how does the word math represent a four digit number

Question

I just don't get it had it on a test can you help me? It was a test for math olympiads this is how the question looked exactly:

                                    MATH
                                     x 4 
                                  ----------
                                    HTAM ?

Answer 1

My guess is that you're working in base $10$. In this case we have $$\begin{matrix} & M & A & T & H \\ \times & & & & 4 \\ \hline & H & T & A & M \\ \hline ^{m} & ^{a} & ^{t} & ^{h} \end{matrix}$$ where the small letters $m,a,t,h$ refer to carried numbers.

Converting this to equations: $$\begin{alignat}{2} 4H &\,= 10h+M & \quad & \leftarrow \text{units}\\ 4T + h &\,= 10t + A && \leftarrow \text{tens}\\ 4A + t &\,= 10a + T && \leftarrow \text{hundreds}\\ 4M + a &\,= H && \leftarrow \text{thousands}\end{alignat}$$ This comes from simply carrying remainders.

We know $0 \le H \le 9$ because $H$ is a single digit, so by the fourth equation, $M = 0$, $1$ or $2$ (otherwise $4M+a$ is too big).

We can't have $M=1$ because the first equation tells us that $M$ is even, and presumably we can't have $M=0$ since then it wouldn't be a four-digit number, so $\boxed{M=2}$.

Now $a=0$ or $1$. Lets look what happens if $a=0$. (If this fails then we'll go back and try $a=1$.) Then the fourth equation gives $\boxed{H = 8}$.

Plugging this into the first equation gives $32 = 10h+2$ and hence $h=3$.

The second equation then gives $4T+3=10t+A$.

Since $a=0$, the third equation gives $4A+t=T$ (i.e. $t=T-4A$), so substituting $t=T-4A$ into the second equation yields $3=6T-39A$, i.e. $2T-13A=1$. This has the solution $\boxed{A=1}$ and $\boxed{T=7}$. (I solved this just by guessing.)

And sure enough $$\begin{matrix} 2 & 1 & 7 & 8 \\ \times & & & 4 \\ \hline 8 & 7 & 1 & 2 \end{matrix}$$

Answer 2

Think of MATH as a 4-digit number, with each letter corresponding to a digit. Since $4*MATH = HTAM$, this would imply that $MATH<2500$. Otherwise we would get a 5-digit number.

So M is either 1 or 2. M cannot be 1, since $4H$ never yields a number ending in 1. It must be that $M = 2$. Since $4M =H$, it must be that $H=4M = 4*2 = 8$. So we have $M=2$ and $H=8$.

So,

MATH          2AT8  
x  4  ---->   x  4 
----          ----  
HTAM          8TA2

We can see that $8*4=32$. So carrying the 3 to T, we get that $4T+3$ must end in A. In order to not mess up what we already have, A cannot be more that 2; otherwise, it would ruin our M, since we would have to carry a 1. So, $A=1$ or $A=2$. Lets try these:

If $A=1$, then $4T+3$ must end in a $1$. This happens when $T=2$ and $T=7$.

If $T=2$, then we have $MATH = 2128$. But $4*2128 = 8512$. So let's try $T=7$.

$MATH = 2178$. So, $4*2178 = 8712$. And we're done!

Answer 3

First Note that $4*MATH$ is a four digit number an less than $10000$ so $M$ must be $1$ or $2$. But $M$ cannot be 1 (Why?). So $M=2$.

$4*H$ ends with 2. So $H$ must be $3$ or 8. But $H$ cannot be 3 because $4*MATH>4000$(Why?). So $H=8$

Now $4*T+3$ ends with $A$. $4*A$ must be less than $10$(Why?) so it should be $1$ or $2$. But $A$ is odd(Why?) so $A=1$ so $T=2$ or $7$. But it is 7. How can we know it?