Ledge's Introduction To Cryptarithms III
CLASSICAL CRYPTOGRAPHY COURSE
September 11, 1996
ALL RIGHTS RESERVED
LEDGE'S INTRODUCTION TO CRYPTARITHMS III
It is again my distinct pleasure to present our guest
lecturer LEDGE's (Dr. Gerhard D. Linz) third and final
lecture on the interesting topic of Cryptarithms. In
this lecture, he covers Multiplication, Multiplicative
Structures, Base 11 and Base 12 calculations. LEDGE
natural writing style, and talent for making under-
standable some difficult concepts, makes this lecture
strong indeed. LEDGE has already produced one of our
better references on novice cryptography, and I
sincerely appreciate his assistance in our course.
NOMENCLATURE AND SYMBOLS
Lecture 15 included addition and multiplication tables
as well as digital squares and cubes for bases 10
through 16. For the additional numerical symbols
required for these bases above ten, it used A to
represent ten, B for eleven, C for twelve, D for
thirteen, E for fourteen and F for fifteen as needed. In
lecture 14 we used t for ten and e or E for eleven, the
t for bases 11 and 12 and e for base 12. That has been
the custom in the Cryptarithm column in The Cryptogram.
We will continue the latter usage in this lecture. The
usage in lecture 15 has the virtue of consistency as,
for instance, A is used for ten in all the higher bases.
Once understood, the tables should occasion no
difficulty. Furthermore, base 16 was called
"Sexdecimal." Those of you knowing some computer
programming recognize it as "Hexadecimal."
As we are restricted to ASCII symbols, we will be using
"*" as the symbol for multiplying and "**" for exponen-
tiation. Thus 3 * 4 is three times four and 4**3 is four
raised to the third power or four cubed.
In this lecture we will be looking at some more complex
cryptarithms: those involving roots of 2 and higher in
bases higher than 10, exponentiation, and base 10
problems that give minimal clues and require more of
what is called brute force methods. To aid our
understanding of cube roots we will first revisit square
root arithmetic to gain a deeper understanding of what
that procedure involves.
First, let's look at the extraction of a square root
using numbers rather than letters but presented in the
same form as a cryptarithm problem.
1 9 4 1
The difference in this presentation as compared with
that in the first cryptarithm lecture is that we do not
have the numbers at each level that were multiplied by
their respective digits in the answer. Thus after the
first level we see that 261 is to be subtracted from
276, but we do not know that it resulted from the
product of 9 times (20 * 1) + 9 or as we pointed out
before, b * ((20 * a) + b).
If you look closely at the process of extracting this
square root, you will see that it is a process of
continual refinement of the trial square root by
subtracting the increment added to the square of the
trial root successively from the original number. Having
marked off every two digits starting at the decimal
point, the process starts off with an approximation
using only the leftmost or highest order digits of the
original number and subtracts the highest number that
could be the square root of that digit or digits. In
this case the highest order digit(s) in the number is
the digit 3. It's square root is between 1 and 2.
Because the square of the root should not exceed the 3,
we choose the number 1 as the first digit of the
root and subtract it's square from 3.
Then we pull down the next pair of digits, 76. Now we
need to estimate the root of 376. For that we need a
second digit to the left of the 1. If we call the
first digit "a" and the second digit "b", we want the
highest possible number such that (a + b)**2 does not
Unless you are aware of it, you may not have recognized
that the number 1 in the quotient is no longer just
itself. It has become the highest order digit of a two
digit number. That means that it has become a ten. The
square that we are looking for has become:
(10a + b)**2
If you remember your algebra, you will remember that
when we expand this expression we get:
100a**2 + 20a*b + b**2
But 100a**2 is the square of the first number in the
trial root. We have already subtracted it from the
number for which we computing the square root and we
don't want to subtract it again. Hence we need the
number (10a + b)**2 - 10a**2 the incremental difference
b makes. In this case since b = 9, we would need to
compute 19**2 - 10**2 giving us 361 - 100 = 261,
and that is just the number below the 276. If you have a
calculator you can use (no, it isn't cheating), you can
perform that arithmetic process quickly and painlessly.
Having subtracted the 261 from 276, we bring down the
next pair of digits, the 85. Now we need the highest
root of 37685. It's at least 190 and no more than 199.
