What is the 1393th number? What is the 1393th number in this string??

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5........

Could you explain me how to answer this kind of questions .
 A: Note: The $(\frac{n(n+1)}{2})^{th}$ number is $n$ and the $(\frac{n(n+1)}{2}+1)^{th}$ number is $n+1$.
What we want to do is find the inverse of $x=\frac{n(n+1)}{2}$.
After solving the quadratic, you get $n=\frac{\sqrt{1 + 8 x}-1}{2}$ (take the positive root, of course).
Now plug in $x=1393$. and we get $n=52.2$. So the $1393^{th}$ number is a $53$ (round up).
So in general the $x^{th}$ number is $\lceil \frac{\sqrt{1 + 8 x}-1}{2} \rceil$.
A: Hint: Keep on subtracting $1, 2, 3, \dots$ from $1393$ until the remainder is larger than or equal to the next number to be subtracted. The next number to be subtracted is the $1393$-th number.
That is, find the largest $n$ such that
$$\sum_{i = 1}^n i < 1393$$
A: The last $1$ is at position $1$.
The last $2$ is at position $1+2 = 3$.
...
The last $n$ is at position $1 + 2 + ... + n = \frac 12 n(n+1)$.
Solve $\frac 12 n(n+1) = 1393$. All you need is an approximation.
Since for larger $n$, $n\approx n+1$, you can state $n^2 \approx (1393)(2) = 2786$ giving $n \approx 52$.
In fact $n$ is slighty over $52$. The final $52$ will be at position $\frac 12 \cdot 52 \cdot 53 = 1378$.
And the remaining $53$ positions (up to position $1431$) will be repeats of the number $53$.
So the answer is $53$.
A: It will be a bit easier if we start counting at 0.

start  0  1     3        6          10             15
index  0, 1  2  3  4  5  6  7  8  9 10 11 12 12 14 15 16
number 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6,........

 The number 1 starts at index  0 = 0
 The number 2 starts at index  1 = 0+1.
 The number 3 starts at index  3 = 0+1+2
 The number 4 starts at index  6 = 0+1+2+3
 The number 5 starts at index 10 = 0+1+2+3+4
 The number 6 starts at index 15 = 0+1+2+3+4+5
 The number n starts at index      n(n-1)/2.

So we need to find the largest n such that
    n(n-1)/2 ≤ 1392 (because we started counting at 0 instead of 1)
    n(n-1) ≤ 2784

Since  $n(n-1) = 2784$ is about $n^2 = 2784$, a good guess for $n$ would be $\lfloor \sqrt{2784} \rfloor = 52$
We compute
\begin{align}
   52\times 51 &= 2652\\
   53 \times 52 &= 2756\\
   54 \times 53 &= 2862\\
\end{align}
 So  the number 53 starts at index 53(53-1)/2 = 1378
 and the number 54 starts at index 54(54-1)/2 = 1431.

 That means that the 1379th number is 53
 and the 1432th number is 54.

 So the 1393th number is 53.

