Calendar Trick: Uncover the Secret!

[soroban]

. . . . . . . Calendar TrickGive your volunteer an old calendar and a marker.

Have him/her select any month.

On that month, draw a square around
. . any 4-by-4 array of numbers.

You turn your back and instruct him/her:

1. Circle any number in the square.
. . X-out all numbers in its row and column.

2. Circle any remaining number in the square.
. . X-out all numbers in its row and column.

3. Circle any remaining number in the square.
. . X-out all numbers in its row and column.

4. Circle the remaining number in the square.

5. Add the four numbers.Without turning back, you announce the total.

Spoiler

Before turning away, note the smallest number
in the square, $$x$$, in the upper-left.

Add 12 and multiply by 4.

The total is $$4(x+12).$$Alternate method:

Add the smallest and largest numbers,
and multiply by 2.

Can you explain why this works?

 

[Deveno]

As long as you know the upper-left hand number ($x$), we know we have one from each row, so that the number is at least:

$x + (x+7) + (x+14) + (x + 21) = 4x + 42$.

By our method of choice, we have have exactly one number from each column, as well, so the number in the first column contributes nothing additional to our minimum sum, the number from column 2 contributes an additional 1, the number from column 3 contributes an additional 2, and the number from the last column contributes an additional 3 (since these number are either:

$x,x+1,x+2,x+3$
$x+7,(x+7)+1,(x+7)+2,(x+7)+3$
$x+14,(x+14)+1,(x+14)+2,(x+14)+3$
$x+21,(x+21)+1,(x+21)+2,(x+21)+3$

respectively).

Thus we have $4x + 42 + 1 + 2 + 3 = 4(x+12)$ as our total.

 

[soroban]

Well done, Deveno!I thought I would explain the result of the instructions.

Consider the followng addition table:

. . $$\begin{array}{c|cccc|}<br /> <br /> & 0 & 1 & 2 & 3 \\ \hline <br /> x & x & x+1 & x+2 & x+3 \\<br /> x+7 & x+7 & x+8 & x+9 & x+10 \\<br /> x+14 & x+14 & x+15 & x+16 & x+17 \\<br /> x+21 & x+21 & x+22 & x+23 & x+25 \\ \hline \end{array}$$

The instructions force us to select four numbers whose sum
is the sum of the eight numbers on the perimeter.