(and not so short) problems

Seasonal Problems
A selection of short
(and not so short) problems
• This year, December 1st is on a Saturday, in
which year does this happen next?
• How many times this century will December 1st
fall on a Saturday?
• A perpetual calendar
consists of 2
numbered cubes and
3 cuboids with the
months on. How
should the cubes be
• The ancient Mayans thought there were 360 days in
a year.
• We have 365 days a year (366 in a leap year)
• If the first day of the calendar year (January 1st for
us) were to coincide on both calendars in 2013, how
many years until it will coincide again?
– Ignoring leap years and assuming there are 365
days in our year
– Accounting for the fact that we have leap years?
(assume 365.25 days in a year)
Some old favourites…
• In a class there are 30 pupils. Each pupil gives
a Christmas card to each of the other pupils in
the class. How many cards are sent in total?
• At a New Year’s Eve party there are 12 young
people who all shake hands with each other.
How many hand shakes are there in total?
• At a much larger gathering there are 50 people,
how many handshakes now?
Can you find a formula for this?
Some old favourites…
At a New Year’s Eve party there are some adults
who all shake hands with each other… except
there are a small (unfriendly) group who all refuse
to shake hands with each other, but will shake
hands with everyone else.
There are 135 handshakes in total. How many
people are at the party? How many are there in
the ‘unfriendly’ group?
How many different answers can you find for this?
Super Santa
This year Santa has to deliver to all
the young people of the world, roughly
2 billion under 18’s in total.
Assuming, on average, that there are 3.5 young people per
household, how long does he have for each household to
get down the chimney, eat the mince pie, grab the carrot
for the reindeer and fill the stockings with goodies?
(Because of the earth’s rotation he probably has about 31
hours available).
…and how many households is that per second or per
minute? (whichever seems appropriate)
Class Calendar
A class of 25 pupils have an advent calendar. The first
pupil decides to open all the windows on the calendar. The
second pupil goes and closes all of the windows that are a
multiple of 2. The third pupil changes all the multiples of 3
– if they are open then she closes them, if they are closed
then she opens them. The fourth pupil changes all the
multiples of 4, if they are open then he closes them, if they
are closed then he opens them. The fifth pupil changes the
multiples of 5 and so on until the 25th pupil changes the
multiples of 25.
When the teacher arrives – which windows on the calendar
are open?
Can you explain why?
Cake Dilemma
Mrs Claus was delighted to find an unusual shaped
Christmas cake in her local supermarket; a regular
hexagonal one instead of the usual circular or square
ones. She bought one immediately and took it home
to show to Santa.
Santa also loved the new cake shape but pointed out
one small problem… “There are 5 of us eating and
we’ll need to share the cake equally, how are you
going to cut it up to give 5 equal portions?”
(the same volume)
“Easy”, said Mrs Claus…
How did she do it?
Answers & notes
Calendar problems
• December 1st is next on a Saturday in 2018
How many Saturday, December 1st are there this century? Because of leap years, the
answer is not ‘simply divide by 7’.
The day of the week that any particular date falls on runs in a 28 year cycle, where
each day of the week will occur 4 times. Hence beginning with there being a Saturday
December 1st this year: 2012 to 2039 (4), 2040 to 2067 (4), 2068 to 2095 (4). 2096 is
the start of the next 28 year cycle (another Saturday) and then think about 2000 to
2012. (2001 and 2007 had a Saturday Dec 1st).
Perpetual calendar: 1, 2, 3, 4, 5, 6 and 0, 1, 2, 7, 8, 9 works. Both cubes must
have 1 and 2 on, the 3 and 0 must be on different cubes, the others can be on either.
Mayan Calendar: If we assume there are 365 days in our year then we are looking for
the lowest common multiple of 360 and 365, which will tell us how many days it will
be until they coincide. This is 26280 days… which is 72 of our years… so 2085
Answers & notes
Calendar problems
Mayan Calendar
• If we have 365 days a year and the Mayans have 360, then one way to solve the
problem is to find the Lowest Common Multiple (LCM) of 365 and 360. This can be
done using factor trees to identify the prime factors for each number and then finding
the Highest Common Factor.
