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\centerline
{
\bf  Solutions to Quiz \# 1 for Dr. Z.'s Number Theory
Course for Sept. 26, 2013
}
\bigskip
{\bf 1.} (3 points) Convert $51$ (written in the usual decimal notation) to
to binary, in (i) sparse notation (ii) dense notation (iii) positional notation.

{\bf Sol. to 1.}: $2^5 \leq 51$ but $2^6>51$ so
$$
B(51)=2^5 + B(51-2^5)= 2^5 + B(51-32)= 2^5 + B(19)\quad .
$$
$2^4 \leq 19$ but $2^5>19$ so
$$
B(19)=2^4 + B(19-2^4)= 2^4 + B(3) \quad .
$$
$2^1 \leq 3$ but $2^2>3$ so 
$$
B(3)=2^1 + B(3-2^1)= 2^1 + B(1)= 2^1 + 2^0 \quad.
$$
Going back:
$$
B(19)=2^4+2^1+2^0 \quad ,
$$
$$
B(51)= 2^5 + 2^4+2^1+2^0 \quad .
$$

{\bf Ans. to 1(i)}: $51= 2^5 + 2^4+2^1+2^0$

Including {\bf all} the powers of $2$ up to $2^5$ and putting a $1$ in front of those that show up, and a $0$ in front of those
that don't

{\bf Ans. to 1(ii)}: $51= 1\cdot 2^5 + 1 \cdot 2^4+ 0 \cdot 2^3 + 0 \cdot 2^2 +  1 \cdot 2^1 + 1 \cdot 2^0$

Listing the coefficients, we get

{\bf Ans. to 1(iii)}: $51= 110011$ (base $2$).

{\bf Comment}: Most people got it right.

{\bf 2.} (3 points) Find, by only using paper-and-pencil, the first three terms of the Euclid-Mullin sequence.

{\bf Sol. to 2}: $M_1=2$, $M_2$ is the smallest  prime dividing $M_1 + 1 =2+1=3$, so $M_2=3$.
$M_3$ is the smallest prime dividing $M_1 M_2 +1 = 2 \cdot 3 +1 =7$, so $M_3=7$.

{\bf Ans. to 2}: The first three terms in the Euclid-Mullin series are $2,3,7$.

{\bf Comment}: Most people got it right, {\bf except}, some people listed {\bf four terms}.
Knowing how to follow instructions is {\bf more} important than number theory.
If you {\bf must} to on, please write:

``Dear Dr. Z.: I know that all you asked for were three terms, so the answer to your question is: $2,3,7$.
Since I {\bf love} doing it, I can't resisit the temptation, just for fun, to do one more term:
$2 \cdot 3 \cdot 7 +1 =43$, the smallest prime divding $43$ is $43$, so the fourth term is $43$.

{\bf 3.} (4 points) Prove  that there are infinitely many primes.

{\bf Sol. to 3}: See wikipedia and thousands of other places on the internet.




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