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Cryptarithmetic Problems; Everyone's favorite mathematical puzzle
Topic Started: 24 Jul 2009, 06:36 AM (70,616 Views)
VirtualSerendipity
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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+(N-1)=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
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uberking
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. . 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
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--made by download 'cuz I has no pro GIMP skills :P
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VirtualSerendipity
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Feel free to post a new problem when you solve one. Here's another problem:

. TAKE
. . . . A
+CAKE
. KATE
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download
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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.
-download | the administration

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chezemaster
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Cheesy Responses requires response from the Cheesy Master of Doom
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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>
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download
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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
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Please help me :(

TWO + THREE = FIVE ????
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download
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markomitr
10 Apr 2010, 11:34 AM
Please help me :(

TWO + THREE = FIVE ????
Your problem isn't possible, as "THREE" has five characters, while the sum, "FIVE," only has four.
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ashish
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cross+roads=danger
cross+roads=danger
ashish
28 Oct 2010, 01:49 PM
cross+roads=danger
cross+roads=danger
Edited by ashish, 28 Oct 2010, 01:55 PM.
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anal
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Link 4 all the solution ::: B)
http://bach.istc.kobe-u.ac.jp/cgi-bin/crypt/crypt.cgi?crypt=no%2Bno%3Dyes
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Red Daisuke
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anal
15 Dec 2010, 01:38 PM
That pretty much takes the fun out of it. :huh:
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Amethyst
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This is very cool stuff....unfortunately I'm not sure how to solve these.
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Jak Atackka
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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.
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download
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Jak Atackka
21 Oct 2011, 03:49 PM
I've got one - CARBON + DIOXIDE = WARMING
I've got one - CARBON + DIOXIDE = WARMING
Try solving without the tool.
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
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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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