### Like Powers

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.

### Two Many

What is the least square number which commences with six two's?

### Power Crazy

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?

# Diggits

##### Stage: 3 Challenge Level:

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).
To prove this rigorously needs a proof that 4 10k+1 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.