The example suggests 4 as the next trial digit. Now a =
190 and b = 4. We have to calculate the value of 194**2
- 190**2. You can see the value of having a calculator
here. It computes to 1536 which we can subtract from
1585 nicely. It's not too large or too small.
Now look what's happened from this viewpoint. We have
subtracted successively 10,000, 26,100, and 1,536 from
37,685. Those first three subtractions total 37,636
which, when subtracted from 37,685, leaves a remainder
of 49. You might also have noticed that 37,636 is the
square of 194.
There is one more detail to notice. In each subtraction
the units digit of the subtracter is the same as the
units digit of the square of the trial root digit b with
which we are working. 9**2 = 81 or 1 mod 10, the units
digit of 251. 4**2 = 16 or 6 mod 10, the units digit of
1536. If you are puzzled by that, look at how we came to
those subtracters. Except for the square of the trial
digit, all other products involve "a" which ends
A DUODECIMAL SQUARE ROOT
Now let's solve the duodecimal square root problem, C-6,
in the May-June, 1996, issue of The Cryptogram. It's by
ARIES and has a key that is two words, 0 - E.
Here is the problem:
N N C
I SL IF
I RB OT
1) Try to spot zero. Failing that, list all the letters
that cannot represent zero. The highest order digits
of the numbers cannot be zero. Numbers in the
quotient that produce non-zero subtracters cannot be
zero. So far, then, N, C, O and I are not zero. Next
look for numbers. either units digits that are not
zero or differences of zero. That adds T, L and A to
the list. Finally, add S from the last subtrahend,
ISLIF, since R is subtracted from it and does not
(cannot produce a carry to the next higher digit.
That leaves B, F, R and Y as the only candidates for
zero. Although the letter representing zero has not
been identified conclusively, the information so far
recovered will prove useful.
2) Next notice the units digit of the squares in the
root, The units digits of N**2 and C**2 are both T.
None of digits are zero. None of the squares unit
digits are N or C. Finally, both squares have the
same units digit. We know that N**2 is a two digit
number. Considering the length of the last of the
last subtracter, five digits, it is reasonable to
hypothesize that C**2 is also a two digit number. Now
look at the table of duodecimal squares given in
Lecture 15 with special attention to the two-digit
N 5 6 7 8 9 t e
n**2 21 30 41 54 69 84 t1
The squares of 6 and 9 do not meet the conditions - T is
not zero and a square cannot end with the digit that is
3) Now the subtracter, IBTT, can be calculated.
N(12) NN(12) NN(10) NN**2(10) N0**2(10)diff(10) diff(12)
5 55 65 4225 3600 625 441
8 88 104 10816 9216 1600 e14
t tt 130 16900 14400 2500 1544
The first column is the estimated value of the digit N.
The number in the parentheses is in each instance the
base, here 12. The next column is the entire trial root
base 12, NN, here 55. That's converted to base 10 in
the next column (5*12 + 5). In the next column that last
number is squared. The fifth column reports the base 10
square of a, 50 base 12 or 60 base 10. The next column
reports the difference of the two squares base 10. The
final column is the base 12 equivalent of the
difference. We compute the base 12 equivalent by
successive division of the base 10 number by 12 as
12/52 r1 625 = 52*12 + 1
4 r4 625 = 4*12**2 + 4*12 + 1
Starting with the last quotient and appending each
remainder from the last in turn to the first produces
441 as the proposed base 12 value of IBTT. As can be
seen, that value is much too small by one digit. When N
is 8, the value is still too small. N = 9 was
eliminated (remember why?). When t is used as the value
for N, the number IBTT becomes 1544. The repeated 4
clinches it as it matches the repeated T. Now I = 1, B =
5, T = 4, N = t, and C = 8. As a check, N**2 = CT or 84.
That's consistent with our result.
4) To find the value of F, we note that F - T = C base
12. substituting known values F - 8 = 4; hence F =
12 (base 10!!!) or F = 10 base 12 or 0 mod 12.
5) Knowing the value of the root, NNC, the value of the
last subtracter, IRBOT, is determined by computing
NNC**2 -NN0**2 as in the above tabular method. Do it.
You should get 12594 (1568**2-1560**2 base 10).
Remember tt8 base 12 converts to base 10 as 10*144 +
10 * 12 + 8.
