Brain Teasers
Language Math
Cryptography
In Cryptography teasers, a phrase or expressions has been encoded in some way (frequently by replacing letters with other letters). You need to figure out the encoding method and then decode the message to find the answer.Cryptography
Professor Abacus has earned degrees in both Mathematics and English. With this education, he has made a study of language math. In today's lesson, Professor Abacus has stated that not only does "3 + 8 + 9 = 20", but also that "THREE + EIGHT + NINE = TWENTY". Substitute the numerals 0 through 9 for the letters to make the equation work. Each letter can represent only one numeral, and each numeral can only be represented by one letter. There are no leading 0's. Since there are only nine letters, one numeral will not be used. So as to produce only one possible solution, G is greater than R.
THREE + EIGHT + NINE = TWENTY
THREE + EIGHT + NINE = TWENTY
Answer
19288 + 85491 + 3538 = 108317Hide Answer Show Answer
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Comments
Jan 12, 2005
i did'nt get it . can u plz explain it properly.
This is just a test message don't know if HTML tags show in comments... trying to figure out how to format a long comment... Moderators can feel free to delete this comment... Sorry
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You asked for it, vamsee
Unfortunately, I can't make newlines in teaser comments, so this won't be pretty, but I'll explain how I arrived
at the solution, in case there are others who wonder, as well. There may be easier ways about it, but this is one
approach...
Just to explain my use of it (since I don't know what it may actually mean in real math), if I'm refering to
multiple possibilities for a letter, I'll list the options in square brackets, separated by commas. For example,
if A can be either 1 or 5, I would say something like A+B=[1,5]+B. Also, I'll refer to the sum of the corresponding
digits as being in columns - i.e. the rightmost column is E+T+E=Y. Soooo, on to the solution..
First, a couple basic observations - adding together 3 different, individual letters, the highest number we can have
is 9+8+7=24, so the greatest amount that can carry is 2, and the greatest sum of a given column of 3 unique letters is
26 (with the carry over).
Looking at the left-most column in the equation, we see that T+E+[0,1,2]=TW. Even assuming the maximum,
9+8+2 = 19, so T = 1 (no leading 0). From there, 1+E+[0,1,2] = TW, so 10
Unfortunately, I can't make newlines in teaser comments, so this won't be pretty, but I'll explain how I arrived
at the solution, in case there are others who wonder, as well. There may be easier ways about it, but this is one
approach...
Just to explain my use of it (since I don't know what it may actually mean in real math), if I'm refering to
multiple possibilities for a letter, I'll list the options in square brackets, separated by commas. For example,
if A can be either 1 or 5, I would say something like A+B=[1,5]+B. Also, I'll refer to the sum of the corresponding
digits as being in columns - i.e. the rightmost column is E+T+E=Y. Soooo, on to the solution..
First, a couple basic observations - adding together 3 different, individual letters, the highest number we can have
is 9+8+7=24, so the greatest amount that can carry is 2, and the greatest sum of a given column of 3 unique letters is
26 (with the carry over).
Looking at the left-most column in the equation, we see that T+E+[0,1,2]=TW. Even assuming the maximum,
9+8+2 = 19, so T = 1 (no leading 0). From there, 1+E+[0,1,2] = TW, so 10
Hmm... Got cut off... Continuing on, from the beginning of that sentence... From there, 1+E+[0,1,2] = TW, so 10
Hmm... Okay, I can't post my explanation, because of html issues... Lemme try this - You asked for it, vamsee
Unfortunately, I can't make newlines in teaser comments, so this won't be pretty, but I'll explain how I arrived
at the solution, in case there are others who wonder, as well. There may be easier ways about it, but this is one
approach...
Just to explain my use of it (since I don't know what it may actually mean in real math), if I'm refering to
multiple possibilities for a letter, I'll list the options in square brackets, separated by commas. For example,
if A can be either 1 or 5, I would say something like A+B=[1,5]+B. Also, I'll refer to the sum of the corresponding
digits as being in columns - i.e. the rightmost column is E+T+E=Y. Soooo, on to the solution..
First, a couple basic observations - adding together 3 different, individual letters, the highest number we can have
is 9+8+7=24, so the greatest amount that can carry is 2, and the greatest sum of a given column of 3 unique letters is
26 (with the carry over).
Looking at the left-most column in the equation, we see that T+E+[0,1,2]=TW. Even assuming the maximum,
9+8+2 = 19, so T = 1 (no leading 0). From there, 1+E+[0,1,2] = TW, so 10 <= TW <= 12 (since 1+9+2=12),
W = [0,2] (Since T=1), and 7 <= E <= 9 (since 1+7+2=10).
