1. **Stating the problem:** We have a new sequence formed by taking the natural numbers and repeating every number divisible by 3 once more immediately after it. For example, the first 17 terms are: 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 10, 11, 12, 12, 13. We need to find: - (1) The position of the number 100 in the new sequence. - (2) The number at the 100th position in the new sequence. - (3) How many even numbers appear between 1 and 101 in the new sequence. --- 2. **Understanding the sequence:** - Every number divisible by 3 appears twice. - Other numbers appear once. 3. **Key formula:** Let $n$ be a natural number. - The count of numbers divisible by 3 up to $n$ is $\left\lfloor \frac{n}{3} \right\rfloor$. - The length of the new sequence up to $n$ is: $$L(n) = n + \left\lfloor \frac{n}{3} \right\rfloor$$ because each multiple of 3 is counted twice. --- ### 1) Find the position of number 100 in the new sequence We want to find $L(100)$: $$L(100) = 100 + \left\lfloor \frac{100}{3} \right\rfloor = 100 + 33 = 133$$ So, the number 100 appears at position 133 in the new sequence. --- ### 2) Find the number at the 100th position in the new sequence We want to find $n$ such that: $$L(n-1) < 100 \leq L(n)$$ Calculate $L(n)$ for $n$ near 100: - For $n=75$: $$L(75) = 75 + \left\lfloor \frac{75}{3} \right\rfloor = 75 + 25 = 100$$ - For $n=74$: $$L(74) = 74 + \left\lfloor \frac{74}{3} \right\rfloor = 74 + 24 = 98$$ Since $L(74) = 98 < 100 \leq 100 = L(75)$, the 100th term corresponds to $n=75$. Check if 75 is divisible by 3: - Yes, $75 \div 3 = 25$. Since multiples of 3 appear twice, the positions for 75 are $L(74)+1=99$ and $L(75)=100$. Therefore, the 100th term is **75**. --- ### 3) Count how many even numbers appear between 1 and 101 in the new sequence We consider numbers from 1 to 101. - Count of even numbers from 1 to 101: $$\left\lfloor \frac{101}{2} \right\rfloor = 50$$ - Count of even numbers divisible by 3 (i.e., divisible by 6) from 1 to 101: $$\left\lfloor \frac{101}{6} \right\rfloor = 16$$ Each even number appears once, except those divisible by 3 (i.e., divisible by 6), which appear twice. So total appearances of even numbers: $$50 + 16 = 66$$ Because the 16 even numbers divisible by 3 are counted twice, adding 16 more to the 50 unique even numbers. --- **Final answers:** 1. Number 100 is at position **133**. 2. The number at position 100 is **75**. 3. Even numbers appear **66** times between 1 and 101 in the new sequence.