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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 ?
lili
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  • Consider each of $a,h,m,t\in [0,9]$ are single-digit integers. Does the equation make more sense now? – abiessu Nov 13 '13 at 00:15
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    I hope you don't mind me offering some general advice for tackling questions that look totally alien to you. The truth is, it doesn't matter how much you've been taught, you'll always find some problem that you have absolutely no idea how to answer. I'm 23 studying math and it happens to me all the time! (maybe even more than it happens to you) and the best thing to do is not panic but instead start playing! Just start asking yourself easier questions than the one you've been given "What happens if I just try some random numbers - where does that plan go wrong?" "what happens if $M$ is a $1$?" – Dan Rust Nov 13 '13 at 01:00
  • "How about if $M$ is a $2$?" "How about $3$?" Just try stuff. there's nothing wrong with writing down something that's wrong, because then at least you know that it's wrong - you've learnt something about the problem that you didn't know before. – Dan Rust Nov 13 '13 at 01:00
  • I really enjoyed math Olympiads when I was ten. I was never very good at it, but I thought it was fun all the same. – abiessu Nov 13 '13 at 01:42

3 Answers3

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

Clive Newstead
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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!

Bonnaduck
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  • thanks for your help but for me when it comes to math i suck at it – lili Nov 13 '13 at 00:47
  • @lili is there anything I could elaborate on? – Bonnaduck Nov 13 '13 at 00:48
  • no some how i was chosen in class for the highest score in math i was shocked but scared and today was my first test. Bonnaduck what do you mean? – lili Nov 13 '13 at 00:49
  • clive im so sorry for not explaining math olympiads better – lili Nov 13 '13 at 00:54
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    I'm sure you don't suck at math @lili ! Sometimes, you have to move on to the next thing before the thing you were looking at before starts to make sense! I bet there was a time you didn't know how to add and subtract, and you might have struggled with it (it's actually pretty hard stuff when you've not seen it before!) but now you can probably add and subtract as well as any of your teachers. – Dan Rust Nov 13 '13 at 00:55
  • @lili I mean is there something that I wrote that you don't understand? – Bonnaduck Nov 13 '13 at 01:00
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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?

hhsaffar
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