6) From the last subtraction the values of S and L can
be found. From the other subtractions the values of
A and Y can now be identified. Putting all known
values in a key table produces ???.
The square root process can be extended to the
extraction of any higher order root, in the present
instance to cube roots. The process is again extending
trial cube roots one digit at a time for a closer and
closer approximation to the root. Since cube roots are
involved, the number whose root is to be extracted is
marked after every third digit from the decimal point.
The digit or digits before the last mark (the highest
order digits) provide the means of estimating a single
digit root. That digit should produce the highest cube
possible without exceeding the number made by the
highest digit(s). The cube of that digit, a, is then
subtracted from the number, and the next group of
letters is brought down. A second digit, b, is then
selected such that the cube of ab (not a * b) is as high
as possible without exceeding the number. That process
is continued until the units digit of the original
number has been brought down and the last increment
Since a is the first digit and b the second we need the
difference of (10*a + b)**3 - a**3. (Remember that the
10 in this instance represents the value of the base,
not decimal 10. 10 base 12 = 12 base 10.) Expanding the
above expression yields a longer expression to evaluate:
1000*(a**3) + (300a**2)*b + (300a) *b**2 + b**3 -
1000*(a**3). "b" can be factored from the result giving:
b*(300a**2 + 300a*b + b**2). Knowing a and b, the value
of that expression can be computed and then subtracted.
But it's easier to compute the unexpanded form as was
done with square roots.
A UNIDECIMAL CUBE ROOT
Now let's tackle an undecimal cube root presented in the
May-June issue of The Cryptogram by FIBBER. It has a key
of two words, 1-0. Here's the problem:
E L I
W AYA EST
W TIL PLA
1) Following the same steps as before try to identify
the letter representing zero, or at least the non-
zero letters. Here we are more fortunate than before.
I - Y = I. If Y were = t, borrowing from W of WIE
would be necessary. The evidence indicates no such
borrowing could have taken place; thus Y = 0. Along
the way we might notice that W - S = W. Since Y = 0
and we're working in base 11, S = t.
2) Now to identify the value for E. E**3 = WYT, a three
digit number whose second digit is zero and ends in a
digit different from E. In the table of unidecimal
cubes from lecture 15 we get:
N 5 9
N**3 104 603
These two are the only ones that meet all the discovered
criteria. Can we find other evidence to be able to
decide between them? It turns out we can.
When the T of WYT is subtracted from the E above it, the
remainder is W, i.e., E - T = W mod 11. We can make a
E T E-T W
5 4 1 1
9 3 6 6
Both values are consistent with the evidence. Hence E =
5 or 9. Since I - W -1 = 0 on the next subtraction, I =
W + 1. If E =5, W = 1 and I = 2. If E = 9, then W = 6
and I = 7. We'll carry both possibilities to the next
3) The next task is to identify L, if we can. L**3 ends
in E as we can determine from the second subtracter,
WSDEE. If E = 5, L**3 ends in 5. We look in the table
of cubes again and find only one cube that ends in 5,
namely 25, the cube of 3. So L = 3. If E = 9 the
cube of L must end in 9. There is again only one
such cube: 4**3 = 59, thus L = 4. So if E = 9, then L
= 4. There is no conflict.
4) Now it's possible to calculate WSDEE. It is EL**3 -
EL(11) EL(10) E0(11) E0(10) EL**3(10) E0**3(10) diff(10)
53 50 58 55 195,112 166,375 28737
94 103 90 99 970,299 729,000 241,299
You probably know the process involved for each step,
but here's the explanation if you don't understand it
all. Since E = 5 and L = 3, EL is 53 base 11. That's
converted to base 10 by computing 5*11 + 3 = 55 + 3 or
58 base 10. Similarly for E0 base 11 becomes 5 * 11 + 0
= 55. The cubes and the difference should be self-
explanatory. To compute the base 11 value of 28737 base
10 repeated division by 11 is necessary as follows:
11/2612 r5 28,737 = 2612*11 + 5
11/237 r5 28,737 = 237*11**2 + 5*11 + 5
11/21 r6 28,737 = 21*11**3 + 6*11**2 + 5*11 + 5
1 rt 28,727 = 1*11**4 + t*11**3 + 6*11**2 +
5*11 + 5 or 1t655
Hence, WSDEE = (starting with the last quotient and
going up the remainders) 1t655. Since W, S and E have
already been identified, D = 6 can be added to the list.
