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

[anemone]

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.

 

[Opalg]

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 $63-14=49$.

 

[anemone]

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.

 

[melese]

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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