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I bet you can't figure this out...

Z

Z24

Senior member
It's a math question. It's pretty tough, but the answer doesn't require any crazy mathematics. Okay, here it is:

Show that for every positive integer k, there exists a positive integer n such that k|n, and every digit of n is either a 0 or a 3.

[note: k|n means that k divides n evenly with no remainder. it would be true to say that 3|6, but untrue to say that 4|6.]

Good luck. If you need clarification, just ask.
 
isnt adding 0 or 3 makes the number to be either odd or even...so whatever the numerator is , the denominator can be divided into cauess its even or odd...dunno what you want like a equation or something
 
Z24,

Looks to me that your example of k|n => 3|6 is a contradiction to the hypothesis. Therefore, since n=6 is neither a 0 or a 3, the hypothesis is false and cannot be proved to be true.

How's that hold up?

Later,
RonC 😉
 
Sorry lnguyen, but for k = 7, 3 * 10^7 is 30000000, which is not evenly divisible by 7.

Currently working on the theory that any number that is divisible by 3 has a multiplior that creates a number composed of 0's and 3's. Can't figure out the proof on that one. Might have to pull my Number Theory and Numerical Analysis books of the shelves. I am missing something simple here.
 
Check here. If someone can explain the notation these guys are using - N = 0 (mod 2^(k+1)) - we'll have the solution in no time 😉

Does every number which is not divisible by 5 have a multiple whose
only digits are 6 and 7?

Yes. My proof follows:

Claim 1: For every k, there is a k-digit number whose only digits
are 6 and 7, which is divisible by 2^k.

The proof is by induction. Suppose N is a k-digit number
satisfying the above condition. Then either N = 0 (mod 2^(k+1))
or N = 2^k (mod 2^(k+1)). Note that 6(10^k) = 0 (mod 2^(k+1)),
and 7(10^k) = 2^k (mod 2^(k+1)). So, either 6*10^k + N or
7*10^k + N is divisible by 2^(k+1).

Claim 2: If m and 10 are relatively prime, then for any r,
there is a number N whose only digits are 6 and 7 such that
N = r (mod m).

Proof: Let K be the (m^2)-digit number whose only digit is 6.
There is an s, 0 <= s < m, so that K + s = r (mod m).
Let N = K + 10^(m - 1) + 10^(2m - 2) + . . . + 10^(sm - s).
Since 10^(im - i) = 1 (mod m), N = K + s (mod m) = r (mod m).
Clearly, every digit of N is either 6 or 7.

Claim 3: If n is not divisible by 5, then there is a number N whose
only digits are 6 and 7, so that N is divisible by n.

Proof: We can write n = (2^k)m, with gcd(m,10)=1.
Use claim 1 to find a k-digit number M, whose only digits are 6 and 7,
which is divisible by 2^k. Choose an integer r so that
(10^k)r + M = 0 (mod m). Use claim 2 to find a number K whose
only digits are 6 and 7, so that K = r (mod m). Let N = 10^k K + M.
Then N = 0 (mod m) and N = 0 (mod 2^k), so N is divisible by n.
Finally, the only digits of N are 6 and 7, so we are done.

 
Lets add a reward... winner (first person to answer correctly, with no edits in the answer) gets my RC5 blocks (PIII-933) for the rest of the month.

 
Curse you for making me think on a Friday. It's been almost 10 years since I was last doing mathematical proofs. I tried to cheat and find an answer on the 'net, but all I could find was this link to a homework assignment that contained the question. No solution was provided, however. This is starting to bug me, and if I don't find a solution before tonight, it's gonna bug me over the weekend and I will be really pissed.
 
Take the Square Root of Pie and add 3% then divide by 4 then round off to the nearest k\n. Now if the number equals 0 it would then be safe to say its a true Noun! IF the number equals 3 start all over and divide by 9. Reguardless you answer will end up with a positive integer. Now please start doing your own homework!





























😀
 
thEnEuRoMancER, i didn't finish reading the threads, but &quot;mod&quot; i believe refers to the modulus... off hand, i don;t remember exactly what the modulus function does... but having the name may help in looking it up if your really into solving this...
 
I don't understand.

Lets say k=1. Any number you use for n is going to make a fraction, assuming we're doing it k/n. I.E. k=1, n=3, k/n=1/3.

Am I missing something?
 
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