From the rightmost column, we see that Y = (2*E+1) MOD 10. If E = [7,8,9], then Y = [5,7,9] respectively.
Both can't be 9, so neither Y nor E is 9, and one or the other is 7 (i.e. 7=[Y,E]).
Going back to the earlier equation, 1+E+[0,1,2] = 1+[7,8]+[0,1,2] = [8,9,10,11].
Based on this, W=0 (since 2 was the only other option).
Moving one column to the right, H+I+N = [E,E+10,E+20]. Since 2+3+4 = 9, and E+20 > 26, that must mean H+I+N = E+10.
Therefore, (first column again) E = 10-1-1 = 8. From earlier, this means that Y = 7.
In the second to last column, we now have H+N+8+1=21 (can't be 11, since H & N would both have to be 1), so H+N = 12.
This means [H,N] = [3,9] (either letter is either digit). At this point, 2, 4, 5, & 6 are left for R, G, and I. Since
R+G+I+2=?N, and 2+4+5+2 = 13 <= R+G+I+2 <= 17 = 4+5+6+2, it follows (from R+G+I+2=N+10) that N = 3, H = 9, and
[R,G,I] = [2,4,5] (6 is unused).
Since R+G+I+2=13 (the one carries), that means I = 18-H-N-1 = 18-9-3-1 = 5. According to the original problem
text, R < G, so R=2, and G=4.
So, in the end we have (in order of discover) T=1, W=0, E=8, Y=7, N=3, H=9, I=5, R=2, G=4, and
19288 + 85491 + 3538 = 108317
Unfortunately, I can't make newlines in teaser comments, so this won't be pretty, but I'll explain how I arrived
at the solution, in case there are others who wonder, as well. There may be easier ways about it, but this is one
approach...
Just to explain my use of it (since I don't know what it may actually mean in real math), if I'm refering to
multiple possibilities for a letter, I'll list the options in square brackets, separated by commas. For example,
if A can be either 1 or 5, I would say something like A+B=[1,5]+B. Also, I'll refer to the sum of the corresponding
digits as being in columns - i.e. the rightmost column is E+T+E=Y. Soooo, on to the solution..
First, a couple basic observations - adding together 3 different, individual letters, the highest number we can have
is 9+8+7=24, so the greatest amount that can carry is 2, and the greatest sum of a given column of 3 unique letters is
26 (with the carry over).
Looking at the left-most column in the equation, we see that T+E+[0,1,2]=TW. Even assuming the maximum,
9+8+2 = 19, so T = 1 (no leading 0). From there, 1+E+[0,1,2] = TW, so 10 <= TW <= 12 (since 1+9+2=12),
W = [0,2] (Since T=1), and 7 <= E <= 9 (since 1+7+2=10).
From the rightmost column, we see that Y = (2*E+1) MOD 10. If E = [7,8,9], then Y = [5,7,9] respectively.
Both can't be 9, so neither Y nor E is 9, and one or the other is 7 (i.e. 7=[Y,E]).
Going back to the earlier equation, 1+E+[0,1,2] = 1+[7,8]+[0,1,2] = [8,9,10,11].
Based on this, W=0 (since 2 was the only other option).
Moving one column to the right, H+I+N = [E,E+10,E+20]. Since 2+3+4 = 9, and E+20 > 26, that must mean H+I+N = E+10.
Therefore, (first column again) E = 10-1-1 = 8. From earlier, this means that Y = 7.
In the second to last column, we now have H+N+8+1=21 (can't be 11, since H & N would both have to be 1), so H+N = 12.
This means [H,N] = [3,9] (either letter is either digit). At this point, 2, 4, 5, & 6 are left for R, G, and I. Since
R+G+I+2=?N, and 2+4+5+2 = 13 <= R+G+I+2 <= 17 = 4+5+6+2, it follows (from R+G+I+2=N+10) that N = 3, H = 9, and
[R,G,I] = [2,4,5] (6 is unused).
Since R+G+I+2=13 (the one carries), that means I = 18-H-N-1 = 18-9-3-1 = 5. According to the original problem
text, R < G, so R=2, and G=4.
So, in the end we have (in order of discover) T=1, W=0, E=8, Y=7, N=3, H=9, I=5, R=2, G=4, and
19288 + 85491 + 3538 = 108317
Can a moderator/editor/whoever delete this message, as well as my first 3 on this teaser, which are incomplete?
Jan 19, 2005
....wow....very impressive.... yet interesting in a weird sense.
Jan 21, 2005
Wow GTregay....You put alot of time into that......
?????????????????????
GTregay, you sound VERY smart.
Sooooooooo confusin man!!!
Well this was a "math-class" type of teaser, and a easy one too Keep up the good work
Okay, I still don't get it.....
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