Each letter represents a different digit.m-a-t-h. Show how you solved it.M A T H x 4= H T A M?

Question

Teacher said 4 isn't used in the problem. Please help, although she lies so try it....

Answer 1

First thing we know is that if we multiply by 4, we must have an even number, so M must be 0,2,4,6,8. 0 will not work as a leading digit for a 4-digit number. 4 through 8 will make the result too big, so M must be 2.

Next H must be 8 or 9. And 4 x 9 is 36 (doesn't end in 2).

So H must be 8.

2 - - 8
.... x 4
--------
8 - - 2

Now you can just try values of A (for example and see if it makes sense. We know we can't have a carry, so A must be 0, 1 or 2. Since the digits are different, that eliminates 2. If we try A = 0, then we need a multiple of 4, that you can add 3 (the carry from 32) to get 0 at the end. But we know that a multiple of 4 can't end in 7. So A must be 1.

2 1 - 8
.... x 4
---------
8 - 1 2

And then we need a multiple of 4 that ends in 8 (added to the carry of 3 makes a number that ends in 1...) That means T = 2 or 7, but again we have used 2. So let's try T = 7

2 1 7 8
.... x 4
---------
8 7 1 2

That works!

Answer 2

ive finally worked it out!!
the answer is 2178*4 = 8712

m = 2 as 1 is a bit low and 3 would mostly give u 5 digit
h = 8 or 9 as 2*4 is 8 but some could carry over

just write u experimented wid the numbers in between

wen u times it by 4 the answer (HTAM)need to end in a 2 as m=2

Answer 3

MATH x 4 = HTAM

M should be 2 or 1 only for H to be a single digit..thus M = 2 to obtain H=8;

2 _ _ 8 x 4 = 8 _ _ 2

trial and error method for the A and T digit obtains
A= 1 and T = 7;

2178 x 4 = 8712;

Whew! nice question..quite challenging..

Answer 4

Trial and improvement is the way to work it out:

Puzzling's way is very good!