Winter Puzzle Solutions

By: Yuliya Nesterova

Solutions to the Winter Puzzles in Issue 15 of Notes from the Margin.

Puzzle 1

  1. Looking at the tens’ digit column, two evens, an odd, and a carried digit must sum to an odd digit. Therefore, the carried digit is a 2, and the ones’ column’s sum is between 21 and 29, inclusive.
  2. Likewise, in the ten thousdands’ digit column, the carried digit must be a 1, so (in the thousands’ column), 5\leq N \leq 9
  3. Notice that the hundreds’ digit column has no carried digits from tens’ column addition: that is, T + E + U + 2 \, (carried) <10 \, \iff T + E + U < 8, with T,U \in \{ 0,2,4,6,8\} \setminus \left\{ A,M,R \right\}.
  4. Working case-by-case (see below), we conclude that A,M,R \in \left\{ 4,6,8\right\}, T,U \in \left\{0,2\right\}, and (N,E,I) is either (9,3,7) or (7,1,5).
  5. Looking at the thousands’ and ten thousands’ digit column, N=9 implies R=8, and since I=7, A must be 7+1\,(carried) = 8. This is a contradiction, since A and R should be distinct digits.
  6. On the other hand, N=7 implies R=4 and, given that I=5,A must be 5+1\, (carried)=6}. E=1.
  7. Completing the rest of the puzzle, M=8, so F=\begin{cases}\frac{8-2}{2}=3\\ or\\ \frac{8-0}{2}=4 \end{cases}, and since all letters are distinct, F=3 and T=2. From here, we know that U=0.
  8. By the process of elimination, W=9.

Puzzle 2

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Puzzle 3

 \begin{matrix}\textcolor{blue!50!black}{\frac{x}{4}-2} &\textcolor{blue!50!black}{ x \in (-5,8)}\\ \textcolor{blue!50!black}{3-x }& \textcolor{blue!50!black}{ x \in (3,4)}\\ \textcolor{blue!50!black}{-2-x } & \textcolor{blue!50!black}{ x\in (-1,0)}\\ \textcolor{blue!50!black}{(x-7)^2 +(y-1)^2 = 2} & \textcolor{blue!50!black}{\text{major arc from }(6,2)\text{ to }(8,0)}\\ \textcolor{red}{\frac{x}{4}-\frac{3}{4}} & \textcolor{red}{x \in (-2.6,3)}\\ \textcolor{red}{(x-2.5)^2 + (y-1.35)^2=2} & \textcolor{red}{\text{major arc between }(1.3,0.6)\text{ and }(3,0)}\\ \textcolor{red}{\frac{-3}{5}x-3} & \textcolor{red}{x \in (-9,-3)}\\ \textcolor{red}{(x-4)^2 + (y-11)^2=117.8} & \textcolor{red}{\text{minor arc from} (-5,5)\text{ to }(1.25,0.5)}\\ \textcolor{red}{(x+8)^2 + (y-5)^2=9.3} & \textcolor{red}{\text{major arc from} (-8,1.95)\text{ to }(-5,5)}\\ \textcolor{violet}{4-\sin(x)+\cos\left(\frac{x}{4}\right)} & \textcolor{violet}{x \in (-20,-1)}\\ \textcolor{blue!50!white}{(x-1)^2 + (y-6.9)^2 = 3.77} & \textcolor{blue!50!white}{x \in (-\infty, \infty)}\\ \textcolor{blue!50!white}{(x+0.96)^2 + (y-3.56)^2 = 3.82} & \textcolor{blue!50!white}{x \in (-\infty, \infty)}\\ \textcolor{blue!50!white}{(x-1.13)^2 + (y-3.64)^2 = 1.1} & \textcolor{blue!50!white}{x \in (-\infty, \infty)}\\ \textcolor{brown}{\frac{17}{2}-\frac{3}{8}x} & \textcolor{brown}{x \in (-4,0)}\\ \textcolor{brown}{\frac{23}{4}-\frac{1}{2}x} & \textcolor{brown}{x \in (-5,-1)}\\ \textcolor{brown}{1.75x+17} & \textcolor{brown}{x \in (-5,-4)}\end{matrix}