- Hi Every one.

I need to solve this problem.please help me.

Prove that for each prime number (that is not 2 or 5) and each

natural number N ,there is a power of P that ends with 000...001,

where the number of zeroes is N-1 - --- omidmazaheri <omidmazaheri@...> wrote:
> Hi Every one.

Look at the finite fields F_(2^N) and F_(5^N). What are the orders,

> I need to solve this problem.please help me.

>

> Prove that for each prime number (that is not 2 or 5) and each

> natural number N ,there is a power of P that ends with 000...001,

> where the number of zeroes is N-1

of the element p in the multiplicative groups of this field?

Call these orders o5 and o2. What is p^lcm(o2,o5) in this field?

Phil

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http://health.yahoo.com - --- "omidmazaheri" wrote:
> Hi Every one.

Wow, I solved this problem at a national maths contest when I was

> I need to solve this problem.please help me.

>

> Prove that for each prime number (that is not 2 or 5) and each

> natural number N ,there is a power of P that ends with 000...001,

> where the number of zeroes is N-1

about 13 y.o. My solution was something like this (it actually works

for every odd number that doesn't end in 5):

step 1) each number P of this kind ends in 1, 3, 7 or 9; then P^4

obviously ends in 1, let q=P^4

step 2) if q=r*10^s+1 (s>0) then q^2=2*t*10^s+1

proof: q=r*10^s+1 => q^2=2*(r^2*5*10^(s-1)+r)*10^s+1

let u=q^2

step 3) if u=2*t*10^s+1 then u^5=v*10^(s+1)+1

proof: let w=2*t*10^s; u=w+1 => u^5=w^5+5*w^4+10*w^3+10*w^2+5*w+1;

we note that w^2 and 5*w are divisible with 10^(s+1), therefore we

can write u^5=v*10^(s+1)+1 (v is integer)

step 4) from 2) and 3) we have: q=r*10^s+1 => q^10=v*10^(s+1)+1;

through induction we can prove that q^(10^N)=r_N*10^(s+N)+1 (where

r_N are integers) for every N>=0

step 5) replacing q=P^4 and s=1 we get P^(4*10^N)=r_N*10^(N+1)+1, so

P^(4*10^N) ends in N zeroes followed by an 1

final note: this is probably not the shortest solution, but it's an

elementary one

Adi - --- "S.R.Sudarshan Iyengar" <gayathrisr@...> wrote:
> >> Prove that for each prime number (that is not 2 or 5) and each

Just 'Phil', no need for 'Mr.' :-)

> >> natural number N ,there is a power of P that ends with

> 000...001,

> >> where the number of zeroes is N-1

>

> >Look at the finite fields F_(2^N) and F_(5^N). What are the

> orders,

> >of the element p in the multiplicative groups of this field?

> >Call these orders o5 and o2. What is p^lcm(o2,o5) in this field?

> Well Mr. Phil

> what do you mean by F_(2^N) do you mean that it is

Nope, I mean the finite field with p^N elements, where in this case

> the field

> containing all elements >=1 which are relatively prime to 2^N?

p=2 and p=5.

The statement, seen a few posts back, that p \in {1,3,7,9} (mod 10)

implies p^4 == 1 (mod 10) is a statement that Order(p) divides 4 for

such p in (Z/10Z)* . The statement that if q == 1 (mod p^n), then

q^10 == 1 (mod p^(n+1)) for p=2,5 effectively says that the order of

q in F_(2^(n+1)) divides 10 times the order of q in F_(2^n).

I think my hints get to the root of the problem, but the

demonstrative proof we saw was probably simpler as it didn't care

what the order actually was, it didn't matter - as long as you raise

a number by a multiple of its order you get the result you need.

(e.g. raising 10n+1 to the power 4 in order to get a number of the

form 10n+1 is a tad unnecessary, but it works.)

Phil

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