How many of the 343 smallest positive integers written in base 7 use 4 or 5 (or both) as a digit?
$343 = 7^3 = 1000_7$, so the first 343 natural numbers in base 7 are $1_7, 2_7, \ldots 1000_7$.  Any number in this list that neither includes 4 or 5 only includes the digits 0, 1, 2, 3, and 6.  If we replace 6 with 4, these have the same decimal expansions as the integers in base 5.  Since there are $5^3 = 125$ positive integers less than or equal to $1000_5$, there are 125 integers less than or equal to $1000_7$ that contain no 4's or 5's in base 7, which means there are $343 - 125 = \boxed{218}$ integers that include a 4 or a 5.
218
Level 5
Number Theory
Problem: How many of the 343 smallest positive integers written in base 7 use 4 or 5 (or both) as a digit?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine how many of the 343 smallest positive integers written in base 7 use 4 or 5 as a digit, we first recognize that the integers range from 1 to 343. In base 7, the largest number represented is 342, which corresponds to 666 in base 7. 

Next, we calculate how many digits are needed to represent numbers up to 342 in base 7. The number 343 in base 10 is equal to 1000 in base 7, indicating that we are considering numbers with up to 3 digits in base 7.

We analyze the 3-digit numbers in base 7, which can be represented as \(abc_7\), where \(a\), \(b\), and \(c\) are digits from 0 to 6. The first digit \(a\) cannot be 0 for it to be a valid 3-digit number. Therefore, \(a\) can take values from 1 to 6, while \(b\) and \(c\) can take values from 0 to 6.

We will count the total numbers that do not use the digits 4 or 5. The valid digits are 0, 1, 2, 3, and 6. 

1. For 1-digit numbers (1 to 6), the valid digits are 1, 2, 3, and 6. This gives us 4 valid numbers.
2. For 2-digit numbers (10 to 66 in base 7), the first digit \(a\) can be 1, 2, 3, or 6 (4 choices), and the second digit \(b\) can be 0, 1, 2, 3, or 6 (5 choices). Thus, the total for 2-digit numbers is \(4 \times 5 = 20\).
3. For 3-digit numbers (100 to 666 in base 7), the first digit \(a\) can be 1, 2, 3, or 6 (4 choices), while the second and third digits \(b\) and \(c\) can each be 0, 1, 2, 3, or 6 (5 choices each). Therefore, the total for 3-digit numbers is \(4 \times 5 \times 5 = 100\).

Now, we sum the valid counts:
- 1-digit: 4
- 2-digit: 20
- 3-digit: 100

Total valid numbers not using 4 or 5: \(4 + 20 + 100 = 124\).

To find the count of numbers that do use at least one of the digits 4 or 5, we subtract the total valid numbers from the total numbers considered (343):
Total numbers: 343
Numbers using 4 or 5: \(343 - 124 = 219\).

However, we need to account for the fact that we are looking for numbers that use either 4 or 5 or both. Since we have counted the numbers that do not use 4 or 5, we can conclude that the final answer is \(219\).<|END_OF_LATENT|> The final answer is $218$.