Friday, March 9, 2012

Confusion Play

Things are not always what they seem. Take this:

Red  Blue Green Yellow

Is the Red red, Blue blue, Green green and Yellow yellow?
Now, consider this statement:


Definitely false!
Believe it or not but the statement is true if each letter is replaced with a different digit. In fact there are two different ways to decode this subtraction. Can you find both solutions?

If the above statement can be true, what about this:


Turns out this one is always wrong. No matter how you substitute the digits for letters. Can you tell why?

(both arithmetic operations come from a book by C.R.Wylie Jr. "101 Puzzles in thought and logic")

Your answers are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon.


Thad said...

In the ones column, N + E = T. In the tens column, E + N = H. This is possible if E + N > 9 causing a carry. But in the thousands column we have E + N = I and there's no way for this sum to have a third solution. Therefore it is impossible to replace the letters in SEVEN + NINE = EIGHT with distinct digits and get a valid arithmetic problem.

Jerome said...

First the two that you can do


Please don't ask me to explain this. I did it with brute force (ie a computer program). Not so the one you cannot do.

Consider the "en" in seven and the "ne" in nine. The n in seven and the e in nine cannot have a carry 1, but they must produce a carry. In other words e + n must be greater than 10. If they did not produce a carry or be greater than 10 then the h and t in eight could not be different.

Now we come to the second letter in seven and the first letter in nine. The most that the carry can be is a 1. So the i in eight would have to have the same value as the h in 8.

Not today, I'm afraid.

Jerome said...

Oh I forgot about the colors. If you allow chemistry answers, you can make a statement like

When in an acid, Thymol Blue is Red,
Bromophenol Blue is yellow, [you have me for green - can't make that into red and for the last one as well].

I used to drive students nuts with statements like that until they realized the difference between a noun and an adjective -- stuff we don't teach anymore, like long division.

Jerome said...

There is a simpler way of explaining E+N.

Only two answers are possible.
E+N = T There is no carry added onto that one.
E+N = H There must be a carry here because E+N does not equal T.
E + N = I This cannot be done. There are just not three possibilities.

TracyZ said...

Question 1:
solutions for SEVEN - NINE = EIGHT
This problem is labeled "easy", but it took me a long time to figure out. I kept writing down the formulas and hoping I could somehow solve it completely algebraically, but never managed to. In the end I used the formulas I came up and then plugged in possible numbers for each variable until I found the two combinations that worked and that had unique values for each letter in the problem.

The formulas I used were the following:
(i) S = E + 1
(ii) N - E = T
(iii) 10 + E - N = H
10 - H = N - E
10 - H = T (combined with ii)
H = 10 - T

(iv) 10 + E - 1 - N = I
9 - N + E = I
I = 9 - T (combined with ii)

(v) V - 1 + 10 - I = G
9 + V - (9 - N + E) = G (combined with iv)
G = V - E + N

With these equations in hand, I starting solving the addition problem by plugging in possibilities for H, T, and I, and then looking at possibilities for N, E, and S given the constraints. I looked at possible values for G and V last.

Here are the two solutions I came up with:
54146 - 6764 = 47382
21514 - 4641 = 16873

Question 2:

Why isn't SEVEN + NINE = EIGHT possible?

I think the impossibility arises from the need to have E + N result in different numbers to satisfy the requirement that each letter in the problem have a unique value.

If E + N < 10 then the problem fails with the first parts of the arithmetic, for the ones and tens columns.
- In the ones column, N + E = T, but
in the tens column, N + E = H
and therefore H = T, and there are not unique values for both H and T.

If E + N >= 10,
with the addition in the ones column, E + N = T + 10
and with the addition in the tens column,
E + N + 1 = H + 10

then in the thousands column, E + N = I + 10 or
E + N + 1 = I + 10 (depending on the value V + I)

With this last addition, it becomes clear that I, H, and T do not all have unique values, which is required by the problem and that, therefore, there is no solution.

Maria said...

Some of you pointed out that the first puzzle (subtraction) is hard. Feel free to solve only the addition.

Ilya said...

coming in slightly late, but I hope that we have a one hour allowance given the DST change :-).

here it goes. two subtraction answers:
1) 21514-4641=16873
2) 54146-6764=47382

Why addition is impossible: in the 5th column, E+N give us T, while the 4th column gives us H, which means E+N>10. But in the second column we again have E+N and this time it's yet a third number. That's impossible, since we can either carry an extra 1 or 0 from the third column and so we should get either T or H, but not an I.

Maria said...

Dear all,
I admit that I screwed up with this puzzle. The addition part of the puzzle is nice and doable but the subtraction is boringly hard. There are a few interesting logical observations that one can start solving with (E=S+1, N>E, H=I+1 etc) but eventually you need a brute force trail-and-fail substitution that is not fun at all.

Jerome - writing a program to solve this was a great idea. TracyZ and Ilya - you are brilliant for cracking this with the open hands. Each gets one puzzle point for the subtraction problem and one for addition. Thad - you get one for addition. By the way, Thad - can you please see my comment for the Billiard puzzle. I am waiting for your answer.

Ilya said...

Boringly hard parts teach us (and our kids!) perseverance and the answer is no less sweet to get in the end :-)

Maria said...

You are absolutely right Ilya. In our age of fast gratification, short attention span and constant rewards we ( and our kids) are loosing the ability of deep long concentration and steady persistence.

TracyZ said...

Hi Maria,
When I was feeling stumped on the subtraction part of the puzzle, I went online for inspiration and uncovered some interesting web sites on alphammetric puzzles aka cryptorithms which is what these types of puzzles are called by those who study this kind of thing.

It was fascinating to learn more about them. In my search, I also found a site with links to a a few different solvers for crypotorithms
, and these solvers quickly provided the answer to your puzzle. They didn't show their work though so I plodded on until I figured out how to solve the puzzle myself with pen and paper. It was fascinating to learn more about these puzzles - thank you for introducing me to them! (though it was also a bit frusrating to see a computer solve a oroblem in less than a second that took me much, much longer... )

Jerome said...

That program #4 that you sent us to is incredibly fast. I wonder what the algorithm is.

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