[regfortruegoo]

Does 1 (1sf) + 1(1sf) = 2 (0dp) ?

[bobibnoxx]

Congrats. New(winning) thread.

[MilenaJaf]

We need a vote, DR. Do you think?

[Heopretg2006]

Originally posted by civman2000
Here's a fun little math challenge: Find uncountably many sets of integers such that the intersections of any two of them is finite. Does this work?

Consider

{a_1, a_2, a_3, ... : 0

[aceriscoolon]

Assuming that my previous reply is correct, here's another challenge:

Given any integer N, show that some integer multiple of N consists entirely of the digits 0 and 1.

For example,

2 x 5 = 10
3 x 37 = 111
4 x 25 = 100

etc.

[Ikrleprl]

Originally posted by Petek


Does this work?

Consider

{a_1, a_2, a_3, ... : 0

[Tam04xa]

If it stopped making sense after that then you probably have a decent enough math background...

[55TRATTERENRY]

I don't understand your objection to my solution.

There are uncountably many real numbers between 0 and 1. Each such number corresponds to a decimal expansion of the form 0.a_1 a_2 a_3 ... , where each a_i is an integer equal to 0, 1, 2, ..., or 9. (There is a slight ambiguity whether the decimal expansion is finite, or ends with 999..., but this doesn't affect the argument).

Thus, there are uncountably many sets of integers {a_1, a_2, a_3, ...} where each a_i is one of the integers 0, 1, ..., 9. Any two such sets have a finite intersection since each set consists of a subset of {0, 1, ..., 9}.

Petek

[gortusbig]

{0,1,1}={0,1} in terms of sets.

[Babposa]

Now, technically you could raise the same objection to my solution, but it only applies when two sequences converge to rationals (say you have R1 = 1, R2 = 2, qn = 2,1,1,1,..., vn = 1,2,2,2,2,....). In this case the set {qn} and the set {vn} are identical (namely {1,2}). However, we can avoid this by making all of our sequences strictly monotonically increasing. This ensures that we never double count. You could also note that the irrationals are uncountable and simply restrict yourself to them. In either case, we force all sets {qRn} to always be infinite, with a finite intersection between any two sets, showing that {qR1n} dne {qR2n} (assuming R1 dne R2)

[ibiDb4uu]

Find uncountably many sets of integers such that the intersections of any two of them is finite.

[Bemapayople]

Originally posted by Petek
Assuming that my previous reply is correct, here's another challenge:

Given any integer N, show that some integer multiple of N consists entirely of the digits 0 and 1.

For example,

2 x 5 = 10
3 x 37 = 111
4 x 25 = 100

etc. Let x(N) be the number we're looking for.

Write N=N'*gcd (10,N); so gcd(N',10)=1.

Use Euler's totient theorem so that
10^(phi(N'))=1 (mod N')
So take x(N')=sum_{i=1}^N' (10^(phi(N'))i).
Clearly x(N') is divisible by N' and also of the required form.

Now take x(N)=x(N')*lcm(10,N).
Clearly x(N) is divisible by N and also of the required form.

[Enladalusange]

A somewhat more elegant (IMHO) solution to Petek's problem: consider the numbers 1, 11, 111, 1111, etc mod n. By the pigeonhole principle, two of them must be congruent mod n; the difference is then a multiple of n consisting of 0s and 1s.