Fluency exercise

Solution

Sudoku 1

Here is one way we could solve this puzzle.

Here are the 16 equations:

1. $c+m+h=19$

2. $f+e=10$

3. $k+g+m+c=23$

4. $g+p=11$

5. $h+f=14$

6. $g+m=13$

7. $a+e+k+h=11$

8. $k+c+f+a=22$

9. $c+g+k=17$

10. $f+g+a=19$

11. $m+k+c=16$

12. $p+e=5$

13. $g+m+f=22$

14. $h+m+a=14$

15. $e+c+h+k=16$

16. $f+h=14$

We first look at the most simple equations with only two unknowns.

Equation 12 says that $p+e=5$. This means that both $p$ and $e$ must be less than or equal to $4$.

Equation 2 says that $f+e=10$, and since we know that $e$ is less than or equal to $4$, this means that $f\ge6$.

Equation 4 says that $g+p=11$, so we must have $g\ge7$.

Equation 5/16 tells us that $h+f=14$, so $h\le 8$.

Now look at equation 7, $a+e+k+h=11$. The only four different positive whole numbers that can add up to $11$ are $1+2+3+5$. So we must have that each of $a$, $e$, $k$, $h$ must be one of $1$, $2$, $3$, $5$. We know that $e\le4$, but we now know that $e\le3$, and so $2\le p\le4$. Also, since $h\le 8$ and must add up with a number less than or equal to $9$ to make $14$ by equation 5, we have that $h=5.$ So by equation 5, $f=9.$ But now by equation 2 $e=1,$ and by equation 12 $p=4,$ which we can put into equation 4 to find that $g=7.$ We now use equation 6 to deduce that $m=6.$

It remains to deduce the values of the letters $a$ and $k$ (each of which we know must take one of the values $2$ and $3$) and $c$. The only number left for $c$ is therefore $8$—we can check using equation 1 $c+m+h=c+6+5=19 \implies c=8 \;\checkmark$

We use equation 14 to check what value $a$ takes: $h+m+a=5+6+a=14 \implies a=3.$

So we must have $k=2$, but we can double check: by equation 11, $m+k+c=6+k+8=16 \implies k=2 \checkmark$

Letter Number
$a$ $3$
$c$ $8$
$e$ $1$
$f$ $9$
$g$ $7$
$h$ $5$
$k$ $2$
$m$ $6$
$p$ $4$

So the Sudoku now looks like

The solution to the Sudoku is

Sudoku 2

For the second Sudoku, we have the following 10 equations:

1. $m+n=a$

2. $g+n+f=g+c\implies n+f=c$

3. $g+d=n$

4. $g+m=m+f+d \implies g=f+d$

5. $g+c+e=a+e\implies g+c=a$

6. $b+g+f=a+g\implies b+f=a$

7. $n+g+c=b+c\implies n+g=b$

8. $a+d+e=2a\implies d+e=a$

9. $g+f+a=m+a\implies g+f=m$

10. $e+n+m=a+b+d$

Equations 3 and 4 share two of the same unknowns, so this seems like a good place to start. Eliminating $g$ yields the equation $f+d=n-d \implies n=2d+f.$

From equation 2, $n=c-f$, so we further see that $c-f=2d+f \implies c=2d+2f.$ From this equation we see that $c$ is even, and $c\ge 6$ (since the smallest values $d$ and $f$ can take are $1$ and $2$), so we have that either $c=6$ and the $d$ and $f$ are equal to $1$ and two or $c=8$ and then $d$ and $f$ are equal to $1$ and $3$.

Now look at equation 5, $g+c=a$. The largest value that $g$ can take is $3$, if $c=6$. The only other option therefore, is that $g=2$, since we have already established that one of $d$ and $f$ must be $1$. But since we must have that $a\le9$, this means that we must have $c=6,$ and therefore each of $d$ and $f$ must take one of the values $1$ and $2$. This excludes the possibility of $g$ taking the value $2$, so we must have $g=3,$ which tells us that $a=9$ by equation 5.

Equation 1 says that $m+n=a=9$. Since the values $1$ to $3$ are already taken, the only way we can make $9$ with two numbers is with $4$ and $5$. So each of $m$ and $n$ must take one of the values $4$ and $5$. The other letters and numbers we have left are $b$ and $e$ and $7$ and $8$. We will look at equation 10, $e+n+m=a+b+d$ to determine which values they should take. We know that $n+m=a=9$, so we can simplify this equation to consider $e=b+d$, which says that $b < e$, so since we knew each of $b$ and $e$ were one of $7$ and $8$, we now know that $b=7 \text{ and } e=8,$ which further tells us that we must have $d=1,$ and so we must also have $f=2.$ Equation 3 now says that $3+1=n$, so we finally know that $n=4 \text{ and } m=5.$

Letter Number
$a$ $9$
$b$ $7$
$c$ $6$
$d$ $1$
$e$ $8$
$f$ $2$
$g$ $3$
$m$ $5$
$n$ $4$

So the Sudoku to solve is

The solution to the Sudoku is