5) Now the remaining letters can be identified. They are
A, P and N and can be computed in that order from the
subtractions. It remains only to write out the key
6) Could we now compute the last subtracter even without
knowing the values of S, D, P, A, or N. The answer is
yes, of course, as we need only the values of the
digits in the root, 532. The subtracter is 532**3 -
530**3 base 11
ELI(11) ELI(10) EL0(10) ELI**3(10) EL0**3(10) diff(10)
532 640 638 262144000 259694072 2449928
1423738 checks out with the numerical equivalent of
WTILPLA. Remember to use successive division by the
base or 11 on the base 10 difference to recover the base
A FOURTH ROOT PROBLEM, BASE 15
The methods used on the square root and cube root
problems will work quite as well on higher order roots
and higher bases. To demonstrate the truth of that let's
look at the C-Sp-1 in the March-April, 1996, issue of
The Cryptogram by CROTALUS, the capable editor of the
Cryptarithm column. It's a fourth root problem in base
15 with a key consisting of three words, 1-0. You will
remember that base 15 requires 15 different numerical
symbols. The first ten are the digits from 0 to 9. The
other five are A, B, C, D, and E representing
respectively 10, 11, 12, 13, and 14. 10 base 15 = 15
base 10. Addition and multiplication tables for base 15
are contained in Lecture 15 as are the squares and cubes
of each of the digits. The digits to the fourth power
are not presented and will have to be calculated. That's
a little chore but not intrinsically difficult. The
simplest method is to raise the base 10 equivalent of
the digit to the 4th power and convert the result to
base 15 using successive division by 15 as was done for
bases 11 and 12. The resulting table is as follows:
N 1 2 3 4 5 6 7 8 9 A B C
N**4 1 11 56 121 2BA 5B6 AA1 1331 1E26 2E6A 4511 68C6
Here's the problem:
S L B
1) The non-zero letters are S, L, B, N, W, H, Y, G, and
2) We can spot the letter representing 1. It has to be
the B as the highest order digit of BWBAUHIPS.
3) When S is raised to the fourth power, the result is a
two-digit number, WH. Looking at the table above,
there is only one such two-digit number with two
different letters, namely 56. 3**4 = 56. Hence S =
3, W = 5, and H = 6.
4) Since W - Y = zero, there must have been a borrowing
in the previous column's subtraction and Y = W - 1 =
5 - 1 = 4.
5) B - R = B. R cannot = 0 else there would be no
necessary borrowing from the W in the next column. So
R = (base -1) or 14 or E.
6) In the first subtraction A - H = B or A - 6 = 1;
hence A = 7.
7) in the units place of the last subtraction S - B = U
or 3 - 1 = U; therefore U = 2.
8) We still have not identified the digit for L. The
subtracter associated with it is YRPOPB. It's unit
digit is B or 1. Hence L**4 end in 1. Looking at the
table, there are eight digits whose 4th power ends in
1. We have to look more deeply to determine the
correct one. We know the values of the first two
digits and the last digit of the subtracter, YRPOPB.
Substituting their values we obtain 4EPOP1. We can
approximate the base 10 value of that number by
expanding it: 4*15**5 + 14*15**4 + P*15**3 + O*15**2
+ P*15 + 1. The two highest terms of that expansion
are the most significant. They become 3,037,500 +
708,750 = 3,746,250. Following the model used
previously we know that the subtracter can be
calculate as SL**4 - S0**4. Since we do not know the
value of L we must assume one and try it out. Let's
take a number from the middle of the pack whose 4th
power ends in 1 as does the subtracter. L = 7 will do
as a first approximation.
9) Now for the calculation:
SL(15) SL(10) SL**4(10) S0(10) S0**4(10) diff(10)
37 52 7311616 45 4100625 3210991 too
38 53 7890481 45 4100625 3789856 4ECDC1
The first trial difference (base 10) was much below
3,746,250. The second trial difference, with L = 8, is
slightly more than the estimated subtracter as can be
expected since the less significant digits were ignored
in the estimation. Notice also the pattern of the
result. The C repeats as expected to match the repeat of
the P. P = C and O = D.
