# Mathematics Extension QuestionWatch

This discussion is closed.
#1
Have a go at these questions (see attachement) especially number 4.
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14 years ago
#2
Question 4.

1Km + delta

Row a small distance in any direction, use the echo locator, see whether you are getting closer to, or not from the shore. By making the "small distance" as small as possible you can minimise delta.

That is the lateral thinking answer, though there is probally a more mathematical answer if you couldn't use the echo locator more than once. But I don't see why you couldn't logically .
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14 years ago
#3
call the number, in the form 11111...1110000...000
(10/9)*10^a - (10/9) + 10^b

Where there are a ones, and (a+b-1) zeroes on the end.

We must show that for every (10/9)*10^a - (10/9) + 10^b, there is an n such that

... erm... lol...

(10/9)*10^a - (10/9) + 10^b = n.k, where k is a positive integer...

er...
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14 years ago
#4
Surely for every n you must find an a and a b, not the other way around. Because for every a and b, there exists a number n (namely 1) that satisfies the property .
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#5
(Original post by AntiMagicMan)
Surely for every n you must find an a and a b, not the other way around. Because for every a and b, there exists a number n (namely 1) that satisfies the property .
true
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14 years ago
#6
Yeah that's the one.. lol
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14 years ago
#7
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.
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14 years ago
#8
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.
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14 years ago
#9
Oooohhh. Thought I had Q1 for a second there but then realized a fatal flaw in my proof. Then I moved to these lines:
Assume there exists a smallest integer n for which it doesn't hold. Therefore it must be true for n-1.
I don't know if it'll work...

BTW. From that, guess what my original poo proof was
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14 years ago
#10
Q2) What is f(2000) ?

I've done this question before , in the mathematical society thread. I refer you to here
f(n) = [2/n(n+1)].f(1)

f(1) = 2000,
f(2000) = [2/2000.2001].2000
f(2000) = 2/2001
=============
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14 years ago
#11
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
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14 years ago
#12
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
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14 years ago
#13
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.
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14 years ago
#14
(Original post by sephonline)
For number 4 - Is it √2 + 2 + √(2(2 +√2)) ?
What path do you have in mind of this length?

BTW liked your solution to 3.
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14 years ago
#15
Hmm. I realised that my previous guess was incorrect - my revised answer is now √ 2 +2 + pi.
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14 years ago
#16
(Original post by sephonline)
Hmm. I realised that my previous guess was incorrect - my revised answer is now √ 2 +2 + pi.
Yea, thats what i got too .

This was the path:
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14 years ago
#17
(Original post by RichE)
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
Hmm just noticed that, that complicates things a bit
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14 years ago
#18
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.
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14 years ago
#19
Very imaginative - and nicer than my earlier proof for not having to rely on Euler's Theorem.
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14 years ago
#20
(Original post by IntegralNeo)
Have a go at these questions (see attachement) especially number 4.
nicked from Imperial
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