Author: Emily Fleming

Secret Santa

santa

The rules of a secret santa are that each person’s name is put in a hat and the names are mixed. Then each person must choose 1 name from the hat. If you choose your own name, you must put it back in the hat.

If 2 people do a secret santa there is only one solution: Person A gives to person B and person B gives to person A.

Stage 1: With 3 people, there are 2 possible ways. Can you think why?

Stage 2: Now how many different ways are there with 4 people?

The final challenge is to find the number of different scenarios with 5 people.

Solutions can be found here.

Merry Christmas from SAMI!

SAMI (Supporting African Maths Initiatives) supports mathematics work across Africa, including coordinating volunteers and resources to help run yearly maths camps in Kenya, Ghana, Ethiopia and Tanzania, as well as weekly maths clubs in over 70 schools. SAMI has also helped to raise money and lend expertise to developing computer programs for statistics and education, as well as providing tablets and training to help improve livelihoods for people living in rural African villages. With some extra funding SAMI could extend their projects to more schools and communities in more countries, and travel to the most isolated areas where their support is needed most.

See here for more information about SAMI. Donations are very welcome here.

Enjoy the puzzles, Merry Christmas.

Christmas circle

Children are sitting in a circle, and a teacher walks around the circle, giving presents. The teacher gives a blue present to every second child, and a green present every third child.

After going around four times, the sixth child from where the teacher started has two blue present and two green presents.

What is the fewest amount of children could there be?

Solution is here.

Christmas circle solution

The teacher goes round the circle four times. If there was an even number of children, the sixth child would always receive a blue present and would have four blue presents at the end. Since they only have two blue presents, there must be an odd number of children in the circle.

With the same logic for the green presents the answer can’t be a multiple of 3.

The sixth child receives a green present in the first round because 6 is a multiple of 3. She only receives one more green present so the answer can’t be a multiple of 3.

The options are 7, 11, 13, 17, 19 … and so 7 is the fewest number of children.