HOW TO SOLVE A PUZZLE
1. Preparation
For
example, the puzzle SEND + MORE = MONEY, after solving, will appear like this:
S E N D
9 5 6 7
+ M O R E
1 0 8 5
---------
M O N E Y
1 0 6 5 2
2. Remember cryptarithmetic conventions
Each letter or symbol represents only one digit throughout the
problem;
When letters are replaced by their digits, the resultant arithmetical operation
must be correct;
The numerical base, unless specifically stated, is 10;
Numbers must not begin with a zero;
There must be only one
solution to the problem.
3. See subtractions as "upside-down"
additions
Ease the analysis of subtractions by
reading them as upside-down additions. Remember that you can check a
subtraction by adding the difference and the subtracter to get the subtrahend:
it's the same thing. This subtraction:
C O U N T
- C O I N
---------
S N U B
Must be read from the bottom to
the top and from the right to the left, as if it were this series of additions:
B + N = T + C1
U + I = N + C2
N + O = U + C3
S + C = O + C4
4. Search for
"0" and "9" in additions or subtractions
A good hint to find zero or 9 is to look
for columns containing two or three identical letters.
Look at these additions:
* * * A * * * B
+ * * * A + * * * A
------- -------
* * * A * * * B
The columns A+A=A and B+A=B
indicate that A=zero. In math this is called the "additive identity
property of zero"; it says that you add "0" to anything and it
doesn't change, therefore it stays the same.
Now look at those same additions in the
body of the cryptarithm:
* A * * * B * *
+ * A * * + * A * *
------- -------
* A * * * B * *
In these cases, we may have
A=zero or A=9. It depends whether or not "carry 1" is received from
the previous column. In other words, the "9" mimics zero every time
it gets a carry-over of "1".
5. Search for "1" in additions or
subtractions
Look for left hand digits. If single, they
are probably "1".
Take the world's most famous cryptarithm:
S E N D
+ M O R E
---------
M O N E Y
In
this Madachy's subtraction problem, "C" stands for the digit
"1":
C O U N T
- C O I N
---------
S N U B
6. Search for "1" in multiplications or
divisions
In this multiplication:
M A D
B E
-------
M A D
R A E
-------
A M I D
The first partial product is E
x MAD = MAD. Hence "E" must equal "1". In math jargon this
is called the "identity" property of "1" in multiplication;
you multiply anything by "1" and it doesn't change, therefore it
remains the same.
Look this division:
K T
--------
N E T / L I N K
N E T
-------
K E K K
K T E C
-------
K E Y
In the first subtraction, we
see K x NET = NET. Then K=1.
7. Search for "1" and "6" in
multiplications or divisions
Any number multiplied by "1" is
the number itself. Also, any even number multiplied by "6" is the
number itself:
4 x 1 = 4
7 x 1 = 7
2 x 6 = 2 (+10)
Looking
at right hand digits of multiplications and divisions, can help you spot digits
"1" and "6". Those findings will show like these ones:
C B
----------
* * A * * A / * * * * *
B C * * * C
------ ---------
* * * C * * * *
* * * B * * * B
--------- -------
* * * * * * * *
The logic is: if
C x * * A = * * * C
B x * * A = * * * B
then A=1 or A=6.
8. Search for "0" and "5" in
multiplications or divisions
Any number multiplied by zero is zero.
Also, any odd number multiplied by "5" is "5":
3 x 0 = 0
6 x 0 = 0
7 x 5 = 5 (+30)
9 x 5 = 5 (+40)
Looking at right hand digits of
multiplications and divisions, can help you spot digits "0" and
"5". Those findings will show like these ones:
C B
----------
* * A * * A / * * * * *
B C * * * A
------- ---------
* * * A * * * *
* * * A * * * A
--------- -------
* * * * * * * *
The logic is: if
C x * * A = * * * A
B x * * A = * * * A
9. Match to
make progress
Matching is the process of assigning
potential values to a variable and testing whether they match the current state
of the problem.
To see how this works, let's attack this
long-hand division:
K M
----------
A K A / D A D D Y
D Y N A
---------
A R M Y
A R K A
-------
R A
To facilitate the analysis,
let's break it down to its basic components, i.e., 2 multiplications and 2
subtractions:
I. K x A K A = D Y N A
II. M x A K A = A R K A
III. D A D D
- D Y N A
---------
A R M
IV. A R M Y
- A R K A
---------
R A
From I and II we get:
K x * * A = * * * A
M x * * A = * * * A
This pattern suggests A=0 or
A=5. But a look at the divisor "A K A" reveals that A=0 is
impossible, because leading letters cannot be zero. Hence A=5.
Replacing all A's with "5",
subtraction IV becomes:
5 R M Y
- 5 R K 5
---------
R 5
From column Y-5=5 we get Y=0.
Replacing all Y's with zero, multiplication I will be:
Now,
matching can help us make some progress. Digits 1, 2, 3, 4, 6, 7, 8 and 9 are
still unidentified. Let's assign all these values to the variable K, one by
one, and check which of them matches the above pattern. Tabulating all data, we
would come to:
K x 5K5 = D0N5
----------------------
1 515 515
2 525 1050
3 535 1605
4 545 2180
6 565 3390
SOLUTION --> 7 575 4025 <-- SOLUTION
8 585 4680
9 595 5355
----------------------
You can see that K=7 is the
only viable solution that matches the current pattern of multiplication I,
yielding:
K x A K A = D Y N A
7 5 7 5 4 0 2 5
This solution also identifies
two other variables: D=4 and N=2.
10. When stuck, generate-and-test
Usually we start solving a cryptarithm by
searching for 0, 1, and 9. Then if we are dealing with an easy problem there is
enough material to proceed decoding the other digits until a solution is found.
This is the exception and not the rule.
Most frequently after decoding 1 or 2 letters (and sometimes none) you get
stuck. To make progress we must apply the generate-and-test method,
which consists of the following procedures:
1. List all digits still unidentified;
2. Select a base variable (letter) to start generation;
To
demonstrate how this method works, let's tackle this J. A. H. Hunter's
addition:
T A K E
A
+ C A K E
----------
K A T E
The column AAA suggests A=0 or
A=9. But column EAEE indicates that A+E=10, hence the only acceptable value for
"A" is 9, with E=1. Replacing all "A's" with 9 and all
"E's" with 1, we get
T 9 K 1
9
+ C 9 K 1
----------
K 9 T 1
Letter repetition in columns
KKT and TCK allows us to set up the following algebraic system of equations:
C1 + K + K = T + 10
C3 + T + C = K
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