in number 370}}$

Step 2 — Finding the number containing the millionth digit

We continue counting digits for larger number ranges. Digits for 1-digit numbers (1-9): $9 \times 1 = 9$. Total digits up to 9: 9. Digits for 2-digit numbers (10-99): $90 \times 2 = 180$. Total digits up to 99: $9 + 180 = 189$. Digits for 3-digit numbers (100-999): $900 \times 3 = 2700$. Total digits up to 999: $189 + 2700 = 2889$. Digits for 4-digit numbers (1000-9999): $9000 \times 4 = 36000$. Total digits up to 9999: $2889 + 36000 = 38889$. Digits for 5-digit numbers (10000-99999): $90000 \times 5 = 450000$. Total digits up to 99999: $38889 + 450000 = 488889$.

We need to find the millionth digit (1,000,000th digit). Since 488889 is less than 1,000,000, the millionth digit must be in a six-digit number. Let us find how many digits are left to count. We subtract 488889 from 1,000,000.

$1000000 - 488889 = 511111$

These 511111 digits belong to six-digit numbers. Each six-digit number uses 6 digits. Let us find how many full six-digit numbers these 511111 digits cover. We divide 511111 by 6.

$511111 \div 6 = 85185 \text{ with a remainder of } 1$

This means we use 85185 full six-digit numbers. Then we use 1 more digit from the next six-digit number. The first six-digit number is 100000. The 85185th six-digit number is $100000 + (85185 - 1)$.

$100000 + 85184 = 185184$

So, we have written all digits up to the number 185184. The next number is 185185. The millionth digit is the first digit of 185185. The first digit of 185185 is '1'. It occurs in the number 185185.

$\boxed{\text{The millionth digit is in the number 185185}}$

Step 3 — Finding when the digit '5' is written for the 5000th time

Let us count the occurrences of the digit '5' in different number ranges. For numbers from 1 to $10^k - 1$, the digit '5' appears $k \times 10^{k-1}$ times.

Occurrences of '5' in 1-digit numbers (1-9): Here $k=1$. So, $1 \times 10^{1-1} = 1 \times 10^0 = 1$. Total '5's up to 9: 1.

Occurrences of '5' in 1-digit and 2-digit numbers (1-99): Here $k=2$. So, $2 \times 10^{2-1} = 2 \times 10^1 = 20$. Total '5's up to 99: 20.

Occurrences of '5' in 1-digit, 2-digit, and 3-digit numbers (1-999): Here $k=3$. So, $3 \times 10^{3-1} = 3 \times 10^2 = 300$. Total '5's up to 999: 300.

Occurrences of '5' in 1-digit, 2-digit, 3-digit, and 4-digit numbers (1-9999): Here $k=4$. So, $4 \times 10^{4-1} = 4 \times 10^3 = 4000$. Total '5's up to 9999: 4000.

We need to find the 5000th occurrence of the digit '5'. Since 4000 is less than 5000, the 5000th '5' must be in a five-digit number. We need $5000 - 4000 = 1000$ more '5's. These 1000 '5's will come from numbers starting from 10000.

Let us count '5's in blocks of 1000 numbers, starting from 10000. Consider numbers from 10000 to 10999. The first digit is '1'. The remaining three digits are like numbers from 000 to 999. The number of '5's in numbers from 0 to 999 is $3 \times 10^{3-1} = 3 \times 10^2 = 300$. So, in 10000-10999, there are 300 '5's. Total '5's up to 10999: $4000 + 300 = 4300$. We still need $5000 - 4300 = 700$ more '5's.

Consider numbers from 11000 to 11999. Similarly, there are 300 '5's in this range. Total '5's up to 11999: $4300 + 300 = 4600$. We still need $5000 - 4600 = 400$ more '5's.

Consider numbers from 12000 to 12999. Similarly, there are 300 '5's in this range. Total '5's up to 12999: $4600 + 300 = 4900$. We still need $5000 - 4900 = 100$ more '5's.

These 100 '5's will come from numbers starting from 13000. Let us count '5's in blocks of 100 numbers, starting from 13000. Consider numbers from 13000 to 13099. The first two digits are '13'. The remaining two digits are like numbers from 00 to 99. The number of '5's in numbers from 0 to 99 is $2 \times 10^{2-1} = 2 \times 10^1 = 20$. So, in 13000-13099, there are 20 '5's. Total '5's up to 13099: $4900 + 20 = 4920$. We still need $5000 - 4920 = 80$ more '5's.

Consider numbers from 13100 to 13199. Similarly, there are 20 '5's in this range. Total '5's up to 13199: $4920 + 20 = 4940$. We still need $5000 - 4940 = 60$ more '5's.

Consider numbers from 13200 to 13299. Similarly, there are 20 '5's in this range. Total '5's up to 13299: $4940 + 20 = 4960$. We still need $5000 - 4960 = 40$ more '5's.

Consider numbers from 13300 to 13399. Similarly, there are 20 '5's in this range. Total '5's up to 13399: $4960 + 20 = 4980$. We still need $5000 - 4980 = 20$ more '5's.

Consider numbers from 13400 to 13499. Similarly, there are 20 '5's in this range. Total '5's up to 13499: $4980 + 20 = 5000$.

So, the 5000th '5' is written when we finish counting all '5's up to the number 13499.

$\boxed{\text{The digit '5' is written for the 5000th time at number 13499}}$

Answer

(a) The 1000th digit would be '3'. It would occur in the number 370. (b) The number that would contain the millionth digit is 185185. (c) You would have written the digit '5' for the 5000th time when you reach the number 13499.