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Algebra Difficulty 2.1 Junior Prove it Canada

Consider the following arrangement of positive integers.

$1$ $2$ $4$ $5$ $7$ $9$ $10$ $12$ $14$ $16$ $\vdots$ The 1st row includes the odd integer $1$ , and the 2nd row includes the two even integers $2$ and $4$ . For $k\geq2$ , the $k$ th row begins with the integer that is one more than the last integer in the previous row, includes, in increasing order, $k$ consecutive odd integers when $k$ is odd, and includes, in increasing order, $k$ consecutive even integers when $k$ is even. A useful fact about this arrangement is that the integer in the $k$ th row and $k$ th position (that is, the last position in the $k$ th row) is $k^2$ . For example, $4^2=16$ and 16 is the integer in the 4th position of the 4th row. What is the average of the integers in the 5th row? Which row has the integer 145 in the 1st position? Determine the row and the position in which the integer $1598$ appears. The average of the integers in row $r$ is 241. Determine the value of $r$ .

Solution

The 5th row includes the integers 17, 19, 21, 23, and 25, and so
the average of the integers

in the 5th row is $\dfrac{17+19+21+23+25}{5}=21$ . The row that has the integer 145 in the 1st position must immediately follow the row that has 144 in the last position. Since $12^2=144$ , then the integer 144 is in the last position of the 12th row, and so 145 is in the first position of the 13th row. Since $40^2=1600$ , then the integer in the last position (the 40th position) of the 40th row is 1600. When moving from right to left along each row, the integers decrease by 2, and so the integer in the 39th position of the 40th row is $1600-2=1598$ . Solution 1 Moving from left to right along a row, the integers increase by a constant (namely 2), and so the average of the integers in a row is equal to the average of the integer in the first position of the row and the integer in the last position of the row. Can you see why this is true? Since $15^2=225$ , then the integer in the last position of the 15th row is 225, and so the integer in the first position of the 16th row is 226. Since $16^2=256$ , then the integer in the last position of the 16th row is 256. Thus, the average of the integers in the 16th row is $\dfrac{226+256}{2}=241$ , and so $r=16$ . Solution 2 Since $15^2 = 225$ , then each of the entries in the first 15 rows is at most 225. This means that the average of the entries in each row up to and including the 15th row must be at most 225. Since $16^2 = 256$ , then each of the entries in the rows after the 16th row is greater than 256. This means that the average of the entries in each row after the 16th must be greater than 256. This means $r$ must be greater than 15 and must be smaller than 17. In other words, $r=16$ . We can check that the entries in row 16 are $226, 228, 230, 232, 234, 236, 238, 240, 242, 244, 246, 248, 250, 252, 254, 256$ and the average of these integers
is indeed 241.

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