Choose any three by three square of dates on a calendar page.
Circle any number on the top row, put a line through the other
numbers that are in the same row and column as your circled number.
Repeat this for a number of your choice from the second row. You
should now have just one number left on the bottom row, circle it.
Find the total for the three numbers circled. Compare this total
with the number in the centre of the square. What do you find? Can
you explain why this happens?
Make a set of numbers that use all the digits from 1 to 9, once and
once only. Add them up. The result is divisible by 9. Add each of
the digits in the new number. What is their sum? Now try some other
possibilities for yourself!
What happens to the perimeter of triangle ABC as the two smaller
circles change size and roll around inside the bigger circle?
Originally no correct solutions were received for this problem
in the month after its publication. It is a nice one to think about
and is not too taxing if you know just a little about the
properties of opposite numbers on a die.
Almost simultaneously, solutions to this toughnut were received
from Cameron who is home-schooled (whose solution is given below)
and Joel of Raffles Junior College.
Three dice cannot be arranged in any way that makes the sum of
the top dots equal the sums of the front, back, and bottom
There are six faces on a die: 1, 2, 3, 4, 5, and 6. These total
Three dice have a total of 63 dots. When three dice are arranged in
a line, and we are concerned with only the top, bottom, front, and
back sides, the end faces and hidden faces do not count.
If we were to pull apart the row of dice so we could see the
hidden faces, we would see that each hidden face is opposite from
either another hidden face or an end.
Opposite faces of a die always total 7, so the two opposite
faces of each die that we aren't concerned with total 7 each,
The remaining dots (top, front, bottom, back) equal 63 (total of
all faces) - 21 (total of faces we aren't dealing with) = 42.
Since 1/4 of these dots must be on the front, 1/4 on the back,
1/4 on the top, and 1/4 on the bottom, we can divide 42 by 4 to see
how many dots are facing one direction. 42 ÷ 4 = 10.5, which
cannot make a solution because there are no half dots on regular
Therefore 3 dice will never satisfy the requirements.
Using the same formula (n = number of dice, t = total: 14n
÷ 4 = t, t must be an integer), only even numbers of dice can
be arranged so it is possible that sum of dots on top faces = sum
of dots on bottom faces = sum of dots on front faces = sum of dots
on back faces.