Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in 10 000! and 100 000! or even 1 000 000!
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
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.
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
30
00011
C
Sum
p
5
2+3
01100
l
4+1
10010
r
a
14
2+3+4+5
01111
o
4
1+3
10100
t
00010
b
8
1+2+5
11001
y
1+3+4
10110
v
3+5
00101
e
10
1+4+5
s
1+2+3+4
11110
-
2+3+5
01101
m
7
1+2+4
11-1-
z
3+4
00110
f
2+5
01001
i
9
1+3+5
10101
u
4+5
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:
The only solution that I can find is: "problematic".