10) The key table has become 1 2 3 4 5 6 7 8 9 A B C D E
B U S Y W H A L P O R
The value of the rest of the letters can be computed
from the various subtractions in the problem. That's
left for you to finish.
Raising a number to a higher power, such as squaring
(2nd power), cubing (3rd power) or more has some facets
that can be helpful to a solution of a problem involving
integer exponents. Generally, such problems are
relegated to specials in the Cryptarithm section,
although problems involving the extraction of a root are
generally not unless they involve other complications.
JE SAURAIS contributed an exponentiation problem that
was published as a special in the March-April issue of
The Cryptogram. It was a base 10 problem. It's key was
one word, 0-1. At worst it could be solved by
anagraming, but that is a non-mathematical approach.
Here is the problem:
(ELT)**I = SLENTSGNI. (PRA)**N = NPARIA,
Problems like this can involve considerable amounts of
trial and error. A calculator (or a computer) can be
very helpful. The calculator need not be fancy. One
that can handle normal arithmetic operations of
addition, subtraction, multiplication and division is
adequate. Having one memory to store numbers can
make the process simpler. Such calculators are very
The problem, while it will involve some trial and error,
has much less of it than might be imagined at first
glance. There are more clues than initially meet the
eye. First we notice that the exponents, I and N, are
digits, i.e., integers having a value of 2 to 9. Next we
could count the number of digits in each number. In
each case the number to be raised to a power has three
digits. In the first equation the result is a nine
digit number. In the second a six digit one. Let's
examine that more closely.
A two digit number can be as small as 10 and as large as
99. When squared (or raised to the 2nd power) they
result in 100 and 9801, Either three or four digits. No
square of a two digit number can have fewer or more
digits. A three digit number can be as small as 100 and
as large as 999. Their squares are 10,000 and 998,001,
either five or six-digits. Notice that there is no
overlap on the number of digits in the length between
powers. We find a similar situation with the cube (3rd
power) of those four numbers. 10**3 = 1000 and 99**3 =
970,279: from four to six digits long. 100**3 =
1,000,000 and 999**3 = 997,002,999: from seven to nine
digits long. Again there is no overlap between powers.
A six-digit number must be the square of a three-digit
number or the cube of two-digit number. There is in fact
a general rule about the number of digits in the result
when a number of known length, L, is raised to a power,
P. The maximum length of the result, R-max, is P*L. The
minimum length, R-min, is L*(P-1) + 1.
We can apply that information to the above problem. In
the second equation, L = 3, R = 6, and power = N. Using
the equation for R-max, 6 = 3 * N; hence N = 2. For the
first equation, L = 3, R = 9 and power = I. Again using
the equation for R-max, 9 = 3 * I or I = 3. If we had
seven digits in the result of the second equation and
ignored the first equation, we would have solved the
equation 7 = 3 * N and N would be greater than 2 (2.33)
but not more than 3.
We could then safely deduce that N = 3. If we wanted to
check on the lower bound of N, we could have used the
equation for R-min. The above formulas work for any
integer power and any length of the original number.
The second equation contains the letters P, R, and A in
both numbers and I in the result, a known number (3).
PRA is a number that when squared produces a number
whose highest order digit is 2. Another way of saying
that is it produces a number between 200,000 and
299,999. The square roots of these numbers extend
from 447 to 547. That's 101 numbers to try, if we need
to. But we don't. We can narrow the search much more
than that. The six-digit result starts NP and three
digit base stars with P. We have just found out that P
must be 4 or 5 (447-547). Hence the range of the six-
digit result is from 240,000 to 259,999. The
square roots of these two numbers extend from 490 to
509, a range of only twenty numbers, quite a reduction
from 101. Yet, we can even do better than that. Both
numbers, base and result end in A so that A**2 = A mod
10. If A were zero, the result of squaring the number
would have two zeros at the end. It does not. So A = 1,
5 or 6. Now we have only six numbers to try that are in
the correct range and end with a possibly correct digit:
491, 501, 495, 505, 496 and 506. We are looking for a
number that has the pattern of NPARIA or 24AR3A OR
25AR3A. Here are the results:
ELT 491 501 495 505 496 506
ELT**3 241081 251001 245025 illegal 246016 256036
Only the last square gives the correct pattern. Now we
know that R = 0, P = 5, and A = 6.
