Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let D be a BIBD with parameters (b,v,r,3,1). Prove that either v=6i+1 or v=6i+3 for some non-negative integer i. BIBD stands for a Balanced Incomplete Block Design

My work, theorem states that bk=vr in BIBD, which I can use to solve the problem. my basic step would be solving the equation with given v=6i+1.

Any suggestion on how to go about doing it?

share|cite|improve this question
You should not substitute anything for $v$. The question asks you to prove that $v$ must be congruent to either $1$ or $3$ modulo $6$. I think that you will find the other basic theorem tying the parameters together also very useful here. – Jyrki Lahtonen Apr 18 '12 at 3:55

I take it your parameters are set up so $k=3$ and $\lambda=1$. In addition to $bk=vr$, there is another relation that holds among the parameters. Do you know what relation I'm referring to? If not, you can find it in the wikipedia article on BIBDs. Then see whether the two relations, along with $k=3$ and $\lambda=1$, aren't enough to get the result on $v$.

EDIT: Note that BIBD with $k=3$, $\lambda=1$ are called "Steiner triple systems," and there is a lot of information about them available on the web and in texts.

FURTHER EDIT: First, use those two relations to get $r=(v-1)/2$ and $b=v(v-1)/6$. Then, note that $r$ is an integer, so $v$ must be odd. So, what are the possible remainders when you divide $v$ (which is odd) by 6? And what gets ruled out by knowing that $b$ has to be an integer?

share|cite|improve this answer
another proposition would be r(k-1) = λ(v-1) correct? – user28699 Apr 18 '12 at 4:46
Yes. ${}{}{}{}$ – Gerry Myerson Apr 18 '12 at 6:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.