In the ones place you are adding 5 H's. That is 5 × H.

To get a result which ends with H also, only the following is possible :

5 × 5 = 25
5 × 0 = 0

(Basically multiplying 5 by an odd digit/number always results in a 5 for the ones place.
5 × 1 = 5
5 × 3 = 15
5 × 5 = 25
5 × 7 = 35
5 × 9 = 45

Multiplying 5 by an even digit/number always results in a 0 for the ones place.

5 × 0 = 0
5 × 2 = 10
5 × 4 = 20
5 × 6 = 30
5 × 8 = 40


But only 5 × 5 and 5 × 0 fits such that we get a result ending with the same digit as the ones place.


Smallest possible for H = 0

In the tens place you are adding 4 T's. That is 4 × T.

Look at the following :

4 × 0 = 0
4 × 1 = 4
4 × 2 = 8
4 × 3 = 12
4 × 4 = 16
4 × 5 = 20
4 × 6 = 24
4 × 7 = 28
4 × 8 = 32
4 × 9 = 36


The only possible value of T such that the 4 × T also ends with a T , is 0

But we cannot put 0 since that would mean TH is 00. That is not considered a 2-digit number. 2-digit numbers range from from 10 to 99.

We cannot have T and H being the same digit also, since the question says all the letters represent different digits.

So H cannot be 0 either.

This also tells us that H can only be the other possibility, 5.

So 5 × H = 5 × 5 = 25.

The digit 2 is carried over from 5 × H.


This means that adding 2 to the result of 4 × T will increase its ones place by 2, to give us T again.

So we should look for an option above where the result has ones place which is 2 less than T.


The only possibility is 4 × 6 = 24


So T = 6


Check :

H + TH + TH + TH + TH
= 5 + 65 + 65 + 65 + 65
= 265

So there is a 2 that is carried over to the hundreds place.

We see 3 A's.

A + A + A = 3 × A

So 3 × A plus the 2 that is carried over, will give a result that also ends with A.


Using the same logic as before, the ones digit of 3 × A must be smaller than A by 2. So when we add the 2, we get A again.


Looking at multiples of 3 ,


3 × 1 = 3
3 × 2 = 6
3 × 3 = 9
3 × 4 = 12
3 × 5 = 15
3 × 6 = 18
3 × 7 = 21
3 × 8 = 24
3 × 9 = 27


There are two possibilities :

3 × 4 = 12
3 × 9 = 27


So A can be 4 or 9

Next we see that M + M = 2 × M

Since A could be 4 or 9,

it means that either 1 or 2 is carried over to the thousands place.


So 2 × M plus the carried over digit will give a result that ends with M.

This tells us that the ones place of 2 × M is smaller than M by either 2 or 1.


And also , in the tens place there will be a 1 that is carried over to the ten thousands place, such that 1 + 1 = 2


Multiplies of 2 :


2 × 1 = 2
2 × 2 = 4
2 × 3 = 6
2 × 4 = 8
2 × 5 = 10
2 × 6 = 12
2 × 7 = 14
2 × 8 = 16
2 × 9 = 18


Only 2 × 8 and 2 × 9 can be considered, since we need the 1 to carry over, and also need the ones place of 2 × M to be smaller than M by either 1 or 2.

So M is either 8 or 9.

As we want the smallest 4-digit number, we will pick the smaller one.


M = 8, which means A = 9

(If you're asked for largest 4-digit number then it will be M = 9 and A = 4)

So M A T H → 8 9 6 5

Check :


5 + 65 + 965 + 8965 + 18965 = 28965

The other possibility (bigger)

M A T H → 9 4 6 5

Check :


5 + 65 + 465 + 9465 + 19465 = 29465

Another way :

Work backwards first.

Realise that subtracting 1 M A T H from 2 M A T H will give you 10000.


So H + T H + A T H + M A T H = 10000


Then,


① 4 × H ends with a 0.
The only possibility is 4 × 5 = 20 or 4 × 0 = 0

So H = 0 or 5


② either no carry over or 2 is carried over.

We cannot have no carry over as 3 × T can never end with a 0.

So H can't be 0 and it has to be 5.

Then, 3 × T + 2 ends with a 0.

Subtracting that 2 means the ones place has to be 8. Such that adding 2 results in 10, and a 1 is carried over.


Only possible 3 × T :

3 × 6 = 18
18 + 2 = 20

So T = 6

2 is carried over. (1 comes from adding 2 to 8 and the other 1 comes from 3 × 6)


③ 2 × A + 2 ends with 0.


Again, 2 × A must end with 8


Two possibilities :

2 × 4 = 8
8 + 2 = 10

Or

2 × 9 = 18
18 + 2 = 20


So A is either 4 or 9

Either 1 or 2 is carried over.


④ The result of adding up the thousands place is 0.


So M + 1 ends with 0 or M + 2 ends with 0

This means that:

M = 9 such that M + 1 = 10
Or
M = 8 such that 8 + 2 = 10


So M = 9, A = 4 or M = 8, A = 9


MATH → 8965 (smaller) or 9465 (bigger)



8965 fits the requirements.

Algebra way :

5 × H + 10 × T × 4 + 100 × A × 3 + 1000 × M × 2 + 10000 = 20000 + 1000 × M + 100 × A + 10 × T + H

5H + 40T + 300A + 2000M + 10000 = 20000 + 1000M + 100A + 10T + H


4H + 30T + 200A + 1000M = 10000

Divide by 2 :

2H + 15T + 100A + 500M = 5000



100A and 500M will give multiples of 100 and 500 respectively. Their tens and ones digits are all 0's.

Since we need to get to 5000, 500M should be close to 5000 .

So we can let M = 9 or 8 such that we get 4500 or 4000

(It cannot be 7 or lower as the maximum possible value of 100A is only 900 (100×9).

7 × 500 = 3500 and 3500 + 900 = 4500 only.

Max value of 15T is only 15 × 9 = 135 and max value of 2H is 2 × 9 = 18 so the max we can get is 4500 + 135 + 18 = 4653 only.)


①
If M = 9, then 2H + 15T + 100A + 4500 = 5000

2H + 15T + 100A = 500

Since the maximum possible of 2H + 15T is only 18 + 135 = 153, then A cannot be 3 and below since that would mean 100A = 300

300 + 153 = 453 only , not enough.

A can't be 5 either since 500A is already 500. Adding 2H and 15T will cause the sum to exceed 500.

So A can be 4 only.

So 2H + 15T + 400 = 500

2H + 15T = 100

T has to be 6 and below, since if T is 7 and above, 15T would be 105 and above.


15T ends with either 5 or 0.

2H ends with either 0,2,4,6,8


In order to have 100, the ones place for both has to be 0. No other combination can give 0.


2H ends with 0 so H is either 0 or 5.
But H can't be 0 as that would mean 15T = 100, and that is not possible since T is a whole number.

So H = 5. →2H = 10


15T can only be 90.

T = 6