BWM Runde 1 Mock Paper 3: Questions, Hints and Solutions

The given values involve $k+1$, so multiply by $x+1$. Put $Q(x)=(x+1)P(x)-1$. It has degree at most $n+1$, vanishes at all $0,1,\ldots,n$, and is not the zero polynomial because $Q(-1)=-1$. The factor theorem therefore gives

$$Q(x)=c\prod_{k=0}^{n}(x-k).$$

Substituting $x=-1$ yields $-1=c(-1)^{n+1}(n+1)!$, hence $c=(-1)^n/(n+1)!$. Now set $u=x+1$. Then

$$uP(u-1)=1+c\prod_{j=1}^{n+1}(u-j).$$

The coefficient of $u$ on the left is $P(-1)$, since the constant term of $P(u-1)$ is its value at $u=0$. To obtain a term containing exactly one $u$ from the product on the right, select $u$ from one factor $(u-j)$ and select the constants from every other factor. For that choice, the coefficient is $(-1)^n(n+1)!/j$. Adding the choices and multiplying by $c$ gives

$$P(-1)=\sum_{j=1}^{n+1}\frac1j.$$

The conditions are consistent: the numerator $1+c\prod_{k=0}^n(x-k)$ vanishes at $-1$, so dividing it by $x+1$ produces a polynomial of degree $n$ with exactly the prescribed values.

Conclusion: $P(-1)=1+\frac12+\cdots+\frac1{n+1}$.

Review the idea: Remainder and factor theorems · Polynomial functions

If $N=\prod p_i^{a_i}$, a divisor independently selects an exponent from 0 to $a_i$ for each prime. Thus $\tau(N)=\prod(a_i+1)$.

Let $a$ be the exponent of 3, allowing $a=0$. Multiplication by 3 changes only that exponent, so

$$\begin{gathered}\frac{a+2}{a+1}=\frac{18}{12}=\frac32,\\\text{giving }a=1.\end{gathered}$$

Similarly the exponent $b$ of 5 satisfies $(b+2)/(b+1)=24/12=2$, giving $b=0$. Write $N=3M$, where $M$ is coprime to 15 and $\tau(M)=6$.

The possible products of factors $a_i+1\ge2$ equalling 6 are 6 and $3\cdot2$. Therefore $M=p^5$, or $M=p^2q$ for distinct primes $p,q$, neither 3 nor 5.

• In the first case $M\ge2^5=32$, hence $N\ge96$.
• In the second case, for two primes $p<q$, putting the square on the smaller gives the smaller product: $p^2q<pq^2$. The two smallest permitted primes are 2 and 7. Hence $M\ge2^2\cdot7=28$, giving $N\ge84$.

The candidate $84=2^2\cdot3\cdot7$ has divisor count $3\cdot2\cdot2=12$. Multiplying by 3 changes this to $3\cdot3\cdot2=18$; multiplying by 5 changes it to $3\cdot2\cdot2\cdot2=24$. Thus 84 is attained and is least.

Conclusion: $N=84$.

Review the idea: Counting and summing divisors · Unique prime factorisation

We use induction because halving the side length produces the same kind of board.

Base case. For $k=1$, a $2\times2$ board with one cell removed is itself an L-shaped triomino.

Induction step. Suppose every $2^k\times2^k$ board with one cell removed can be tiled. Take a $2^{k+1}\times2^{k+1}$ board with one missing cell. Divide it by its horizontal and vertical midlines into four $2^k\times2^k$ quadrants. Exactly one quadrant contains the missing cell.

The four cells meeting at the centre form a $2\times2$ block. Cover the central cell in each of the other three quadrants with a single L-shaped triomino. Each quadrant now has exactly one unavailable cell: the original missing cell in one quadrant, and a cell occupied by the central triomino in each of the other three.

Apply the induction hypothesis separately to the remaining cells of all four quadrants. These tilings cannot overlap because the quadrants are disjoint, and none uses a cell already covered by the central triomino. Together they cover every required cell. The base case and induction step prove the assertion for all $k\ge1$.

Conclusion: Every such deficient square can be tiled; one central triomino reduces the problem to four smaller deficient squares.

Review the idea: Mathematical induction the first principle · Proof methods

We may view the triangle with $A,B,C$ in counterclockwise order; reflecting the entire drawing if necessary changes no lengths or assumptions.

Because $ABX$ is equilateral and external to $ABC$, the clockwise rotation through $60^\circ$ about $A$ takes $B$ to $X$: it turns the ray $AB$ to the external ray $AX$, and $AB=AX$.

For the other equilateral triangle the external ray $AY$ is $60^\circ$ counterclockwise from $AC$, since $B$ is on the other side of $AC$. Hence that same clockwise rotation takes $Y$ to $C$, with $AY=AC$.

A rotation moves every point by the same angle about its centre and preserves distances between points. Thus it takes segment $BY$ to segment $XC$, proving $BY=XC$. The choice of opposite external sides is essential to using one rotation for both endpoints.

Conclusion: $CX=BY$.

Review the idea: Triangle congruence · Angles