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Cryptarithmetic Problems; Everyone's favorite mathematical puzzle  

Topic Started: 24 Jul 2009, 06:36 AM (72,365 Views)  
VirtualSerendipity  24 Jul 2009, 06:36 AM Post #1 
Established Member

What is a cryptarithmetic problem? It is a mathematical puzzle in which each letter represents a digit (for example, if X=3, then XX=33). The object is to find the value of each letter. No two letters represent the same digit (If X=3, Y cannot be 3). And the first letter cannot be 0 (Given the value ZW, Z cannot be 0). They can be quite challenging, often involving many steps. Here's an example, illustrating how to solve them: . SEND +MORE MONEY M must be 1. If you'll notice, this is an addition problem; the sum of two four digit numbers can't be more than 10,000, and M can't be 0 according to the rules since it's the first letter. So now you have: . SEND +1ORE 1ONEY Now in the column S1O, S+1≥10. S must be 8 (if there is a 1 carried over from the column E0N) or 9. O must be 0 (if S=8 and there is a 1 carried or S=9 and there is no 1 carried) or 1 (if S=9 and there is a 1 carried). But 1 is already taken, so O must be 0. . SEND +10RE 10NEY There can't be a carry from the column E0N, because any digit plus 0 < 10, unless there is a carry from the column NRE and E=9; but this cannot be the case, because then N would be 0, and 0 is already taken. So E<9 and there is no carry from this column. Therefore, S=9, because 9+1=10. . 9END +10RE 10NEY In the column E0N, E cannot be equal to N, so there must be a carry from the column NRE; E+1=N. We now look at the column NRE; we know that E+1=N. Since we know that there is a carry from this column, N+R=1E (if there is no carry from the column DEY) or N+R+1=1E (if there is a carry from the column DEY). Let's try out both cases. No carry: N+R=10+(N1)=N+9 R=9 9 is already taken, so this won't work. Carry: N+R+1=N+9 R=8 This must be the solution for R. . 9END +108E 10NEY The digits we have left are 7, 6, 5, 4, 3, and 2. We know there must be a carry from the column DEY, so D+E>10. N=E+1, so E can't be 7 because then N would be 8 which is already taken. D is at most 7, so E cannot be 2 because then D+E<10, and E cannot be 3 because then D+E=10 and Y=0, but 0 is taken already. Likewise, E cannot be 4 because if D>6, D+E<10, and if D=6 or D=7, then Y=0 or Y=1, which are both taken. So E is 5 or 6. If E=6, then D=7 and Y=3, so this part works. But look at the column N8E. Remember, there is a carry from the column D5Y. N+8+1=16 (because we know there is a carry for this column). But then N=7, and 7 is taken by D. Therefore, E=5. . 95ND +1085 10N5Y Now that we've gotten this important digit, it gets much simpler from here. N+8+1=15, N=6. . 956D +1085 1065Y The digits left are 7, 4, 3, and 2. We know there is a carry from the column D5Y, so the only pair that fits is D=7 and Y=2 . 9567 +1085 10652 And Voila! The problem is solved! These are quite tricky and require some thinking, but are lots of fun. Now we'll take turns posting problems. When a problem is solved, you may post another problem. Be sure to show your work; you don't have to write detailed instructions like I did, but show your steps at least, since figuring out how to solve the problem is the most important part. I'll start things off here: . . EAT +THAT APPLE 
uberking  24 Jul 2009, 08:53 PM Post #2 

. . EAT +THAT APPLE I started off knowing that T is not 0, as T+T is not equal to T. Same case with A. From know on, I'm gonna show the steps to my answer. Don't ask HOW I got them, it's too much work that I did on a piece of paper. And I can't type all that. . . EA9 +9HA9 APPLE . . 8A9 +9HA9 APPL8 . . 819 +9H19 APPL8 . . 819 +9H19 APP38 . . 819 +9219 A0038 . . 819 +9219 10038 FINAL ANSWER: . . EAT . . 819 +THAT +9219 APPLE 10038 
made by download 'cuz I has no pro GIMP skills  
VirtualSerendipity  25 Jul 2009, 03:34 AM Post #3 
Established Member

Feel free to post a new problem when you solve one. Here's another problem: . TAKE . . . . A +CAKE . KATE 
download  29 Nov 2009, 06:47 AM Post #4 
Founder

3961+9+2961=6931 I have to admit i cheated. There's quite a few programs out there that solve cryptarithmetic problems. I wonder how they do it... By the way, here's another problem. BASE+BALL=GAMES Edited by download, 30 Nov 2009, 04:49 AM.

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chezemaster  7 Dec 2009, 02:31 AM Post #5 
Cheesy Responses requires response from the Cheesy Master of Doom

I can't solve that problem NO + NO = YES 
That, by the way, was my Educated Guess Don't try to be witty and write sarcastic things when you write things. If you want to be sarcastic at least do this: <sarcasm>***insert sarcasm here***</sarcasm> <useful tag> just trying to be useful </useful tag> Inkscape is equal to or larger than GIMP  
download  7 Dec 2009, 02:39 AM Post #6 
Founder

There's a lot of possibilities to your problem: 79+79=158 69+69=138 78+78=156 67+67=134 86+86=172 76+76=152 85+85=170 65+65=130 64+64=128 54+54=108 93+93=186 73+73=146 53+53=106 92+92=184 82+82=164 52+52=104 
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markomitr  10 Apr 2010, 11:34 AM Post #7 
New Member

Please help me TWO + THREE = FIVE ???? 
download  10 Apr 2010, 11:29 PM Post #8 
Founder

Your problem isn't possible, as "THREE" has five characters, while the sum, "FIVE," only has four. 
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ashish  28 Oct 2010, 01:49 PM Post #9 
New Member

cross+roads=danger cross+roads=danger
Edited by ashish, 28 Oct 2010, 01:55 PM.

anal  15 Dec 2010, 01:38 PM Post #10 
New Member

Link 4 all the solution ::: http://bach.istc.kobeu.ac.jp/cgibin/crypt/crypt.cgi?crypt=no%2Bno%3Dyes 
Red Daisuke  2 Jan 2011, 01:32 PM Post #11 
The Green Arrow

That pretty much takes the fun out of it. 
Amethyst  20 Oct 2011, 03:11 AM Post #12 
New Member

This is very cool stuff....unfortunately I'm not sure how to solve these. 
Jak Atackka  21 Oct 2011, 03:49 PM Post #13 
Wise Guy

I've got one  CARBON + DIOXIDE = WARMING I've got one  CARBON + DIOXIDE = WARMING Try solving without the tool. Edited by Jak Atackka, 21 Oct 2011, 03:49 PM.

download  21 Oct 2011, 03:52 PM Post #14 
Founder

Very interesting problem  one question first, is there only ONE solution? Seems like there are too many variables for there to be one solution; then again, if all the letters represent a different number that'd work. 
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Jak Atackka  21 Oct 2011, 04:20 PM Post #15 
Wise Guy

By each variable representing a different number, do you mean that, for example, the two Ds in DIOXIDE are two different digits? If so, no. Also, since there are 13 different variables, some variables might have the same value. 
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