Find the number of times the digit $9$ appears in the list of all integers from $1$ to $750$. (The number $ 99 $, for example, is counted twice, because $9$ appears two times in it.)

fjeihqoO Mar 30, 2024

#1**0 **

Here's how to find the number of times the digit 9 appears in the list of all integers from 1 to 750:

Approach 1: Separating by Hundreds Place

We can break down the count into three sections based on the hundreds digit:

Numbers from 1 to 99:

Every unit digit (from 0 to 9) will appear 10 times (once in each number from 10 to 19, 20 to 29, and so on).

Since 9 is a unit digit, it appears 10 times in this range.

Numbers from 100 to 199:

The hundreds digit is always 1, so we only need to consider the tens and units digits.

Following the same logic as before, each digit (including 9) will appear 10 times in the tens place.

Therefore, 9 appears 10 times in this range.

Numbers from 200 to 750:

We again need to consider both the hundreds and tens/units digits.

The hundreds digit can be 2, 3, ..., 7 (6 different values).

For each hundreds digit value, the tens and units digits will each allow the digit 9 to appear 10 times (as explained earlier).

Counting the Occurrences:

Occurrences in 1-99: 10 times

Occurrences in 100-199: 10 times

Occurrences in 200-750: 6 hundreds digits * 10 occurrences/hundreds digit = 60 times

Total Occurrences:

Adding the occurrences from each section:

Total = 10 (from 1-99) + 10 (from 100-199) + 60 (from 200-750) = 80 times

Approach 2: Utilizing Digit Patterns

Another approach is to analyze the digit patterns within the range.

Single-digit numbers (1-9): The digit 9 appears once.

Two-digit numbers ending in 9 (from 19 to 99): Each number contributes the digit 9 once. There are 10 such numbers (9 total for 10s digit and 1 for the unit digit of 99).

Three-digit numbers with 9 in the tens or units place (from 109 to 759, and from 190 to 799): Similar to two-digit numbers, each number contributes the digit 9 once. There are 10 numbers for each placement of 9 (tens or units), resulting in 20 numbers.

Three-digit numbers with 9 in the hundreds place (from 900 to 999): Each number contributes the digit 9 twice. There are 100 such numbers (10 numbers from 900 to 909 and another 90 numbers from 910 to 999).

Counting the Occurrences:

Single digit: 1 time

Two-digit ending in 9: 10 times

Three-digit with 9 in tens/units: 20 times

Three-digit with 9 in hundreds: 100 times (each digit counted twice)

Total Occurrences:

Adding the occurrences from each pattern:

Total = 1 + 10 + 20 + 2 * 100 = 131 times

Addressing the Discrepancy:

The second approach seems to give a higher count (131) compared to the first approach (80). However, there's a double-counting error in the second approach.

Numbers like 199 and 909 are counted twice (once for 9 in tens place and again for 9 in units place).

Correcting the Count:

We need to subtract the number of times a three-digit number is counted twice for both tens and units digits containing 9.

There are 20 such numbers (as counted earlier).

Final Count:

Total occurrences (corrected) = 131 (initial count) - 20 (double-counted) = 111 times

This corrected count (111) is incorrect. The first approach (80) provides the accurate answer.

The mistake in the second approach highlights the importance of careful analysis and avoiding double-counting when dealing with digit patterns.

booboo44 Mar 30, 2024