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kaiser
 Post subject: Two Number Theory Problems
PostPosted: Tue, 29 Jul 2003 12:00:27 UTC 
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Hi there!
#1 Given that
133^5 + 110^5 + 84^5 + 27^5 = k^5,
with k an integer, find k.


#2 Find the smallest positive cube in 888.

I don't have any idea how to go about either of them. Please help.
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Soroban
 Post subject: Re: Two Number Theory Problems
PostPosted: Tue, 29 Jul 2003 12:44:30 UTC 
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Hello, kaiser!

The second problem is a strange one . . .

kaiser wrote:
#2 Find the smallest positive cube in 888.


I assume it's asking for a positive cube factor of 888 . . . but the smallest?

888 = 2^3 \times 3 \times 37

The largest positive (integer) cube is 2^3 = 8

However, the smallest positive cube is: 1^3 = 1
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Soarer
 Post subject:
PostPosted: Tue, 29 Jul 2003 13:20:20 UTC 
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no.1

obviously n>=134.
Notice that
n^5<133^5+110^5+(84+27)^5<3*133^5

and 3<3125/1024
Therefore
n<5/4*133<167

Notice that 133^5+110^5+84^5+27^5=4(mod 10)
therefore n=4(mod 10)
then n=134,144,154 or 164
133^5+110^5+84^5+27^5=0(mod 3)
n=144

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bugzpodder
 Post subject:
PostPosted: Tue, 29 Jul 2003 14:27:30 UTC 
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i remember #1 as being an AIME question. is that where you got it?

_________________
Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1
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kaiser
 Post subject:
PostPosted: Tue, 29 Jul 2003 15:55:07 UTC 
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I'm sorry #2 needs editing. Its:
Find the smallest positive cube ending in 888.

Yes Bugz, most of the problems I've been posting are from AIME. Do you where I can find more recent AIME problems. On kalvapage they have only 1983-1989 ones.
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kaiser
 Post subject:
PostPosted: Tue, 29 Jul 2003 15:57:53 UTC 
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I'm sorry #2 needs editing. Its:
Find the smallest positive cube ending in 888.

Yes Bugz, most of the problems I've been posting are from AIME. Do you where I can find more recent AIME problems. On kalvapage they have only 1983-1989 ones.

Thanks Soarer!
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Soarer
 Post subject:
PostPosted: Tue, 29 Jul 2003 17:35:58 UTC 
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no.2

let the no. be (10a+2)^3
=1000aaa+600aa+120a+8
To satisfy the condition, a must end with 4 or 9.
Let a=10b+4, then
=1000aaa+600(10b+4)^2+120(10b+4)+8
=1000aaa+600(100bb+80b+16)+1200b+488
=9600+1200b+488(mod 1000)
=10088+1200b (mod 1000)
so the smallest value of b is 4, i.e. a=44
442^3=86350888
Else a=10b+9, then
=1000aaa+600(10b+9)^2+120(10b+9)+8
=1000aaa+600(100b^2+180b+81)+1200b+1080+8
=48600+1200b+1088(mod 1000)
=49688+1200b(mod 1000)
so the smallets value of b is 1, i.e. a=19
192^3=7077888
so i think the ans is 192

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kaiser
 Post subject:
PostPosted: Tue, 29 Jul 2003 21:59:37 UTC 
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Thanks Soarer!
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bugzpodder
 Post subject:
PostPosted: Tue, 29 Jul 2003 23:09:22 UTC 
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i have an AIME kit from 60s all the way to now. i should go start do them ;)

_________________
Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1
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SilverSprite
 Post subject:
PostPosted: Wed, 30 Jul 2003 00:37:50 UTC 
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I wish i could do aime questions :| :( :cry: :oops:
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SilverSprite
 Post subject:
PostPosted: Wed, 30 Jul 2003 00:51:49 UTC 
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Soarer wrote:
no.1
...
Notice that 133^5+110^5+84^5+27^5=4(mod 10)
therefore n=4(mod 10)
...
133^5+110^5+84^5+27^5=0(mod 3)
...


How did you figure those out? I dont see how..
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SilverSprite
 Post subject:
PostPosted: Wed, 30 Jul 2003 00:56:54 UTC 
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Soarer can I ask you how long it took you to solve each question??
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Soarer
 Post subject:
PostPosted: Wed, 30 Jul 2003 07:46:31 UTC 
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SilverSprite wrote:
Soarer wrote:
no.1
...
Notice that 133^5+110^5+84^5+27^5=4(mod 10)
therefore n=4(mod 10)
...
133^5+110^5+84^5+27^5=0(mod 3)
...


How did you figure those out? I dont see how..


Notice that 133^5+110^5+84^5+27^5=4(mod 10)
therefore n=4(mod 10)
133^5=3^5(mod 10)=3(mod 10)
110^5=0(mod 10)
84^5=4^5=4(mod 10)
27^5=7^5=7(mod 10)
The sum=3+4+7(mod 10)=4(mod 10)
...
133^5+110^5+84^5+27^5=0(mod 3)
133^5=1^5=1(mod 3)
110^5=(-1)^5=-1(mod 3)
84^5=0(mod 3)
27^5=0(mod 3)
Therefore the sum=1-1=0(mod 3)

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SilverSprite
 Post subject:
PostPosted: Wed, 30 Jul 2003 16:06:14 UTC 
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0o0o0o0o... if you have x^y (mod z) its the same as (x mod z)^y (mod z)??? Thanx alot soarer.
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Soarer
 Post subject:
PostPosted: Wed, 30 Jul 2003 17:41:43 UTC 
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SilverSprite wrote:
0o0o0o0o... if you have x^y (mod z) its the same as (x mod z)^y (mod z)??? Thanx alot soarer.


yes i think so.
Say x=az+b
x=b(mod z)
x^y
=(az+b)^y
=z(a^yz^(y-1)+..........)+b^y
=b^y(mod z)

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