Friday puzzle -- Mirage

Re: Friday puzzle -- Mirage

Cool. Maybe I can answer this one. MIRAGE is an optical phenomenon that creates the illusion of water, often with inverted reflections of distant objects, and results from distortion of light by alternate layers of hot and cool air. Also, a hotel in Vegas.

Hint

The first equation says a lot about M and I, or at least narrows the possibilities.

Try to figure out the carry digit in each place in the second equation.

A start

_******

+MIRROR

=IMAGE*

From the left-most column, we know that I>M

That means that in the second column, the answer must be 10+M, so there is a carry in the first column.

So I >= M+2 (the asterisk is non-zero, plus the carry).

_MIRROR

+IMAGE*

=*EGAMI

Let Cn be the carry digit (0 or 1) in the nth column from the left.

Can C1 = 1? That would mean that in column 2, C2+I+M >= 11, since E is non-zero. That would mean that I+M >= 10, which we know from column 1 is not the case. So C1 = 0.

From column 1, we know that 9 >= M+I

From column 2 and the fact that C1=0, we know that E>M

Use that fact to examine column 5, and determine C4.

I'll stop here for now. If anyone is still working on this, feel free to post the rest, else I will late tonight.

The rest (warning -- tedious to put it mildly!)

From the "start" post, we know that

I>=M+2

I+M<=9

C1>1

E>M>I

Now looking at the fifth column of

_MIRROR

+IMAGE*

=*EGAMI

Since E>M, this column must generate a carry one for column 4.

Columns 3 and 4 are the next keys.

C3+R+A =G or 10+G, and C4+R+G=A or 10+A

only works if Cn+R = 5 and A and G are different by 5, e.g., 5+2=7 and 5+7=12.

So we know that C3=C4=1, R=4, and A and G are different by 5. And since column 4 generates a carry , column three must not, so C2=0, and G=A+5, specifically, (A,G) = (1,6), (2,7), or (3,8)

Now, let's look at what I can be. I>=M+2>=3. If I=3, M=1, and from column 2, E=I+M=4, but R=4. Therefore I>4, and I>R.

That means that column 6 does not produce a carry, so from column 5, O+E=10+M. From column 2 we have I+M=E. Substituting, we have

O+(I+M) = 10+M, or O+I=10.

We already know I>4. It can't be 5 or O would also be 5. It can't be 6, or O would be 4=R. And it can't be 9, since I+M<=9, so I = 7 or 8, with corresponding values of O = 3 or 2.

These alternatives eliminate two of the choices for (A,G), leaving only (1,6)

With A=1, M>1, so I, which is equal to E-M, is LT 8

So we have I=7, O=3, M=2, E=9, A=1, G=6, R=4, and the only possible equations are

447265 + 274434 = &21699, and 274434 + 721693 = 996127, which have no zeros, and

MIRAGE = 274169

This one was painful! :-)