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Replace the same characters by the same numerals so that the mathematical operations are correct.

(AA)B = ABBA

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No. 10 - solution

113 = 1331

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• 5 months later... (AA)(multiplied by itself B times) = ABBA

A=2 B=1

2x2 to the 1st = 2x1x1x2=4

AxA to the B = AxBxBxA

A=2 B=2

2x2 to the 2nd =2x2x2x2=16

AxA to the B = AxBxBxA

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##### Share on other sites The ABBA doesn't imply multiplication as variables next to each other usually do. In these types of problems, each letter is assumed to be in position as if base ten numbers had been transposed with different letters of the alphabet. For example: 16 could be represented by AB, FD, UR, LH, etc. but not by AA. The puzzle is asking you which 2-digit multiple of 11, when raised to a single digit power, yields a 4-digit answer whose first and last digits are the same as those in the multiple of 11 and whose middle two digits are the power it was raised to. For example, raising 11 (represented by "AA") to the 2nd power (represented by "B") would give 121 (or "ABA"). We need a four digit answer though, so we need to either choose a higher multiple of 11, a higher power, or both. Happy solving!

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• 3 weeks later... this one is too easy...

11^3 = 11 * 11 * 11 = 11 * 121 = 1331

##### Share on other sites this one is too easy...

11^3 = 11 * 11 * 11 = 11 * 121 = 1331

And then you beat me to it a second time! I'm impressed

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• 1 year later... I thought the same thing, and that is true, as long as ABBA=A*B*B*A. Then I realized ABBA is a 4-digit number w/ with the 2nd and 3rd being the same and the 1st and 4th being the same.

(AA)(multiplied by itself B times) = ABBA

A=2 B=1

2x2 to the 1st = 2x1x1x2=4

AxA to the B = AxBxBxA

A=2 B=2

2x2 to the 2nd =2x2x2x2=16

AxA to the B = AxBxBxA

THANK YOU THANK YOU

I ACCEPT DONATION

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• 2 months later... No. 10 - Back to the Cryptograms and Algebra Puzzles

Replace the same characters by the same numerals so that the mathematical operations are correct.

(AA)B = ABBA

Now let's say instead of the letters being the positions of digits in the number, let them to multiplied against themselves.

Thus we have (A*A)B = A*A*B*B

taking the log base A2 of both sides:

log(A*A)(A*A)B = log(A*A)(A*A*B*B)

B = log(A*A)(A*A) + log(A*A)(B2)

B = 1 + 2logA*A B

Which can only be true if B = 1 and A can be any number.

Edited by James T
##### Share on other sites This topic is now closed to further replies.