Posted 24 Jan 2013 (edited) Edit. Removed misleading information from the original post. I ran across this puzzle - it's part of a larger puzzle - a week ago. There are twelve letters, each of which corresponds to one of eight digits. The letters are juxtaposed into two-digit numbers, and they occur that way in six equations. Here are the equations - six, in twelve unknowns. Their integer properties provide additional constraints. VG = CV - QP JY = GP - ZY QG = VJ - MV XC = BC - XP XK = KX - QB ZP = CY - GM Can you find consistent values of the 12 letters B C G J K M P Q V X Y Z using just eight digits? Edited 24 Jan 2013 by bonanova Remove incorrect information from original post 0 Share this post Link to post Share on other sites

0 Posted 25 Jan 2013 (edited) I found couple solutions, but not with as many digits, I'm afraid. P=0; Q=Z=1; X=3; J=M=Y=4; K=5; B=G=V=6; C=7. Then: 66 = 76 - 10; 44 = 60 - 14; 16 = 64 - 46; 37 = 67 - 30; 35 = 53 - 16; 10 = 74 - 64; Everything checks out. (In the octal system, of course. Could that fact be used as an argument for using 8 digits?) Another solution where all letters are equal to zero works in any system, but uses just one digit and requires no work to attain. Edited 25 Jan 2013 by Prime 0 Share this post Link to post Share on other sites

0 Posted 24 Jan 2013 Please check the equations for typos. I wrote a program to look for solutions and it found none. I looked for bugs to no avail. It is certainly is easier to check the equations for typos than to find an easily overlooked bug. 0 Share this post Link to post Share on other sites

0 Posted 24 Jan 2013 Edit. Removed misleading information from the original post. I ran across this puzzle - it's part of a larger puzzle - a week ago. There are twelve letters, each of which corresponds to one of eight digits. The letters are juxtaposed into two-digit numbers, and they occur that way in six equations. Here are the equations - six, in twelve unknowns. Their integer properties provide additional constraints. VG = CV - QP JY = GP - ZY QG = VJ - MV XC = BC - XP XK = KX - QB ZP = CY - GM Can you find consistent values of the 12 letters B C G J K M P Q V X Y Z using just eight digits? Can we assume, there are no leading zeroes in any of those 2-digit numbers? 0 Share this post Link to post Share on other sites

0 Posted 24 Jan 2013 When I tried to work this out, I hit a dead end pretty quickly. Makes me think this isn't solvable. Maybe I'm missing something obvious... BC - XP = XC tells me that P=0 If P=0, CV - QP = VG means V=G and GP-ZY=JY means Y=5 and then CY - GM = ZP means M=5 So...P=0, Y=5, M=5, V=G VJ - MV = QG becomes GJ - 5G = QG...that means J = 2G, therefore G must equal 1, 2, 3, or 4 But if G<5, we can't do the subtraction problem GJ - 5G = QG unless Q is a negative integer. 0 Share this post Link to post Share on other sites

0 Posted 24 Jan 2013 (edited) ... So...P=0, Y=5, M=5, V=G VJ - MV = QG becomes GJ - 5G = QG...that means J = 2G, therefore G must equal 1, 2, 3, or 4 But if G<5, we can't do the subtraction problem GJ - 5G = QG unless Q is a negative integer. G doesn't NEED to be 1, 2, 3, or 4. If G < 5 Then J = 2G Else J = 2G-10 In the same way that 14 - 7 = 7 If, for example, G = 7, J = 4, and Q = 1, then the equation GJ - 5G = QG (74 - 57 = 17) would work. Sorry, had trouble with the spoilers. Edited 24 Jan 2013 by BobbyGo 0 Share this post Link to post Share on other sites

0 Posted 25 Jan 2013 Now that I took few minutes (enough to prohibit editing previous post) to study the statement of the problem closely, I find my solution (above) answers OP's requirements in full. 0 Share this post Link to post Share on other sites

0 Posted 25 Jan 2013 Now that I took few minutes (enough to prohibit editing previous post) to study the statement of the problem closely, I find my solution (above) answers OP's requirements in full. Yes. It does. Great solve. To clarify, the OP might better have read: Each letter has a value selected from just eight digits. The actual wording, and the title, suggests (wrongly) that eight digits are used. 0 Share this post Link to post Share on other sites

0 Posted 25 Jan 2013 Nothing wrong with the wording. You didn't give away the trick, but you didn't hide it. It's just that many of us (*palm*) weren't open minded. I, too, couldn't get anything but all zeroes. Good puzzle! Thanks again for the education! 0 Share this post Link to post Share on other sites

0 Posted 25 Jan 2013 I like the original wording. It is fair and accurate. It does not actively mislead, rather allows solvers to mislead themselves. It could use no leading zeroes stipulation to prohibit all zero solution, but then it would simplify solving process quite a bit. The fact that some people suspected an error in the problem statement, while Bonanova remained silent on the subject served as a hint to me. The title could be “6 equations, 12 unknowns,” or even more catchy “6 against 12,” to attract more attention. Great puzzle! 0 Share this post Link to post Share on other sites

Posted (edited)

Edit. Removed misleading information from the original post.

I ran across this puzzle - it's part of a larger puzzle - a week ago.

There are twelve letters, each of which corresponds to one of eight digits.

The letters are juxtaposed into two-digit numbers, and they occur that way in six equations.

Here are the equations - six, in twelve unknowns.

Their integer properties provide additional constraints.

VG = CV - QP

JY = GP - ZY

QG = VJ - MV

XC = BC - XP

XK = KX - QB

ZP = CY - GM

Can you find consistent values of the 12 letters B C G J K M P Q V X Y Z

using just eight digits?

Edited by bonanovaRemove incorrect information from original post

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