MHB Find the number of line and column where the number 2002 stays.

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The discussion revolves around determining the position of the number 2002 in a triangular number arrangement. It is established that the nth element in the top row corresponds to the nth triangular number, and calculations show that 2002 falls within the 63rd group. Specifically, it is found that 2002 is located in row 15 and column 49. The method involves calculating triangular numbers and their relationships to derive the position accurately. Ultimately, the conclusion confirms that 2002 is at (15, 49) in the arrangement.
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Positive real numbers are arranged in the form:

$$1 \;\;\; 3 \;\;\; 6 \;\;\;10 \;\;\; \cdots$$
$$2 \;\;\;5 \;\;\;9 \;\;\; \cdots$$
$$4 \;\;\;8 \;\;\; \cdots$$
$$7 \;\;\; \cdots$$

Find the number of the line and column where the number 2002 stays.
 
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anemone said:
Positive real numbers are arranged in the form:

$$1 \;\;\; 3 \;\;\; 6 \;\;\;10 \;\;\; \cdots$$
$$2 \;\;\;5 \;\;\;9 \;\;\; \cdots$$
$$4 \;\;\;8 \;\;\; \cdots$$
$$7 \;\;\; \cdots$$

Find the number of the line and column where the number 2002 stays.
You will have noticed that the $n$th element in the top row is the $n$th triangular number $\frac12n(n+1)$. Then working back from that, each time you subtract $1$ you move back one column and down one row (until you reach the first column).

The $63$rd triangular number is $2016$, so that will be in row $1$, column $63$. Working back $14$ places from that, you find that $2002$ will be in row $1+14 =15$ and column $63-14=49$.
 
Thanks for your solution, Opalg! :)

My solution:

[TABLE="class: grid, width: 200"]
[TR]
[TD="align: center"]Number[/TD]
[TD="align: center"]Row[/TD]
[TD="align: center"]Column[/TD]
[TD="align: center"]Group[/TD]
[/TR]
[TR]
[TD="align: center"]1[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]1st[/TD]
[/TR]
[TR]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]2nd[/TD]
[/TR]
[TR]
[TD="align: center"]3[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2nd[/TD]
[/TR]
[TR]
[TD="align: center"]4[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]3rd[/TD]
[/TR]
[TR]
[TD="align: center"]5[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3rd[/TD]
[/TR]
[TR]
[TD="align: center"]6[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]3rd[/TD]
[/TR]
[TR]
[TD="align: center"]7[/TD]
[TD="align: center"]4[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]4th[/TD]
[/TR]
[TR]
[TD="align: center"]8[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]4th[/TD]
[/TR]
[TR]
[TD="align: center"]9[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]4th[/TD]
[/TR]
[TR]
[TD="align: center"]10[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]4[/TD]
[TD="align: center"]4th[/TD]
[/TR]
[TR]
[TD="align: center"]11[/TD]
[TD="align: center"]5[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]5th[/TD]
[/TR]
[TR]
[TD="align: center"]12[/TD]
[TD="align: center"]4[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]5th[/TD]
[/TR]
[TR]
[TD="align: center"]13[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]3[/TD]
[TD="align: center"]5th[/TD]
[/TR]
[TR]
[TD="align: center"]14[/TD]
[TD="align: center"]2[/TD]
[TD="align: center"]4[/TD]
[TD="align: center"]5th[/TD]
[/TR]
[TR]
[TD="align: center"]15[/TD]
[TD="align: center"]1[/TD]
[TD="align: center"]5[/TD]
[TD="align: center"]5th[/TD]
[/TR]
[/TABLE]

To see where the number 2002 stays in the table above, we first need to investigate at which group that it lies, and this could be done in the following manner:

$$S_n=\frac{n}{2}\left(n+1\right)$$

$$2002=\frac{n}{2}\left(n+1\right)$$

$$n=62.78$$

We know that 2002 must lie in the 63th group, and to know what the previous number in the 62th group is, we do the following:

$$S_62=\frac{62}{2}\left(62+1\right)=1953$$

Thus, we have:

[TABLE="class: grid, width: 200"]
[TR]
[TD]Number[/TD]
[TD]Row[/TD]
[TD]Column[/TD]
[TD]Group[/TD]
[/TR]
[TR]
[TD]1953[/TD]
[TD]1[/TD]
[TD]62[/TD]
[TD]62th[/TD]
[/TR]
[TR]
[TD]1954[/TD]
[TD]63[/TD]
[TD]1[/TD]
[TD]63th[/TD]
[/TR]
[TR]
[TD]...[/TD]
[TD]...[/TD]
[TD]...[/TD]
[TD]...[/TD]
[/TR]
[/TABLE]

To find where the number 2002 lies in the 63th group, i.e. how far it is from 1954, we compute:

$$2002-1954=48$$

Hence, 2002 lies in the $$63-48=15th$$ row and the $$1+48=49th$$ column.
 
I tried to find a formula depending only on $k$, namely, given a positive integer $k$, to find its location $(C_k,R_k)$ in the table. For brevity of reference, we say that $k=(C_k,R_k)$ (abuse of notation)

Considering the obvious diagonals, it's clear that '$k$ is in the $n$'th diagonal' is equivalent to '$T_{n-1}<k\leq T_n$(*) for some integer $n\geq1$'.

Notice that for the diagonal $\{(1,n),(2,n-1),...,(1+x,n-x),...(n,1)\}$ we have $(1+x,n-x)=(1,n)-x=T_n-x$, because the numbers are consecutive.
Since $k$ lies on this diagonal, we may set $k=T_n-x$ and obtain $(C_k,R_k)=(1+(T_n-k),n-(T_n-k))$ (I expect the ambiguity of the notaion to be confusing, but I hope I'm wrong)

Now we want to "eliminate" $n$. For this, we go back to one of(*): $T_{n-1}<k\Leftrightarrow T_{n-1}+1\leq k\Leftrightarrow n^2-n+2\leq 2k\Leftrightarrow (n-1/2)^2+7/4\leq2k\Leftrightarrow n\leq\frac{(8k-7)^{1/2}+1}{2}$

From (*), it clear that $n$ is the greatest integer that satisfies $T_{n-1}<k$, and therefore the greatest to satisfy $n\leq\frac{(8k-7)^{1/2}+1}{2}$; so by definition $n=\lfloor\frac{(8k-7)^{1/2}+1}{2}\rfloor$.

At last, the answer: The location of $k$ is $(T_n-k+1,n+k-T_n)$, where $n=\lfloor\frac{(8k-7)^{1/2}+1}{2}\rfloor$.

Example: Take $k=2002$, then $n=63$ and hence $2002$ lies in $(T_{63}-2002+1,63+2002-T_{63})=(15,49)$.(Whew)

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