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We had a number of solutions to this problem and I would like to mention Karan of Beecroft Primary school and Ross of the Blue Coat School, who both managed to get the first part right but did not quite get to grips with the second.
Below is a very good solution sent by Andrei of School 205 Bucharest. Well done Andrei.
As in the first case I have to deal with a super increasing series, for each coded number there is only one corresponding letter:
Code 
Binary 
Letter 
33 
01011 
K 
18 
01110 
N 
20 
00001 
A 
1 
10000 
P 
31 
10011 
S 
20 
00001 
A 
30 
00011 
C 
33 
01011 
K 
Code 
Sum 
Binary 
Letter 
1 
1 
10000 
p 
5 
2+3 
01100 
l 

4+1 
10010 
r 

5 
00001 
a 
14 
2+3+4+5 
01111 
o 
4 
1+3 
10100 
t 

4 
00010 
b 
5 
2+3 
01100 
l 

4+1 
10010 
r 

5 
00001 
a 
8 
1+2+5 
11001 
y 

1+3+4 
10110 
v 

3+5 
00101 
e 
10 
1+4+5 
10011 
s 

1+2+3+4 
11110 
 

2+3+5 
01101 
m 
5 
2+3 
01100 
l 

4+1 
10010 
r 

5 
00001 
a 
4 
1+3 
10100 
t 

4 
00010 
b 
7 
1+2+4 
111 
z 

3+4 
00110 
f 

2+5 
01001 
i 
9 
1+3+5 
10101 
u 

4+5 
00011 
c 
There are many possibilities to make a word. But I need to find one which has sense. The possibilities are written in the following table:
p 
l 
o 
t 
l 
y 
s 
l 
t 
z 
u 

r 

b 
r 
v 
m 
r 
b 
f 
c 

a 


a 
e 

a 

i 

The only solution that I can find is: "problematic".
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}1 is divisible by 100.