Can You Help with These Integer Algebra Homework Questions?

[Matthollyw00d]

First Question:
Let N_n be the integer whose decimal expansion consists of n consecutive ones. For example, N₂=11 and N₇=1,111,111. Show that N_n|N_m iff n|m.

Second Question:
If (a,c)=1, prove that (a,bc)=(a,b).

On the second question I can see that it is true because a and c are relatively prime, I realize that ax₁+bx₂=1, but I'm having difficulty expressing it in a proof satisfactory way. I think I'm just over looking a fact somewhere in my text.

As for the first one, I'm not very certain as to where at all to start, so any and all help here would be appreciated.

Note that I'm not really wanting the full proof of either, more or less just a few helpful pointers or some key facts that are needed that I'm missing; something to get me started so I can get a better attempt at it.

If anything is unclear, I'll be happy to restate it in a more satisfactory format if possible.

Thanks all.
Thanks all.

 

[tiny-tim]

Hi Matthollyw00d!

You're making this far too complicated …

First Question:
Let N_n be the integer whose decimal expansion consists of n consecutive ones. For example, N₂=11 and N₇=1,111,111. Show that N_n|N_m iff n|m.

Second Question:
If (a,c)=1, prove that (a,bc)=(a,b).

For the first one, just divide … what is the remainder?

For the second, write a = np, b = nq, bc = nqc …

 

[Matthollyw00d]

Would this be sufficient for the first one:

Assume N_n|N_m, then (t)N_n=N_m, t≥1. For all m,n≥1, there exists q,r such that m=qn + r, 0≤r<n.
Next assume m does not divide n, therefore 1≤r<n.
Thus N_m=N_qn+r=10^r(s)N_n+N_r
Then, (t)N_n=N_qn+r=10^r(s)N_n+N_r
Thus, N_r=N_n(t-s10^r)=N_n(d), d≥1, which implies r≥n, which contradicts our hypothesis of r<n.
Thus N_n|N_m iff n|m.

 

[ramsey2879]

Would this be sufficient for the first one:

Assume N_n|N_m, then (t)N_n=N_m, t≥1. For all m,n≥1, there exists q,r such that m=qn + r, 0≤r<n.
Next assume m does not divide n, therefore 1≤r<n.
Thus N_m=N_qn+r=10^r(s)N_n+N_r
Then, (t)N_n=N_qn+r=10^r(s)N_n+N_r
Thus, N_r=N_n(t-s10^r)=N_n(d), d≥1, which implies r≥n, which contradicts our hypothesis of r<n.
Thus N_n|N_m iff n|m.

If you work out a couple of divisions as a grade school student is taught a simpler proof should be apparent. First try $$/N_6/N_2$$ to get 010101. Then Try $$N_8/N_4$$ to get 00010001. Now what would the result of $$N_20/N_5$$ be?

 

[Matthollyw00d]

If you work out a couple of divisions as a grade school student is taught a simpler proof should be apparent. First try $$/N_6/N_2$$ to get 010101. Then Try $$N_8/N_4$$ to get 00010001. Now what would the result of $$N_2_0/N_5$$ be?

00001000010000100001
But I'm not seeing a proof that is simpler than what I did unfortunately. =/

 

[ramsey2879]

00001000010000100001
But I'm not seeing a proof that is simpler than what I did unfortunately. =/

Assume 0 < r < n
$$m = tn + r$$
$$N_{m} = N_{m} - (10^{(t-1)*n + r})*N_{n} \mod N_n$$
$$N_{m} = N_{m-n} \mod N_{n}$$
$$N_{m} = N_{m-n} - (10^{t-2}*n + r})*N_{n} \mod N_n$$
$$0 = N_{m-2n} \mod N_{n}$$
...
$$0 = N_{m - tn} \mod N_n$$
$$0 = N_{r} \mod N_n$$
$$r = 0$$

 

[Matthollyw00d]

That would be why I didn't see it, I've yet to learn Modular Arithmetic.

 

[ramsey2879]

Assume 0 < r < n
$$m = tn + r$$
$$N_{m} = N_{m} - (10^{(t-1)*n + r})*N_{n} \mod N_n$$
$$N_{m} = N_{m-n} \mod N_{n}$$
$$N_{m} = N_{m-n} - (10^{t-2}*n + r})*N_{n} \mod N_n$$
$$0 = N_{m-2n} \mod N_{n}$$
...
$$0 = N_{m - tn} \mod N_n$$
$$0 = N_{r} \mod N_n$$
$$r = 0$$

redoing this
$$m = tn + r$$
If $$N_{n}|N_{m}$$ then $$N_{n}$$ also divides:
$$N_{m} - (10^{(t-1)*n + r})*N_{n}$$ which has the same remainder i.e., zero But that is $$N_{m-n}$$ since you just removed the first n $$1$$'s. This is simply going through the longhand division process as the divisor multiplied by $$10$$ to the appropiiate power removes the n most righthand $$1$$'s. When doing longhand division in this manner you are not changing the ultimate remainder so we can say that the new term $$N_{m-n} = N_{m} \mod N_n$$ (i.e. they have the same remainder if divided by $$N_n$$).
Likewise we continue the longhand division process by removing the next n most righthand $$1$$'s and so on until we removed t*n $$1$$'s. The resulting number comprises r $$1$$'s and has the same remainder as the original number. So we say $$N_{m} = N_{r} \mod N_n$$ iff $$m = r \mod n$$. Since n|m, r = zero is the only possible r less than n.