Our key table is 0 9 8 7 6 5 4 3 2 1
R A P I N
Now let's look at the cube. T**3 must = I mod 10 since I
is the units digit of the result. I = 3; the only digit
that when cubed ends in three is 7 (check the unidecimal
table in Lecture 15); hence T = 7. The largest eight
digit number is 99,999,999. Since the result is a nine
digit number, the base that produced it must be larger
than the cube root of 99,999,999. That cube root < 465.
The highest order digit of the base, E can be 9, 8, or
4. The last digit, T is 7. The second digit of the base
is the same as the second digit of result.
Now let's use intelligent trial and error. For ELT, E =
4, 8, or 9: L = 1, 4 ,8 or 9 and must differ from E; and
T = 7. The possible values for ELT are as follows:
487 497 817 847 897 917 947 987
That's only 8 numbers to try. 947**3 = 849,278,123.
Match that with the pattern of SLE,NTS,GNI. S = 8, L =
4, E = 9, N = 2, T = 7, G = 1, I = 3. Add those which
are new to the key table and read the resulting word. If
you have followed the reasoning and understand it,
congratulations. Perhaps in the future you will say to
yourself, "I can probably do that."
The major lesson to have learned is this: when faced
with trial and error, try to limit as much as you can
the range of the possible. In the last part of this
problem we had identified six of the digits, leaving
only four to choose among. Further we able to determine
that for ELT, E was had only three possible values, L
had four and T was identified. Without any clues, ELT
could range from 102 to 987, a range of 886
possibilities less those numbers that have a repeated
digit. We were able to reduce that number of permissible
values to just eight.
A MORE DIFFICULT ADDITION
Equations and additions can produce more challenges for
the solver because often very few if any of the
numerical equivalents of the letters can be identified.
Algebraic equations involving the digits can be written.
These equations are often helpful, but much trial and
error is still required. Trial and error is often called
brute force, because, while it must be systematically
applied, it does not require much deep thinking. Yet is
does require some, as we discovered in the previous
Here is an addition problem in base 10 provided by THE
RAT as C-11 in the May-June, 1996, issue of The
Cryptogram. The key is two words, 9-0.
+ SNAKE +GO
1) We can identify the following non-zero letters: S, G,
L, O, R, E, and probably T. That leaves K, N, and A
as the candidates for zero.
2) From the second addition, we can identify S = 1.
3) Also from the second addition, since L cannot be zero
(since it's the highest order digit of LO), L = 9.
4) There are two digital sums that could be useful: O +
O = G mod 10 and E + E = K mod 10. In each case the
sum has to produce a carry of 1 so that L + 1 = 0 mod
10. Hence neither E nor O can be less than 5. G
cannot equal zero so O must be 6 or more.
5) Since L + K + 1 = 10 + K, T + A + 1 = G and T + N (+
1?) = K.
6) We now have enough information to produce a table of
known and unknown values to try out, remembering L =
9 and S = 1.
O G E K (T, A) N
6 2 5 0
7 4 5 0
8 6 5 0
In case it's not clear, we start each line with a
possible value of O: 6, 7, or 8. O cannot be 5, nor
9(L). In each case G = O + O mod 10. Then we start with
the smallest possible value for E, 5 and add the
resultant value for K which is 0. It turns out that E
can be 5 for each value of O. When O is 6, E cannot be 6
but it can = 7. Nor can E be 8 as that would make K = 6.
It turns out that E can be 6 or 8 only when O = 7. When
O = 8, E can be 5 or 7.
7) Since K occurs four times in the problem, and the
value of zero works well for it in each place and
occurs in three positions of the table, I have a
preference for trying those places first.
8) On the top line where O = 6 and E = 5, T + A + 1 = G:
2 or 12. Because 2 and 0 are assigned to G and K. T
+ A = 11. Therefore the pair, T, A, can be 8,3; 7,4;
or 6,5. We do not know which letter represents which
digit. 6,5 is not permissible since O = 6 and E = 5.
The two other pairs produce a carry to the next
addition: T + N + 1 = K: 10 or T + N = 9. For the 8,
3 pair N = 1 or 6, since 8 + 1 = 9 and 3 + 6 = 9.
