Problem 164

To identify the 100th term, we must understand the sequence's construction rule. The sequence starts with 1, followed by two 2's, three 3's, four 4's, and so on. The pattern is that the number $n$ is repeated $n$ times.

Let's denote $S(n)$ as the cumulative sum of the first $n$ numbers. Mathematically, $S(n) = 1 + 2 + 3 + ... + n$. This is an arithmetic series and can be calculated by the formula $S(n) = \frac{n(n+1)}{2}$.

Now, we are looking for the smallest $n$ such that $S(n) \geq 100$.

Let's find $S(n)$ for some $n$:
- $S(1) = \frac{1(1+1)}{2} = 1$
- $S(2) = \frac{2(2+1)}{2} = 3$
- $S(3) = \frac{3(3+1)}{2} = 6$

Continuing this way would be inefficient, so let's use the explicit formula to find an $n$ such that $S(n) \geq 100$.

Assuming $S(n) \geq 100$,
$\frac{n(n+1)}{2} \geq 100$,
$n^2 + n \geq 200$,
$n^2 + n - 200 \geq 0$.

Looking for integral solutions of the quadratic equation, $n^2 + n - 200 = 0$, we can estimate the value by trial and error or solve the quadratic equation using the quadratic formula,
$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=1$, and $c=-200$.

$n = \frac{-1 \pm \sqrt{1 + 800}}{2}$,
$n = \frac{-1 \pm \sqrt{801}}{2}$.

We discard the negative root because we are looking for a positive $n$, and we approximate the positive root:
$n \approx \frac{-1 + \sqrt{801}}{2}$,
$n \approx \frac{-1 + 28.3}{2}$,
$n \approx 13.65$.

Since $n$ must be an integer, we consider $n=14$ because $S(13) < 100$ and $S(14)$ is the smallest sum larger than 100.

Calculating $S(14)$,
$S(14) = \frac{14(14+1)}{2} = \frac{14 \cdot 15}{2} = 7 \cdot 15 = 105$.

Thus, the 14 repeated 14 times is the 105th term, which means that the 100th term is also 14, as it falls in the block of fourteen 14's from terms 92 to 105 ($105 - 14 = 91$, so the block starts at term 92).

Therefore, the 100th term of the sequence is $\boxed{14}$.