How do you solve ABCD x 9=DCBA   SOLVE BY ALPHAMETRIC

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Request for Question Clarification by tutuzdad-ga on 15 Dec 2002 15:01 PST

I can solve it for you without explanation if that will suffice as an
answer?

Otherwise, detailing the procedure for reaching the answer to a
mathematical problem such as this would likely take more time than a
question associated with this price normally takes.

Regards;
tutuzdad-ga

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Clarification of Question by fabiana40-ga on 15 Dec 2002 16:45 PST

Then please cancel my question. Thank you.

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Clarification of Question by fabiana40-ga on 15 Dec 2002 16:46 PST

Please cancel this question. Send e-mail that this has been canceled. Thank you.

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Request for Question Clarification by tutuzdad-ga on 15 Dec 2002 20:54 PST

Dear fabiana40-ga;

I hope my suggestion to adjust your price didn’t affect your decision
to proceed with your question, but if you are dead set on canceling
your question, I’d rather give you a correct answer than have you
leave here with information that is less than reliable.

I’m sure the term you intended to use was “alphanumeric”.

The answer to ABCD x 9 = DCBA is: 1089 x 9 = 9801

Why? 

Because in this problem, ABCD x 9 = DCBA, specifically:
A always =1
B always =0
C always =8 
and D always =9

How do I know this? Divide 9801 by 9. It equals 1089.

To verify, compare the list of letters represented by numbers and
apply the same theory to the number 1089:

“A” is still represented by “1” 
“B” is still represented by “0”
“C” is still represented by “8” 
and “D” is still represented by “9”

Thus (ABCD x 9 = DCBA) = (1089 x 9 = 9801). See what I mean? 

Now, you have the correct answer. You can go ahead and cancel your
question if you still wish to do so. You can consider this one an
early Christmas present from me to you. Does this answer it for you?

Thanks for using Google Answers.

Tutuzdad-ga
GOOGLE ANSWERS RESEARCHER

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Request for Question Clarification by tutuzdad-ga on 15 Dec 2002 20:57 PST

In case this was unclear consider this:

A always =1 
B always =0 
C always =8  
and D always =9 

...with both numbers, 1089 and 9801.

Perhaps this would be a better way to explain it.

Regards
tutuzdad-ga

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Clarification of Question by fabiana40-ga on 16 Dec 2002 05:30 PST

Thank you. Have a nice holiday to you to.

Apparently your question has not yet been canceled. Unless "alpha
metric" is a math discipline of which I am not familiar, either A,B,C
or D needs to be zero or infinity for the equation to balance in usual
algebra. Reversing the order DCBA does nothing as multiplication is
implied.   Neil

Hi, neilzero:

In these kinds of problems, ABCD is interpreted as a four digit number
with A,B,C,D representing distinct digits.  Hence reversing the order
to DCBA will generally produced a different value; multiplication is
not implied.  However it is often assumed that different letters
represent different digits.  That additional assumption was not
required in tutuzdad-ga's solution here.

-- mathtalk-ga

However one does need to assume the convention that the leading digits
are not zero (if not that different letters imply different digits),
for otherwise one can have a "trivial" solution with all digits zero.

-- mathtalk

A method of solution:

The digit A cannot be zero and must be less than two, since ABCD * 9
is again a four digit number (it would be five digits if A were two or
more).  Thus A = 1.

Since A = 1, D must be such that D*9 ends in 1.  Only D = 9 satisfies
this.

Now since D = 9 and A = 1, we see that in multiplying ABCD * 9, there
cannot be any carry from B*9 into the fourth column of the result. 
Therefore B is zero or one.  If we assume "different letters are
different digits" (almost a universal condition on such problems),
then immediately B must be zero since already A is one.

For complete generality we can show that there actually is no solution
with B = 1.  For if the two digit number AB were eleven, then this
would force AB * 9 to be 99, and there would be no room for a carry
from CD * 9 into the third column (since that would promote the answer
to five digits).  Since already D = 9, the only choice for C such that
CD * 9 would be a two digit number is C = 0.  But then this would be
false:

ABCD * 9 = (1109) * 9 = 9981, not 9011 = DCBA

Therefore B = 0.

Only the digit C remains to find, such that:

10C9 * 9 = 9C01

which we can treat as a linear equation to solve for C:

90C + 1009*9 = 100C + 9001

10C = 9081 - 9001 = 80

C = 8

tutuzdad has already verified that this "solution" works:

ABCD = 1089, DCBA = 9801 and 1089 * 9 = 9801

regards and holiday best wishes,
mathtalk-ga