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1 2 3 4 5 6 7 8 9 1 0 ⋮ We note that the numbers in the each row is equal to rowth number and every row ends up with sum of r t h rows.
Let the r j be the any row and 2 0 1 0 should also fall in any of the r j row number. Then we note that r = 1 ∑ j − 1 r < 2 0 1 0 ≤ r = 1 ∑ j r 1 9 5 3 < 2 0 1 0 ≤ 2 0 1 6 r = 1 ∑ 6 3 − 1 r < 2 0 1 0 ≤ r = 1 ∑ 6 3 r Shows that 2 0 1 0 falls in 6 3 r d row which has 63 distinct number from 1954 to 2016 and 2010 is at 7th position before from back ( 2016 ) or 56th position after from 1954.
Now the number exactly below the 2010 is in the 64th row at the 56th position after from 2017 and 7th position back from 2080 .
Therefore the number just below 2010 will be 2 0 8 0 − 7 = 2 0 7 3 .