The last two digits of the number 41999 are 44. The fact that this problem focuses on the last two digits suggests we should use 'clock' or modulus arithmetic where we are only interested in the remainders after division by 100.

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
^{1991}is some multiple of 100 (say b) plus 4, i.e. 4^{1991}is congruent to 4 modulo 100, which is written 4^{1991}4 (mod 100). - 4
^{8}is some multiple of 100 (say B) plus 36, i.e. 4^{8}is congruent to 36 modulo 100, which is written 4^{8}36 (mod 100). - 4
^{1999}is 4^{1991}multiplied by 4^{8}, or (100b+4)(100B+36). - Hence 4
^{1999}is some multiple of 100, plus 4 times 36, giving 44 as the last two digits. - 4
^{1999}4^{1991}x 4^{8}4 x 4^{8}4^{9}44 (mod 100).