rookie1ja 14 Posted March 31, 2007 Report Share Posted March 31, 2007 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 This old topic is locked since solution is already provided in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. No. 10 - solution 11^{3} = 1331 Link to post Share on other sites

Guest Posted September 8, 2007 Report Share Posted September 8, 2007 (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 Link to post Share on other sites

Guest Posted September 8, 2007 Report Share Posted September 8, 2007 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! Link to post Share on other sites

Guest Posted September 26, 2007 Report Share Posted September 26, 2007 this one is too easy... 11^3 = 11 * 11 * 11 = 11 * 121 = 1331 Link to post Share on other sites

Guest Posted September 26, 2007 Report Share Posted September 26, 2007 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 Link to post Share on other sites

Guest Posted January 30, 2009 Report Share Posted January 30, 2009 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 Link to post Share on other sites

Guest Posted April 9, 2009 Report Share Posted April 9, 2009 (edited) 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 A^{2} 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)}(B^{2}) B = 1 + 2log_{A*A} B Which can only be true if B = 1 and A can be any number. Edited April 9, 2009 by James T Link to post Share on other sites

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