Find 2016

Algebra Level 3

T 1 T 3 T 6 T 10 T 15 T 21 T 2 T 5 T 9 T 14 T 20 T 27 T 4 T 8 T 13 T 19 T 26 T 34 T 7 T 12 T 18 T 25 T 33 T 42 T 11 T 17 T 24 T 32 T 41 T 51 T 16 T 23 T 31 T 40 T 50 T 61 \begin{array} { c c c c c c c } T_1 & T_3 & T_6 & T_{10} & T_{15} & T_{21} & \cdots \\ T_2 & T_5 & T_9 & T_{14} & T_{20} & T_{27} & \cdots \\ T_4 & T_8 & T_{13} & T_{19} & T_{26} & T_{34} & \cdots \\ T_7 & T_{12} & T_{18} & T_{25} & T_{33} & T_{42} &\cdots \\ T_{11} & T_{17} & T_{24} & T_{32} & T_{41} & T_{51} & \cdots \\ T_{16} & T_{23} & T_{31} & T_{40} & T_{50} & T_{61} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array}

In the table above, T n = n ( n + 1 ) 2 T_n= \dfrac {n(n+1)}{2} is the n n th triangular number. If 2016 lies in c c th column and r r th row, find c + r c+r .

For example: T 9 = 45 T_9=45 lies in column 3 and row 2, hence the answer is 2 + 3 = 5 2+3=5 .


The answer is 12.

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1 solution

It's easy to solve that n=63,so we are looking for T 63 T_{63} . Looking at the figure it means that we have to move 2 columns right and 2 rows up from T 61 T_{61} . This lies on row 6 and column 5, so we need to move to row 4 column 8 which gives the answer 8+4=12.

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