Earlier today I set you three puzzles for 12yearolds, used by the charity Axiom Maths, whose mission is to help top performing children from low incomes continue achieving well throughout secondary school.
1. Backwards multiplication
What fourdigit number reverses itself when multiplied by 4? As in, what are the digits a, b, c and d such that the number abcd x 4 = dcba?
(In this problem, the letters a, b, c and d all stand for different digits.)
Solution 2178 x 4 = 8712
STEP 1 a can only be 1 or 2, because four times a number greater than 3000 will be greater than 12,000, so have five digits, not four;
STEP 2 The solution must be even (since all multiples of 4 are even) and so a must be 2.
STEP 3 We know that 4 x d is a number ending in 2. By working through the four times table we get to d = 3 or 8
STEP 4 d can only be 8 or 9, as 4 x 2 thousandandsomething is at least 8 thousand and something; so d is 8; so 4 x b (possibly plus a carry) is less than 10, otherwise d would have to be 9, so b = 1 or 2, and so must be 1 as it has to be different from a.
STEP 5. c x 4 + carry of 3 has ones digit 1, so c x 4 has units digit 8, so c can only be 2 or 7 and so must be 7 as it has to be different from a.
2. Really secret Santa
A group of nine secret agents: 001, 002, 003, 004, 005, 006, 007, 008 and 009 have organised a Secret Santa. The instructions are coded, to keep the donors secret.

Agent 001 gives a present to the agent who gives a present to agent 002

Agent 002 gives a present to the agent who gives a present to agent 003

Agent 003 gives a present to the agent who gives a present to agent 004

and so on, until

Agent 009 gives a present to the agent who gives a present to agent 001
Which agent will agent 007 get her present from?
Solution 002
The easiest way to do this is draw a circle and then fill it in.