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 Grammar School.
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
0
c
< 99.
The first few terms in the cyclic pattern are:
| Power of 4 ( a ) | Result | b | c |
|---|---|---|---|
| 1 | 4 | 0 | 4 |
| 2 | 16 | 0 | 16 |
| 3 | 64 | 0 | 64 |
| 4 | 256 | 2 | 56 |
| 5 | 1024 | 10 | 24 |
| 6 | 4096 | 40 | 96 |
| 7 | 16384 | 163 | 84 |
| 8 | 65536 | 655 | 36 |
| 9 | 262144 | 2621 | 44 |
| 10 | 1048576 | 10485 | 76 |
| 11 | 4194304 | 41943 | 4 |
| 12 | 16777216 | 167772 | 16 |
| 13 | 67108864 | 671088 | 64 |
| 14 | 268435456 | 2684354 | 56 |
| ... | ... | ||
| 10k | 76 | ||
| 10k+1 | 4 |
The steps in the argument, given in words and in the language of modulus arithmetic, are:
4 (mod 100).
36 (mod 100).
4 1991 x 4 8
4 x 4 8
4
9
44 (mod 100).
4 (mod 100) which can be done using the axiom of
mathematical induction or by other methods. It also requires the
use of some simple algebra to show that 4 1999
4 9 (mod 100) as outlined
in the bullet points above. Some students may like to write a full
account of this argument for publication in the NRICH
'Inspirations' section.