LCM = a x b ÷ HCF
• The answer is 365 x 360 ÷ 5 = 26280 (days) then 26280 ÷ 365 = 72 (years)
When accounting for leap years, using the fact that we actually have 365.25 days a
year, a similar method can be used, but both values need to be multiplied by 4 to
obtain integer values initially to be able to use factor trees (whilst maintaining the
correct ratio), and then dividing the answer by 4 again at the end. (To understand this
it may help to consider what happens with smaller values such as finding the LCM of
1.25 and 3)
365.25 and 360 multiplied by 4 become 1461 and 1440
The LCM of 1461 and 1440 is 701280
Divide this by 4 701280÷ 4 = 175320 (days)
Then 175320 ÷365.25 = 480 (years)
Answers & notes
Old Favourites
• Class cards: Each person sends to 29 people, so 30 x 29 = 870
Handshakes: this is a way into triangle numbers and follows on from the question
above. Each person must shake every other person’s hand. 12 x 11 = 132,
however, the difference here is that if A has shaken B’s hand and then B has shaken
A’s hand, they would have shaken hands twice, so the answer needs to be divided by
2. 132 ÷ 2 = 66. pupils might arrive at this answer through experimentation, or by
thinking about a smaller group of people and searching for a pattern in the numbers
or they may be able to go straight to the calculation.
Pupils might be encouraged towards finding a formula for triangle numbers by
thinking about a very large group of pupils, the example given is 50 people. For this
the calculation is 50 x 49 ÷ 2 = 1225
The formula for the number of handshakes for n people (triangle numbers) is
ℎℎ =
Answers & notes
Old Favourites
Adults shaking hands.
This problem continues the triangle numbers theme, but with a less obvious method of
One way to solve it is to consider the first 20 triangle numbers which are given on the
next slide. This could be displayed to pupils, but it would spoil the opportunity for pupils
to think if it is shown too early on.
To find an answer for the numbers of people at the party and in the ‘unfriendly’ group
simply search for a pair of triangle numbers with a difference of 135.
There are 3 possible answers for this:
• 17 at the party, with 2 who refuse to shake each others’ hand
• 19 at the party, with 9 who all refuse to shake each others’ hands
• 20 at the party, with 11 who all refuse to shake each others’ hands … which doesn’t
sound like much of a fun party to be at!
Handshakes for a certain
number of people
Answers & notes
Super Santa
The first possible issue with this problem is knowing how to write a billion: 1 x 109
or 1 000 000 000
Rounding answers to different degrees of accuracy will give slightly different answers:
He has to deliver to 2 000 000 000 ÷ 3.5 = 285 714 286 households
He has 31 x 60 x 60 = 111 600 seconds to do this
So he has 111 600 ÷ 285 714 286 = 0.0004 seconds to deliver to each household
Or 285 714 286 ÷ 111 600 = 2560 households per second
Or 2560 x 60 = 153 600 households per minute
You might like to pose some ‘localised’ questions, (or ask members of the class to pose
some) such as how long Santa will take to deliver to all the pupils in school, or the local
village, town or city, or the UK.
Answers & notes
Class Calendar
This problem is adapted from a ‘prison door’ problem.
Each door is initially opened by child 1 and each is then subsequently opened or closed
by any child who’s ‘number’ is a factor of it.
The cycle clearly alternates:
open close open close etc.
Any number which has an even number of factors will end up closed; any number with an
odd number of factors will end up open.
Hence, calendar doors with square numbers end up as the only open ones.
Answers & notes
Cake Dilemma
For the problem given of dividing a hexagon into five equal sections,
Firstly divide
each side into
5 equal
Find the
Draw a
segment from
the centre to
any marked
Count round
the hexagon
marking every
sixth segment
Answers & notes
Cake Dilemma
Can you prove that the 5 segments are equal?
Consider each of the small triangles… what is the base and height of each?
Would this idea work for other
polygons and/or number of
required equal sections?
Dynamic geometry would be
helpful to demonstrate that the
sections have an equal area, but
also to show that if equal angles
are taken at the centre, this does
not result in sections of equal

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