Its school and its time for crazy math

[ThunderAeroI]

Algebra is where this is going, but I'm lost as to how to solve this equation without having to manually try 4! numbers.

The equation is math + taak = hmmmm

Each letter is a number, common letters are the same number. I have come close to solving it a few times I thought, but it always fails near the end.

What am I doing wrong to solve this. Once I know how I can try to explain it to the kid.

Thanks.

 

[Leith Jones]

Things I figured out right away:

h = 1
k + 1 = m
a + 1 = t
t must be 9 (or you wouldn't get m + t = m, and there must be a 1 carried)
so a must be 8

Am I on the right track?

 

[Leith Jones]

Got it:

07891
09886
-------
17777

Do you need the rest of the reasoning? I just used pencil and paper and put numbers in when I figured them out.

 

[ThunderAeroI]

Got it:

07891
09886
-------
17777

Do you need the rest of the reasoning? I just used pencil and paper and put numbers in when I figured them out.

Yes, I 'd like to know how you figured it out

 

[Trekker4747]

It's not crazy math until you learn to subtract by adding.

 

[Kommander]

This is what I came up with. Although, I'm pretty terrible at math...



...especially after... why does everything taste like the color blue all of a sudden? I... why are my ears and nose leaking... transmission fluid? This isn't good. I'm not supposed to contain transmission fluid... that means...

Long story short, will using human blood instead of transmission fluid ruin my car? Google is unhelpful as usual...

Anyway, the equation works if all the letters equal zero. There may be other solutions.

 

[Leith Jones]

Here we go. I hope this isn't fuzzy or confusing. I worked it out in Notepad where the font characters are all the same width so it may not look right.

math
taak
-----
hmmmm

h must be 1. Adding 2 4-digit numbers will never be greater than 19,999.

mat1
taak
-----
1mmmm

Now we see that k + 1 = m.

Also a + a = m AND t + a = m. If we assume that each letter substitutes for a different number, then we must have carried a 1. Therefore, a + 1 = t.

We have m + t = m. If we assume that m does not equal 0 ('math' would just be 'ath'), then t = 9 and a 1 is carried. So a = 8.

m891
988k
-----
1mmmm

1 + k will not cause a 1 to be carried. The tens column has 9 added to 8. So m = 7 and a 1 is carried. Since a + 1 = t, a = 8:

7891
988k
-----
17777

Since k + 1 = 7, k = 6:

7891
9886
-----
17777

Ta da! Clear as mud.

Edited to add: I think Kommander got it right. I hope he lived.

 

[ThunderAeroI]

Here we go. I hope this isn't fuzzy or confusing. I worked it out in Notepad where the font characters are all the same width so it may not look right.

math
taak
-----
hmmmm

h must be 1. Adding 2 4-digit numbers will never be greater than 19,999.

mat1
taak
-----
1mmmm

Now we see that k + 1 = m.

Also a + a = m AND t + a = m. If we assume that each letter substitutes for a different number, then we must have carried a 1. Therefore, a + 1 = t.

We have m + t = m. If we assume that m does not equal 0 ('math' would just be 'ath'), then t = 9 and a 1 is carried. So a = 8.

m891
988k
-----
1mmmm

1 + k will not cause a 1 to be carried. The tens column has 9 added to 8. So m = 7 and a 1 is carried. Since a + 1 = t, a = 8:

7891
988k
-----
17777

Since k + 1 = 7, k = 6:

7891
9886
-----
17777

Ta da! Clear as mud.

Edited to add: I think Kommander got it right. I hope he lived.

Thanks, how do you know h=1. I understand the rest of it

 

[Leith Jones]

If you add any 2 4-digit numbers, the result will either be 4 digits or 5 digits. And the largest number it can be is 19,998 (9999 + 9999). Zero wouldn't make much sense (unless this is a tricky puzzle), so it must be a 1.

I made the assumptions that each letter stands for a different number and the first digit would not be zero. Does that make sense?

 

[ThunderAeroI]

If you add any 2 4-digit numbers, the result will either be 4 digits or 5 digits. And the largest number it can be is 19,998 (9999 + 9999). Zero wouldn't make much sense (unless this is a tricky puzzle), so it must be a 1.

I made the assumptions that each letter stands for a different number and the first digit would not be zero. Does that make sense?

yup. thanks.

 

[Kommander]

Edited to add: I think Kommander got it right. I hope he lived.

I'm a little light-headed, but otherwise okay. My engineer friend assures me that everything that happened is perfectly normal and also suggested that, if I insist on trying to divide by zero, that I shouldn't try so hard to visualize something dividing by zero.

It looks like the substitution cipher where each letter is a different digit that everyone else did is what the OP was looking for. I decided to do the "every letter is a variable" thing instead for some variety. Based on the instructions given in the OP, I figured what everyone else did was what he was looking for, but the way I did it also applies.

I probably could have simplified what I did on paper a little more, but I essentially solved it. All variables being zero works, but there are multiple solutions. As long as everything is being multiplied by zero, the other variables could be anything. The easiest way to accomplish this is setting h and a to zero. So...
a = 0
h = 0
k = (-∞, ∞)
m = (-∞, ∞)
t = (-∞, ∞)

If this is a school thing, my solution will either get full credit, or will get a no credit and the student kicked out of the class for being a smart ass, depending on the teacher and the actual instructions given for the assignment.

 

[Alidar Jarok]

Here we go. I hope this isn't fuzzy or confusing. I worked it out in Notepad where the font characters are all the same width so it may not look right.

math
taak
-----
hmmmm

h must be 1. Adding 2 4-digit numbers will never be greater than 19,999.

mat1
taak
-----
1mmmm

Now we see that k + 1 = m.

Also a + a = m AND t + a = m. If we assume that each letter substitutes for a different number, then we must have carried a 1. Therefore, a + 1 = t.

We have m + t = m. If we assume that m does not equal 0 ('math' would just be 'ath'), then t = 9 and a 1 is carried. So a = 8.

m891
988k
-----
1mmmm

1 + k will not cause a 1 to be carried. The tens column has 9 added to 8. So m = 7 and a 1 is carried. Since a + 1 = t, a = 8:

7891
988k
-----
17777

Since k + 1 = 7, k = 6:

7891
9886
-----
17777

Ta da! Clear as mud.

Edited to add: I think Kommander got it right. I hope he lived.

Thanks, how do you know h=1. I understand the rest of it

The reason he said h=1 is that h can't equal anything greater than one. Any number starting with a 2 would be too large (the maximum possible number combination would be 9,999 + 9,999 = 19,998. With that, h would be 1). This assumes h doesn't equal zero, I suppose, but it's not convention to begin a number with zero.

 

[Captain_Nick]

I have a question. Once you got that h = 1, you said that k + 1 = m. How did you know that k + 1 wasn't equal to 1m with a carry over to the next column?

 

[Leith Jones]

I have a question. Once you got that h = 1, you said that k + 1 = m. How did you know that k + 1 wasn't equal to 1m with a carry over to the next column?

I didn't. But it seemed easier at the time to check how the other columns worked out first.

To answer your question directly, the only way that k + 1 = m with a carryover is for m to be zero and k be 9. That would violate the convention of not starting a number with zero ("math"). If you get it to work out with that beginning, you're a better puzzle solver than I ever will be.