Yeah that's the one.. lol

For number 4 - Is it √2 + 2 + √(2(2 +√2)) ? You may have it in a different form. It comes up to approximately 6.027. I think this value works, I'm not sure whether I'm able to prove that this is the minimum required though.

For number 3 - First, draw a circle with diameter 2 inside the circle with diameter 5, with the two circles being concentric. Then, divide the doughnut shaped area which remains into 8 equal pieces. Using the pigeonhole principle, at least 2 of the 10 dots will lie in one of the 9 areas.

Let R be the area covered by 1/8th of a doughnut. It is therefore sufficient to prove that the the furthest distance between 2 points inside R is less than 2. The arc of the circle with diameter 5 has length 1/8*pi*diameter = 1/8*pi*5 = 5/8*pi < 5/8*3.2 = 5*0.4 = 2. Also, the distance between an edge of the longer arc and the opposite edge of the shorter arc is equal to:

√[ (5/2 - 1/√2)^2 + (1/√2)^2 ]

= √[ (5/2 - √2/2)^2 + 1/2]

= √[ (25-10√2 + 2 + 2)/4 ]

= √ (29-10√2)/4)

Which is less than 2.

By making the inner circle *slightly* less than 2 instead of equal to 2, all the previous equations will still hold. Hence, 2 points out of 10 will have a distance less than 2.

Addendum: Proving that √[ (29-10√2)/4 ] is less than 2.

Assume it is true.

then,

√[ (29-10√2)/4 ] < 2

√(29-10√2)/2 < 2

√(29-10√2) < 4

29 - 10√2 < 16

13 < 10√2

169 < 200

Which is obviously true. Hence, the original assertion is also true.

I did this a while ago, but since noone has posted the answer:

q.1 Consider the numbers 1,10, 10^2 ... 10^(n^2).

There are n^2+1 numbers, and there are n congruences mod n. It thus

follows that there must be at least n numbers that share the same

congruence mod n (call it x mod n). Adding these numbers gives a

number that's nx mod n = 0 mod n. But this number is a summation of

distinct powers of ten and thus contains only 1's and 0's and is

divisible by n. QED

But doesn't it need to be a number that begins with a string of 1s and ends in a string of 0s - which is different from just containing 0s and 1s

Here's one way to do question 1.

There's a theorem of Euler's in number theory that if a and n are coprime [have no factors in common] then

a^f = 1 mod n

where f = phi(n)

and phi(n), Euler's function, is the number of k in the range 0< k <n which are coprime with n. Anyway, for us the important point is such an f exists.

Then if n has no factors of 2 or 5 we can say

10^f = 1 mod n

So n divides 10^f -1

and n/9 divides (10^f-1)/(10-1) = 1+10 + 10^2 + ... + 10^(f-1)

The RHS is some number that is a string of 1s when decimally expressed.

Almost there, apart from that awkward 9. One way around that is to multiply the RHS by 1 + 10^f + 10^2f + ... + 10^8f. So instead of f 1s we now have 9f 1s. A number is divisible by 9 if the sum of its digits are divisible by 9 - and so what we just multiplied by had a factor of 9 in it.

So n divides 1+ 10 + 10^2 + 10^3 + ... + 10^(9f-1)

In the general case though n will have factors of 2 (say M of these) and 5 (say N of these). If we put max{M,N} 0s at the end of the number, this will deal with those factors and we only need to divide the other factors of n into a string of 1s, as accomplished above.

Hmm. I realised that my previous guess was incorrect - my revised answer is now √ 2 +2 + pi.

q.1 Consider 10, 10^2 .. 10^(n+1).

Now Consider these numbers mod n, two must share the same congruence,

say 10^j and 10^k with j < k.

It thus follows 10^(j+i) mod n = 10^(k+i) mod n.

Now set i = k-j-1

it follows that (mod n) 10^j + 10^j+1 ... + 10^(k-1) = 10^k + 10^k+1 .. + 10^(2k-j-1) = x mod n

it follows similarly that 10^(2k-j)... + 10^(3k-j-1) = x mod n.

etc

Now add up n such sequences to get a number thats nx mod n = 0 mod n.

This number is a sum of consecutive powers of 10, and so consists of a series of 1's followed by a series of 0s.

Very imaginative - and nicer than my earlier proof for not having to rely on Euler's Theorem.