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Hard Math Problem: Who Can Do it?

N

Nessoldaccount

Senior member
This is from a math challenge book of mine, can anyone solve it??? I've tried soooo long.

Given the following addition problem, determine the one-to-one correspondence between the letters used and the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; i.e., make an association between the letters and the digits in which each letter is associated with a unique digit, and vice-versa, in such a manner that the resulting numerical addition is correct.

THIRTY
TWELVE
TWELVE
TWELVE
TWELVE
TWELVE
------
NINETY

E=
I=
H=
L=
N=
R=
T=
V=
W=
Y=

Good luck!
 
Ness: Spacing is irrelevant. They are all six letter words. You did not specify no duplicates, though unique might have implied that to you. If they used the word unique, I think they mean a unique number letter pair.
 
LOL I'm an idiot for not noticing the V and E thing. Although it is POSSIBLE that with carrying numbers over that they might not be 0 but it's very unlikely.

Also T must be 1.
 
Ness:

T + 5V = T, ergo V must be zero.
Y + 5E = Y, ergo E must be zero.

Nothing else works for those two. No carrying over will work. 🙂 I assume no carrying over to the next column.
 
[EDIT]I'm assuming like chess did that we're not dealing with carrying here, & that we're just dealing with one column being one problem.[/EDIT]

Step 1:

You can easily form six simultaneous equations:

6T = N
5W + H = I
5E + I = N
5L + R = E
5V + T = T => V=0
5E + Y = Y => E=0

Solving six way simultaneous equations is a bitch, so we'll just re-write them in such a way that a TI-92 can solve them:

0E + 0I + 0H + 0L - N + 0R + 6T + 0V + 0W + 0Y= 0
0E - I + H + 0L + 0N + 0R + 0T + 0V + 5W + 0Y= 0
5E + I + 0H + 0L - N + 0R + 0T + 0V + 0W + 0Y= 0
-E + 0I + 0H + 5L + 0N + R + 0T + 0V + 0W + 0Y= 0
0E + 0I + 0H + 0L + 0N + 0R + 0T + 5V + 0W + 0Y= 0
5E + 0I + 0H + 0L + 0N + 0R + 0T + 0V + 0W + 0Y= 0

Somebody else can solve the system, I don't feel like dragging my 92 out right now.

E= 0
I=
H=
L=
N=
R=
T=
V= 0
W=
Y=

[EDIT AGAIN]I make no guarantees as to the accuracy of this, I'm working under the influence of painkillers. 😉[/EDIT AGAIN]

Viper GTS
 
Chess and Viper,

"determine the one-to-one correspondence between the letters used and the digits"

Since statement implies that each letter must be assigned to a unique number.
That is what one-to-one means in function theory.

BTW, so far:

E = Must be even
V = Must be even

still working on it 🙂
 
Mister T...

Solving my set of simultaneous equations should result in one value per letter, thus a one to one correspondence.

That's the beauty of simultaneous equations. If I could find my TI-92, I'd solve it myself, but I can't find the damned thing.

:|

Viper GTS
 
It's not unlikely, in fact, in the 2nd column you are definitely carrying over to the 1st.

Also, there must be no duplicates, i'm sure of this, if they were allowed then you could just make everything 0. Also, there are 10 letters and 10 numbers.
 
Questions like these are the reason I am not dues paying member of MENSA.

Who cares?

Who thinks like this? (besides whats her name in the newspaper) Vos Savant, sounds like a member of the Psycic network to me.

I also agree with Locutus, rugs and merkins.

 
good point ness.

bestMAX , grow up we're just trying to have fun...or satisfy some twisted male desire to be perfect...you know, one of the two.

Chris

Check my post above from now on....
 
Ness: What do you mean in the second column you are carrying over to the first? Are you suggesting reverse carrying? Wierd.
Anyway, assuming no duplicates, carrying would be necessary and the problem is a bit more difficult.

By the way, I remember a question similar to this one and all the vertical letters added up to the result INDIVIDUALLY, and totally, i.e. without carrying. Also, all the horizontal letters added up to the word, i.e. NINE adds up to 9. Otherwise, this problem would be too difficult to do without a calculator or plenty of time.

Also, do you have the exact wording of the problem, or is that it? Perhaps the instructions are poorly worded? 😛 Or we're just too stupid!

Edit: Oh, Ness, I see what you mean about carrying from right to left. That addition, so tough!
 
There would be no carry from the far right hand column to the penultimate right hand column. So the second column would have no carry and what I said about V equalling zero must be true. 😛
 
This is what I have so far. I hope i'm right...

N =
T = 1
E = 0
V = [2,4,6,8]
W =
L =
I =
R =
Y =
H =

still workin' on it

[edit N doesn't have to equal 6]
 
Does anybody else agree with me that E isn't 0? How else can we find Y? One thing i just thought of though. Y can be 0 and E can be even. Keep thinkin'

Chris
 
Ground0:

If Y + 5E = Y, how could E be anything but zero? Also, it's in the far right hand column, so carrying isn't an issue. Since nothing will be carried to column two, because no number is over 9, V must also be zero. Otherwise, the problem makes no sense.
 
chess9 is right.. If E isn't zero there is no integer that would satisfy the eq ( [1,2,3,4] + 5V = a number divisible by 10)
 
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