At Galbraith H.S., the lockers are arranged in banks of 20 lockers. Each bank of lockers consists of six columns of lockers; the first two columns in each bank consist of two larger lockers and the last four columns in each bank consist of four smaller lockers. The lockers are numbered consecutively starting at 1 , moving down each column and then down the next column, and so on. The first twenty-one lockers and their locker numbers are shown in the diagram.

| 1 | 3 | 5 | 9 | 13 | 17 |
| :--- | :--- | ---: | ---: | ---: | ---: |
| 2 | 6 | 10 | 14 | 18 |  |
| 2 | 7 | 11 | 15 | 19 |  |
|  | 8 | 12 | 16 | 20 |  |

(a) What is the sum of the locker numbers of the column of lockers that contain the number 24 ?

(b) The sum of the locker numbers for one column is 123 . What are the locker numbers in this column?

(c) The sum of the locker numbers for another column is 538. What are the locker numbers in this column?

(d) Explain why 2013 cannot be a sum of any column of locker numbers.
(a) In the second bank of lockers, each of the first two columns consists of two lockers.

The first column consists of lockers 21 and 22, and the second column consists of lockers 23 and 24 .

Thus, the sum of the locker numbers in the column containing locker 24 is $23+24=47$.

(b) Suppose that a column contains two lockers, numbered $x$ and $x+1$, for some positive integer $x$.

Then the sum of the locker numbers in this column is $x+(x+1)=2 x+1$, which is odd. Suppose that a column contains four lockers, numbered $y, y+1, y+2$, and $y+3$, for some positive integer $y$.

Then the sum of the locker numbers in this column is

$$
y+(y+1)+(y+2)+(y+3)=4 y+6=2(2 y+3)
$$

which is even.

Since 123 is odd, then there must be two lockers in the column which has locker numbers that add to 123 .

If these locker numbers are $x$ and $x+1$, then we need to solve $2 x+1=123$, which gives $2 x=122$ or $x=61$.

Therefore, the locker numbers in this column are 61 and 62 .

(Note that since there are 20 lockers in each bank, then locker 61 is the first locker in the fourth bank, so does start a column containing two lockers.)

(c) Since 538 is even, then from (b), it must be the sum of four locker numbers.

If these numbers are $y, y+1, y+2$, and $y+3$, then we need to solve $4 y+6=538$, which gives $4 y=532$ or $y=133$.

Therefore, the locker numbers are 133, 134,135, 136.

(Note that these four numbers do appear together in one column, since locker 140 ends the seventh bank of lockers, so lockers 137, 138, 139, 140 form the last column, giving the four lockers 133, 134, 135, 136 as the previous column.)

(d) Since 2013 is odd, then if it were the sum of locker numbers in a column, it would be the sum of two locker numbers, as we saw in (b).

If these two locker numbers are $x$ and $x+1$, then $2 x+1=2013$ or $2 x=2012$, and so $x=1006$.

This would mean that the locker numbers were 1006 and 1007.

But the first locker number in each column is odd. This is because there is an even number of lockers in each column and so the last locker number in each column is even, making the first locker number in the next column odd.

Therefore, lockers 1006 and 1007 do not form a column of two lockers, so there is no column whose locker numbers have a sum of 2013.