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?
Let the blue Flibbins be B1, B2, B3, B4 and B5, and the red
Flibbins be R1, R2, R3, R4 and R5, whereby B1 and R1 are a pair, B2
and R2 are a pair and so on so forth.
Firstly, as Betsy mentioned, 3 Flibbins should go on the first
trip. Thus R1, R2 and R3 set off on the first trip.
R3 then becomes the pilot, and returns to planet Filbert to
fetch R4 and R5. Therefore, R3, R4 and R5 reach the new planet
where R1, and R2 are in.
R5 then becomes the pilot, and returns to fetch B5.
Upon reaching, B5 cannot stay on the new planet with R1, R2, R3,
and R4 as B1, B2, B3 and B4 (respectively) will be jealous. Hence,
B5 becomes the pilot, and returns to planet Filbert to fetch B3 and
B3 and B4 arrive with B5. Thus, R1 and R2 have to leave on the
spacecraft, to avoid B1 and B2 from becoming jealous.
B1, R1 and B2 then set off for the new planet, leaving R2 behind
on planet Filbert. This is possible as B1 will not be jealous (as
he is accompanying R1), and neither will B2 be jealous (as there
are no blue Flibbins left on planet Filbert).
B2 then becomes the pilot, and returns to fetch R2.
Hence, in total, there are 11 trips, out of which 5 are return
Alternatively, on the 5th trip, R5 could also fetch another blue
Fibbin, other than B5. This is possible as B5 will not be jealous
(because it is accompanying R5). However, if say R5 also fetched
B4, on the 6th trip, both B4 and B5 will be pilots. Thus, this will
not change the least amount of the total number of trips made,
which is 11.
Joel of Raffles Junior College also gave a solution involving11
We have solved this puzzle and this is how. By sending three red
Flibbins on the 1st trip, three blue Flibbins on
the 2nd trip. On the 3rd trip we sent one red and one blue Flibbin
and again the same on the 4th. In total 4 trips (plus return
journeys) were needed.
1st= Red, Red, Red
2nd= Blue, Blue, Blue
3rd= Red, Blue
4th= Red, Blue