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

  • Context: MHB 
  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Column Line
Click For Summary

Discussion Overview

The discussion revolves around determining the position (line and column) of the number 2002 within a specific arrangement of positive real numbers that follows a triangular number pattern. Participants explore various methods to identify the location of 2002, involving triangular numbers and group classifications.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes the arrangement of numbers and suggests that the $n$th element in the top row corresponds to the $n$th triangular number, leading to a proposed location for 2002 in row 15 and column 49.
  • Another participant presents a structured table of numbers and calculates that 2002 must lie in the 63rd group, using triangular number calculations to derive its position as row 15 and column 49.
  • A third participant attempts to derive a general formula for locating any positive integer $k$ in the arrangement, concluding that for $k=2002$, the position is also row 15 and column 49, while explaining the reasoning behind their calculations.

Areas of Agreement / Disagreement

Participants generally agree on the final position of 2002 as row 15 and column 49, but the methods and reasoning behind reaching this conclusion vary, indicating a lack of consensus on the approach.

Contextual Notes

Some calculations rely on the properties of triangular numbers and their relationships, which may introduce assumptions about the arrangement and indexing of numbers that are not explicitly stated.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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.
 
Mathematics news on Phys.org
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)

መለሰ
 

Similar threads

MHB How can I get equal columns and rows from a total?
  • · Replies 1 ·
Replies
1
Views
1K
MHB -c5.LCM and Prime Factorization of A,B,C
  • · Replies 3 ·
Replies
3
Views
1K
MHB Apc.1.1.13 what is the prime factorization of 30,030
  • · Replies 3 ·
Replies
3
Views
1K
Undergrad What is the missing number in this number sequence pattern?
  • · Replies 3 ·
Replies
3
Views
4K
Undergrad Math Myth: A prime is only divisible by 1 and itself
  • · Replies 4 ·
Replies
4
Views
2K
Graduate Is this Recursive Pattern in the Collatz Sequence Known?
  • · Replies 3 ·
Replies
3
Views
1K
MHB Number of natural numbers that have primitive roots
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad What is the role of geometry and dimensional operations?
  • · Replies 3 ·
Replies
3
Views
2K
MHB Find solutions in natural numbers
  • · Replies 3 ·
Replies
3
Views
2K
MHB -c6. Find the GCF and the LCF of A and B.
  • · Replies 1 ·
Replies
1
Views
1K