The Student Room Group
Reply 1
can the binary expansion start with a zero ?
Reply 2
the bear
can the binary expansion start with a zero ?


It doesn't say in the question. But I dont think it can.
501 x 4 = 2004

111110101 in binary = 501

501x3 = 1503

= 10111011111

Since the number of ones in each are coprime, there is a way to add the digits next to each other so that 2004 1's are in the binary number. (this is the same as sticking so many zeros on the end of whichever number, i.e multiplying by 2n , then adding the other number. )

The number of zeros in the number will be < 2002 , since with each number the difference between the number of 1's and 0's is greater than 2, so the difference between the amounts of each always grows, with the number of zeros being less. when you reach a number with 2004 1's in it, multiply it by 4 (i.e, adding 2 zeros on the end) this means you will have 4 x(certain amount of 501s) which is the same as a certain number of 2004s, meaning the number is a multiple of 2004 decimal.

Adding zeros to the end of this number will simple multiply it by 2n so you can add as many zeros as you like, prefarably, the amount needed to take the amount of zeros in the number up to 2004.


That isn't a great argument, but I hope it's right lol, spent a long time squirming....attempted it a while ago with absolutely no success
Reply 4
KAISER_MOLE
501 x 4 = 2004

111110101 in binary = 501

501x3 = 1503

= 10111011111

Since the number of ones in each are coprime, there is a way to add the digits next to each other so that 2004 1's are in the binary number. (this is the same as sticking so many zeros on the end of whichever number, i.e multiplying by 2n , then adding the other number. )

The number of zeros in the number will be < 2002 , since with each number the difference between the number of 1's and 0's is greater than 2, so the difference between the amounts of each always grows, with the number of zeros being less. when you reach a number with 2004 1's in it, multiply it by 4 (i.e, adding 2 zeros on the end) this means you will have 4 x(certain amount of 501s) which is the same as a certain number of 2004s, meaning the number is a multiple of 2004 decimal.

Adding zeros to the end of this number will simple multiply it by 2n so you can add as many zeros as you like, prefarably, the amount needed to take the amount of zeros in the number up to 2004.


That isn't a great argument, but I hope it's right lol, spent a long time squirming....attempted it a while ago with absolutely no success


I cant find any false in your argument. I guess it is right. :smile: thanx.



I got stuck on another one in the 2002/3 paper.

x.y.z positive real numbers. x^2 +y^2 +z^2 = 1

to prove: (x^2)yz + (y^2)xz +(z^2)xy <= 1/3
Reply 5
KAISER: I tried it and think I ended up doing the same thing as you.
edit: thanks to mathsexams, proof fixed :biggrin:

2004 = 111110101002

there's 7 ones and 4 zeroes there.

We want 2004 = 286x7 + 2 ones in our number. So, consider the number N, whose binary expansion is formed by repeating the string '11111010100' 286 times. We can write this as:

N = 2004 + 2004x211 + 2004x222 ... + 2004x2285x11

Clearly N = 0 mod 2004. The binary expansion of N contains 286 x 7 = 2002 ones, and 286 x 4 = 1144 zeroes (because it's that of 2004 repeated 286 times). So we need 2004-1144 = 860 more zeroes.

The last 13 digits of N are

0011111010100

(the 00 is from the previous repetition of 11111010100). Here we have 7 ones and 6 zeroes

Now add let M = 2x2004 + N. Consider the last 13 digits in the addition:

0011111010100 | N
0111110101000 | 2x2004
1011101111100 | N + 2x2004

Now there are 9 ones and 4 zeroes. So there are 2 more ones, and 2 less zeroes than in N. So M has 2002+2=2004 ones and 1144-2 = 1142 zeroes. Note that as M = N + 2004x2, and 2004 divides N, 2004 divides M also.

Consider M x 22004-1142. This has 2004 ones and now 2004 zeroes too. As it's a multiple of M, it divides by 2004. So we have found a number divisible by 2004 with exactly 2004 ones and 2004 zeroes.
Reply 6
pure6gapper

I got stuck on another one in the 2002/3 paper.

x.y.z positive real numbers. x^2 +y^2 +z^2 = 1

to prove: (x^2)yz + (y^2)xz +(z^2)xy <= 1/3
haha, this is the greatest question ever. let me try and type up an answer without checking...

3AM = x+y+z
GM^3 = xyz
3AM.GM^3 = (x^2)yz + (y^2)xz +(z^2)xy

AM<GM<RMS

RMS = &#8730;((x^2+y^2+z^2)/3) < &#8730;(1/3)

so AM<&#8730;(1/3), GM<&#8730;(1/3)

3AM.GM^3 < 3*&#8730;(1/3)&#8730;(1/3)&#8730;(1/3)&#8730;(1/3)

(x^2)yz + (y^2)xz +(z^2)xy <= 1/3

wooh!!!
Reply 7
pure6gapper
I cant find any false in your argument. I guess it is right. :smile: thanx.



I got stuck on another one in the 2002/3 paper.

x.y.z positive real numbers. x^2 +y^2 +z^2 = 1

to prove: (x^2)yz + (y^2)xz +(z^2)xy <= 1/3

https://nrich.maths.org/discus/messages/27/67438.html?1133106584 (near the bottom)
Reply 8
AM-RMS
(x+y+z)/3 <= sqrt[(x^2+y^2+z^2)/3] <= 1/sqrt(3)
=> (x+y+z) <= 3/sqrt(3)

Now note that
(x-y)^2 + (y-z)^2 + (z-x)^2 >= 0
=> xy + yz + zx <= x^2 + y^2 + z^2 = 1

HM-AM
3/(1/x + 1/y + 1/z) = 3xyz/(xy+yz+zx) <= (x+y+z)/3
=> xyz <= 1/[3sqrt(3)]

Hence
(x^2)yz + (y^2)xz +(z^2)xy = xyz(x+y+z) <= 1/[3sqrt(3)] * 3/sqrt(3) = 1/3

--------------

Meh.. It took me a good 10mins to post this because my internet is currently capped at 1KB/s. And finally after what seemed like a lifetime of loading I find that two solutions have been posted. Argh.. :p:
Reply 9
meathead

2004 = 111110101002

there's 6 ones and 4 zeroes there.


7 ones and 4 zeros? I think this breaks the rest of your proof :frown:
Reply 10
mathsexams
7 ones and 4 zeros? I think this breaks the rest of your proof :frown:


Ahhh, looks like my counting is on form as usual :redface: I've updated the proof, I think it's OK now. Thanks for the heads up :biggrin:
chewwy
haha, this is the greatest question ever. let me try and type up an answer without checking...

3AM = x+y+z
GM^3 = xyz
3AM.GM^3 = (x^2)yz + (y^2)xz +(z^2)xy

AM<GM<RMS

RMS = &#8730;((x^2+y^2+z^2)/3) < &#8730;(1/3)

so AM<&#8730;(1/3), GM<&#8730;(1/3)

3AM.GM^3 < 3*&#8730;(1/3)&#8730;(1/3)&#8730;(1/3)&#8730;(1/3)

(x^2)yz + (y^2)xz +(z^2)xy <= 1/3

wooh!!!

What do AM, GM and RMS mean?
Reply 12
arithmetic mean, geometric mean, root mean squared.
RMS can also be called QM I think (Quadratic mean) - It was only about 1 month ago I found out what AM-GM was ....saw a proof in one of the Handbooks which started

'If we apply Cauchy-Schwarz to the vectors, whilst using AM-GM....'

I was like, 'Meh?'