Positive integers are put into the following table. \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|} \hline 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & & \\ \hline 2 & 5 & 9 & 14 & 20 & 27 & 35 & 44 & & \\ \hline 4 & 8 & 13 & 19 & 26 & 34 & 43 & 53 & & \\ \hline 7 & 12 & 18 & 25 & 33 & 42 & & & & \\ \hline 11 & 17 & 24 & 32 & 41 & & & & & \\ \hline 16 & 23 & & & & & & & & \\ \hline ... & & & & & & & & & \\ \hline ... & & & & & & & & & \\ \hline \end{tabular} Find the number of the line and column where the number $2015$ stays.
Problem
Source: JBMO Shortlist 2015 C3
Tags: combinatorics, positive integers, Array, table
tait1k27
24.04.2019 16:14
$S(a(line),b(column))=\frac{b(b+1)}{2}+(b+(b+1)+...+(b+a-2))=\frac{b(b+1)}{2}+b(a-1)+\frac{(a-2)(a-1)}{2}=\frac{(a+b)^2-(3a+b)+2}{2}$
Electro__Wizard
24.04.2019 20:13
Observe That the first column $n$th row contains the $n$th triangular number .We simply need to find $n$ such that $\frac{(n)(n+1)}{2} \le 2015 \le \frac{(n+1)(n+2)}{2}$. By taking the difference along the diagonal we can find the exact position of the number.
mszew
24.04.2019 22:30
Notice that $(63 \cdot 64)/2=2016$
$f(n,1)=\frac{n(n+1)}{2}$
$f(n-k,1+k)=f(n,1)-k$
Then $\boxed{f(62,2)=2015}$
Steve12345
24.04.2019 22:39
Wait a minute... https://artofproblemsolving.com/community/c6h239180p1316339 copying problems from other JBMO Shortlists...
PEKKA
28.07.2023 21:59
We notice that $\frac{k(k+1)}{2}$ lies on the $k$th column and first row. After a bit of computation, we see that $2016=\frac{63(64)}{2}$ so 2016 is on row 1, column 63. Based on the problem statement, we conclude that 2015 is 1 column to the left and 1 row below 2016, so it is in column 62, row 2.