Calendar magic
Tonight's idea
Look at any month on a calendar. Step one day to the right and the date goes up by 1. Step one row down and it goes up by 7, because a week has 7 days.
Now box any 3 by 3 square of dates. Call the middle date \(m\). Every other date in the box is a fixed number of steps from the middle, so each one can be written using \(m\).
Add the nine dates a row at a time. The middle row is \((m - 1) + m + (m + 1) = 3m\): the \(-1\) and \(+1\) cancel. The top row is \(3m - 21\). The bottom row is \(3m + 21\). The 21s cancel too.
Sum of a 3 by 3 box of dates \(= 9m\), nine times the middle date.
The letter \(m\) stood for every possible middle date at once. So this is true for every box, in every month, and you never have to check another one.
Watch one: the trick
Ask a friend to box a 3 by 3 square of dates in secret and tell you only the sum. They say 153.
\(153 \div 9 = 17\), so the middle date is 17. The top-left corner is \(17 - 8 = 9\) and the bottom-right is \(17 + 8 = 25\). You can recite the whole box: 9, 10, 11, then 16, 17, 18, then 23, 24, 25.
Check: \(30 + 51 + 72 = 153\).
The 7 in a week also lets you find the weekday of any date in your head: mathdetours.com/demos/weekday.html.
Your turn
- A 3 by 3 box has middle date 12. Write out the nine dates. Add them the long way, and check that you get \(9 \times 12\).
- A friend's 3 by 3 box adds up to 171. What is the middle date? What are the smallest and largest dates in the box?
- (a) Three dates directly under one another add up to 48. Use \(m\) for the middle one to find all three. (b) Use the letters in the picture to show that each diagonal of a 3 by 3 box adds up to \(3m\).
- A 2 by 2 box has no middle, so call its top-left date \(d\). (a) Write the other three dates using \(d\), and show the sum is \(4d + 16\). (b) A 2 by 2 box adds up to 76. Find its four dates. (c) Can a 2 by 2 box add up to 90? Explain.
Stretch: a bigger box
A 4 by 4 box of dates has top-left date \(d\). Write the sum of its sixteen dates using \(d\). Then show that the sum is always eight times the sum of two opposite corners.
Answers
- 4, 5, 6, then 11, 12, 13, then 18, 19, 20. Rows: \(15 + 36 + 57 = 108\), and \(9 \times 12 = 108\).
- \(171 \div 9 = 19\). Smallest: \(19 - 8 = 11\). Largest: \(19 + 8 = 27\).
- (a) The dates are \(m - 7\), \(m\), \(m + 7\), which add to \(3m\). So \(3m = 48\), \(m = 16\), and the dates are 9, 16, 23. (b) One diagonal is \((m - 8) + m + (m + 8) = 3m\). The other is \((m - 6) + m + (m + 6) = 3m\). The numbers cancel each time.
- (a) \(d\), \(d + 1\), \(d + 7\), \(d + 8\). Four \(d\)s and \(1 + 7 + 8 = 16\), so \(4d + 16\). (b) \(4d + 16 = 76\), so \(4d = 60\) and \(d = 15\). The dates are 15, 16, 22, 23. (c) No. \(4d + 16 = 90\) needs \(4d = 74\), and \(d = 18.5\) is not a date. The sum \(4d + 16 = 4(d + 4)\) is always a multiple of 4, and 90 is not.
Stretch. The rows start at \(d\), \(d + 7\), \(d + 14\), \(d + 21\), and each row is its start four times plus \(0 + 1 + 2 + 3 = 6\). Sum: \(16d + 4 \times (7 + 14 + 21) + 4 \times 6 = 16d + 168 + 24 = 16d + 192\). Opposite corners are \(d\) and \(d + 24\), which add to \(2d + 24\). Eight times that is \(16d + 192\). The other pair, \(d + 3\) and \(d + 21\), also adds to \(2d + 24\).