Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n $ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
What is the least square number which commences with six two's?
What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?
Well done all of you who explained carefully
how you found the pattern in the last two digits of powers of 4 and
decided that the answer must be 44. Congratulations to the
following people for their solutions: Claire and Rhona of Madras
College, St Andrew's; Angela, Geoffrey, Rachel, David and James of
Hethersett High School, Norwich; Bithian and Guobin of The Chinese
High School, Singapore; and The Key Stage 3 Maths Club at Strabane
Proving that the pattern really does go on repeating itself
indefinitely amounts to looking at multiples of 100 plus the last
two digits, in other words, using arithmetic modulo 100.
We write 4 a = 100 b + c
where a , b and c are whole numbers and
The first few terms in the cyclic pattern are:
The steps in the argument, given in words and in the language of
modulus arithmetic, are: