Problem 511

The first row contains the integers 1 through 6.

Each successive row contains the next six integers, in order, that follow the largest integer in the previous row.

Thus, the largest integer in any row is six times the row number.

Therefore, the largest integer in row 30 is $6\times 30=180$.
By a similar argument to part (a), it follows that the largest integer in row 2012 is $6\times 2012=12\,072$.

We find the other numbers in the row by counting backwards.

Thus, the six integers in row 2012 are $12\,072, 12\,071, 12\,070, 12\,069, 12\,068, 12\,067$.

The sum of the six integers in row 2012 is $$12\,072+12\,071+12\,070+12\,069+12\,068+12\,067=72\,417.$$
Again from part (a), the largest integer in any row is six times the row number.

Thus to find the approximate row in which 5000 appears, we divide 5000 by 6.

Since $\frac{5000}{6}=833\frac{1}{3}$, and $6\times833=4998$, then the largest integer in row 833 is 4998.

Therefore, row 834 contains the next six consecutive integers from 4999 to 5004.

(We can check this by recognizing that $6\times834=5004$.)

Thus, the integer 5000 appears in row 834.

Next, we recognize that all even numbered rows list the largest integer in the row beginning in column A through to the smallest integer in column F.

Since row 834 is an even numbered row, then the integers are listed in the order

$5004,5003,5002,5001,5000,4999$, with 5004 beginning in column A.

Therefore, the integer 5000 appears in row 834, column E.
The largest integer in row $r$ is $6\times r$ or $6r$.

Since each row contains six consecutive integers, counting backwards the remaining five integers in the row are, $6r-1,6r-2,6r-3,6r-4,6r-5$.

Thus, the sum of the six integers in row $r$ is $$6r+(6r-1)+(6r-2)+(6r-3)+(6r-4)+(6r-5)=36r-15.$$ Since we require the sum of the six integers in the row to be greater than 10 000,

then $36r-15>10\,000$ or $36r>10\,015$ or $r>\frac{10\,015}{36}$, and so $r>278\frac{7}{36}$.

But the row number $r$ must be a whole number, so $r\geq279$.

Since we also require the sum of the six integers in the row to be less than 20 000,

then $36r-15<20\,000$ or $36r<20\,015$ or $r<\frac{20\,015}{36}$, and so $r<555\frac{35}{36}$.

But the row number $r$ must be a whole number, so $r\leq555$.

Therefore, the rows in which the six integers have a sum greater than 10 000 and less than 20 000 are $279,280,281, \dots,555$.

This gives $555-279+1$ or 277 rows that satisfy the requirements.