This tutorial will be very helpful in solving those questions.I myself solved those 3 questions just by reading this tutorial. you won't find any help regarding this topic anywhere else so please read this tutorial.
Rewrite
the problem, expanding the interlinear space to make room for
trial numbers that will be written under the letters.
For example, the puzzle SEND + MORE = MONEY, after solving, will appear like
this:
trial numbers that will be written under the letters.
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.
Ease
the analysis of subtractions by reading them as upsidedown 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
C1,
C2, C3 and C4 are the carryovers of "0" or "1" that are to
be added to the next column to the left.
A
good hint to find zero or 9 is to look for columns containing two or three
identical letters.
Look at these additions:
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 carryover of "1".
Look
for left hand digits. If single, they are probably "1".
Take the world's most famous cryptarithm:
Take the world's most famous cryptarithm:
S E N D
+ M O R E

M O N E Y
"M"
can only equal 1, because it is the "carry 1" from the column S+M=O
(+10). In other words, every time an addition of "n" digits gives a
total of "n+1" digits, the left hand digit of the total must be
"1". 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)
8 x 6 = 8 (+40)
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
then
A=0 or A=5
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 longhand division:
To see how this works, let's attack this longhand 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 Y5=5 we get Y=0. Replacing all Y's with zero, multiplication I will be:
K x 5 K
5 = D 0 N 5
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,
generateandtest
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 generateandtest method, which consists of the following procedures:
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 generateandtest method, which consists of the following procedures:
 1. List all digits still unidentified;
 2. Select a base variable (letter) to start generation;
 3. Do a cycle of
generation and testing: from the list of still
unidentified digits (procedure 1) get one and assign it to
the base variable; eliminate it from the list; proceed guessing values for
the other variables; test consistency; if not consistent, go to perform the
next cycle (procedure 3); if consistent, stop: you have found the solution
to the problem.
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
Obviously
C1=1 and C3=1. Solving the equation system we get K+C=8: not much, but we
discovered a relationship between the values of "K" and "C"
that will help us later. But now we are stuck! It's time to use the "generateandtest"
method. Procedure 1: digits 2,3,4,5,6,7 and 8 are still unidentified;
Procedure 2: we select "K" as the base variable;
Procedure 2: we select "K" as the base variable;
CYCLE #1, procedure 3: column TCK shows that T+C=K and no carry, hence "K" must be a high valued digit. So we enter the list obtained through procedure 1 from the high side, assigning "8" to the base variable "K".
Knowing that K+C=8, if K=8 then C=0. But this is an unacceptable value for "C", because the addend "CAKE" would become "0981" and cryptarithmetic conventions say that no number can start with zero. So, we must close this cycle and begin cycle #2.
By now, the addition layout and the table summarizing current variable data would look like this:
T 9 8 1
CYCLE A E K C T
9 ========================
+ 0 9 8
1 #1 9 1 8
[0]

8 9 T 1
Conflicting
values for variables are noted within square brackets.
CYCLE
#2,
procedure 3: assigning "7" to the letter "K" we get C=1
because K+C=8. This is an unacceptable value for "C" considering that
we have already fixed E=1. Again we have to close the current cycle and go to
cycle #3, with the setup and table showing:
T 9 7 1 CYCLE A
E K C T
9
========================
+ 1 9 7
1
#1 9 1 8 [0]
 #2 9 1 7 [1]
7 9 T 1
CYCLE
#3,
procedure 3: assigning "6" to the letter "K" we get C=2
because K+C=8. Testing these values for "K" and "C" in the
column TCK, we get C3+T+2+=6 making T=3.Now, testing T in column KKT, we would
obtain C1+K+K=T+10 or 1+6+6=T+10, making T=3. This is an acceptable value for
T, confirming the previous value T=3 we had already found.
So, we have got the final solution to the problem, stopping the routine "generateandtest".
The final layout and table would read
So, we have got the final solution to the problem, stopping the routine "generateandtest".
The final layout and table would read
3 9 6 1 CYCLE A E K
C T
9 ========================
+ 2 9 6
1 #1 9 1
8 [0]
 #2 9 1 7[1]
6 9 3
1 #3 9 1
6 2 3
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