3

I found this question on social media, from a math account I follow. $$\begin{align} M+A+T+H &=10 \\ M-A+T+H &=\;6 \\ M+A-T+H &=\;4 \\ M+A+T-H &=\;2 \\ (-M)^{ATH} &=\;\text{??} \end{align}$$

To solve this I set up an augmented matrix that looked like: $$ \left[\begin{array}{rrrr|r} 1 & 1 & 1 & 1 & 10 \\ 1 & -1 & 1 & 1 & 6 \\ 1 & 1 & -1 & 1 & 4 \\ 1 & 1 & 1 & -1 & 2 \\ \end{array}\right] $$ I proceeded with Gaussian Elimination and cleaned it up nicely after doing $R_2-R_1$, $R_3-R_1$, and $R_4-R_1$.

I ended up with $$ M=1 \\ A=2 \\ T=3 \\ H=4 \\ \Rightarrow (-M)^{ATH}=(-1)^{(2\cdot 3\cdot 4)}=1 $$

My question is:

What is a more efficient way of solving this?

I instantly went to Gaussian Elimination, but I feel that setting up a matrix is a bit overkill for a brain teaser on Instagram.

I realize we only need to figure out if the product $ATH$ is odd or even (meaning we only need to know if one of $A$,$T$, or $H$ is odd), but I can’t think of any way to determine that.

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
AllCatsAreGrey
  • 2,388
  • Note that the left-hand side second equation comes from the first by subtracting $2A$, while the right-hand side decreases by $4$; thus, $2A=4$. Etc. – Blue Jul 12 '24 at 18:12
  • 1
    Label the equations $1,2,3,4$. Then find what $1-4$, $1-3$, $1-2$ give you. – Sahaj Jul 12 '24 at 18:13

3 Answers3

6

$$\begin{align} M+A+T+H &=10 \tag1\\ M-A+T+H &=\;6 \tag2\\ M+A-T+H &=\;4 \tag3\\ M+A+T-H &=\;2 \tag4\\ \end{align}$$

Then the symmetry of the coefficients of the variables in each equation suggests pairwise subtraction of the equations. In particular, $(1)-(4)$ gives $2H = 8$, $(1)-(3)$ gives $2T = 6$, $(1)-(2)$ gives $2A = 4$. With the values of each of these variables, one can substitute into $(1)$ to find $M$.

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Sahaj
  • 5,355
2

Yes. Let $L_{i}$ denote the $i$-th equation (counted from bottom to top). Notice that:

$L_{2}=L_{1}-2A$

$L_{3}=L_{1}-2T$

$L_{4}=L_{1}-2H$

So, it's easy to see:

$10-2A = 6$ $10-2T = 4$ $10-2H = 2$

Which gives $A=2$,$T=3$,$H=4$. Looking back to $L_{1}$, we plug our results back in to get $M=1$. So, our answer to $(-M)^{ATH}$ is $1$.

Almeida
  • 550
2

The sum of the last three equations has coefficient 1 on any of the ATH variables. This means $2=(2)+(3)+(4)-(1)= 2M$, that implies $M=1$. In particular, we are only interested in the parity of $ATH$, but $(1)-(2)$ gives $A=2$ and the result is 1.

Let me say, however, that a couple calculations will always be necessary, since the general solution in terms of constants $a,b,c,d$ on the right side is

$$ \left ( \frac{b+c+d-a}{2} \right )^{ \frac{(a-b)(a-c)(a-d)}{8} } $$

seems to have no reason to simplify - and it's not even well defined if the differences in the exponent are all odd and the base is negative.

Andrea Marino
  • 6,166