Second Round

We first note that there is a useful relation between $a$, $b$, $d$, and $v$. The total number of integers on the two pieces of paper is $a+b$, the number of integers on Alicia's piece of paper plus the number of integers on Britt's piece of paper. This, however, also equals $v+d$: the total number of distinct integers, plus the total number of integers that have been written down twice. Hence, we get that $a+b=v+d$.

a. We choose $a=b=2022$ and look for a solution to $a\cdot b=d\cdot (v+d)$. We use the fact that $a+b=v+d$. This means that we are looking for solutions to $a\cdot b=d\cdot (a+b)$. If we substitute $a=b=2022$, then we find that $2022\cdot 2022=d(2022+2022)=d\cdot 2\cdot 2022$, so $d=1011$. Together with $a+b=v+d$, we find that $2022+2022=v+1011$, so $v=3033$. This situation happens for example if Alicia writes down the numbers $1$ to $2022$, and Britt writes down the numbers $1012$ to $3033$.

b. With a little bit of trying, and by choosing $d$ not too large, we find that $a=b=3$, $d=1$, and $v=5$ is a solution: $3\cdot 3=1\cdot (5+4)$. The numbers also satisfy the equation $a+b=v+d$. This situation can occur if Alicia writes down the numbers $1$, $2$, and $3$, and Britt writes down the numbers $3$, $4$, and $5$, for example.

c. Suppose that there are numbers such that $a\cdot b=v\cdot d$. We already deduced that $a+b=v+d$, or $v=a+b-d$. Substituting this yields $$ab=vd=(a+b-d)d=ad+bd-d^2.$$ If we now subtract $ad$ from both sides of this equation, we find $ab-ad=bd-d^2$, so $a(b-d)=d(b-d)$. Because Britt wrote down at least one number that Alicia did not write down, we have $b>d$. Therefore, we can divide the equation $a(b-d)=d(b-d)$ by the positive number $b-d$, and we find that $a=d$. On the other hand, Alicia wrote down at least one number that Britt did not write down, hence $a>d$. This gives a contradiction and hence there cannot exist numbers such that $a\cdot b=v\cdot d$.