Saturday, May 4, 2013

Crypt-arithmetic e litmus questions abc*def


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

. . 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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