If all the numbers from 501 to 700 are written, what is the total number of times the digit 6 appears?

### Concept & Strategy When asked to find how many times a specific digit appears in a given range of numbers, it is most efficient to count the occurrences of that digit in each positional place value independently (hundreds, tens, and units). Because we are counting the literal number of times the digit is written, a number like $666$ will correctly be counted three times using this positional method. ### Step-by-Step Solution * **Given:** The numerical range is from $501$ to $700$. We need to count the frequency of the digit $6$. * **Step 1: Count occurrences in the Hundreds place.** The digit $6$ appears in the hundreds place for all numbers from $600$ to $699$. Total occurrences = $100$ times. * **Step 2: Count occurrences in the Tens place.** The digit $6$ appears in the tens place for the $560$s ($560$ to $569$) and the $660$s ($660$ to $669$). There are $10$ occurrences in the $500$s and $10$ occurrences in the $600$s. Total occurrences = $20$ times. * **Step 3: Count occurrences in the Units place.** The digit $6$ appears in the units place whenever a number ends in $6$ (e.g., $506, 516, 526, \dots, 596$). In the $500$s block, this happens $10$ times. In the $600$s block, this also happens $10$ times ($606, 616, \dots, 696$). Total occurrences = $20$ times. * **Step 4: Sum the occurrences.** Total = $100$ (Hundreds) + $20$ (Tens) + $20$ (Units) = $140$. ### Exam Strategy & Shortcut Memorize the standard distribution of digits: In any full $100$-number block (like $500$ to $599$), any given digit from $0$-$9$ will appear exactly $20$ times in the tens and units places combined. For $501$ to $700$, we have exactly two $100$-number blocks ($500$s and $600$s). Tens/Units count = $20 \times 2 = 40$. Then, just add the $100$ occurrences where $6$ is the leading digit in the $600$s. $100 + 40 = 140$. Solved in 5 seconds. ### Common Pitfall A very common mistake is trying to list out numbers with $6$ and mistakenly treating a number like $666$ as a single occurrence, or conversely, subtracting overlaps. The question asks how many times the *digit* is written, not how many *numbers* contain the digit. Positional counting avoids this trap entirely. ### Final Answer **Therefore, the correct answer is 140.**