Nether value of N is permissible. For the 7,4 pair N
= 2 or 5, neither of which is permissible. So we
abandon O = 6 for the moment. Not permissible
means that the results conflict with assignments
already made to other letters.
For O = 7, T + A + 1 = 4 mod 10 or T + A = 3, 13.
Only 13 is permissible. It is produced by 6, 7; 5,
8; or 4, 9. Since on this line 4, 7, 5, and 9 are
already in use, this solution is not permissible. It
For O = 8. T + A + 1 = 6(G) mod 10 or T + A = 5, 15.
15 can be produced by 8 + 7 but 8 is already
assigned. 5 = 2 + 3. No problem. T, A are 2, 3 but
maybe not in that order. With T + A = 5 there is no
carry to the next addition; hence T + N = 10(K).
Since 8 is already assigned, T = 3 and N = 7. A must
= 2. No contradictions so far.
9) Let's construct our partial key table and then go
back to the problem, if everything looks OK.
9 8 7 6 5 4 3 2 1 0
L O N G E T A S K
The only letter left to place is the letter, R, which
must be 4.
You can substitute all the digits in the problem and
check the answer.
Sometimes it takes courage to tackle a cryptarithm,
particularly if it might take you to less well known
territory. My best advice is to forge ahead. You cannot
lose. Either you will solve the problem and perhaps be
surprised by your competence or ingenuity, or you will
find yourself stumped, needing to learn something new.
Look at a book, or consult with a mathematically
inclined friend or a friendly math teacher, someone who
can point the way or find a fallacy. That way, you learn
and add something you had not known to your
armamentarium of mathematical strategies.
You are of course welcome to contact me with a problem,
a success, or a new wrinkle you've discovered. If there
are problem types you'd like me to write more about, let
me know. Phrase any questions as clearly as you can,
and I'll see what I can do with them. There's no sin in
being stumped - our hobby sees to it that we run into
that situation with regularity.
DOUBLE KEY DIVISION
>From time to time a division problem is presented which
has two sets of substitutes for the digits, one upper
case and one lower case. The sets are complete but keyed
differently. The problem is often done in base ten, but
occasionally in base 11 or twelve. Such a problem was
presented as a special in the September - October, 1994,
issue of The Cryptogram. A base twelve problem with two
words, 0-1, and FOUR WORDS, 0 to E, was propounded by
ARIES. It's presented here written in standard
r h l d b
i l l a d/G O L D E N A G E
i l l a d
A Y Y G L N
l y d t i d
U M U Y Y A
r h i b h y
U L O N A G
r l r e h h
S N Y L B E
d r u c u h
S L D O U
As in any division problem, we have a series of
multiplications, or products, and a series of
subtraction (or additions). The subtractions involve
both sets of letters, but, interestingly, the
multiplications involve only the lower case letters. We
cannot do the subtractions without both sets of letters.
We can, however, attack the multiplications by
considering only the lower case letters. Let's see what
can be done to identify the numerical equivalents of the
lower case letters.
As usual, the first effort is to find the letters
representing 1 and 0. The letter representing 1 is easy
to find: r * illad = illad; so r = 1. The letter for
zero is hidden a little better. In the third
subtraction, A - y = A; so y = 0. Our equivalent table
0 E X 9 8 7 6 5 4 3 2 1
So the first of the two words starts with y and the
second ends with r. A double-key division problem
usually has a lot of products. This one is typical. That
characteristic allows us to build a partial
multiplicative structure to which we can look for the
appropriate diagram (see Lecture 14). Since we are
interested at this point only in the units digits of the
products, we will use modulo 12 multiplication.
h * illad = lyttid or h * d = d or h => d. Likewise l =>
y; d => h; and b => h. We can combine this information
for the following partial structure:
b => h <=> d and l => y. Since we know that y = 0, l * d
= 0; or, finally, l => 0.
Having this much information, we can now look at the
base 12 structures in lecture 14. Look them over
yourself. There are two possible matches. Can you spot
them? It could be a good exercise to stop and try this
on your own.
The two that match produce the following table:
d h b l
3 <=> 9 <= E, 7 4, 8
8 <=> 4 <= 2, 5, E 3, 6, 9
In each case there is only one possible value for d and
h on each line. To narrow the possibilities, choosing a
product with most letter equivalents partially
identified will provide the quickest entry. Such a
product is the second one: h * illad = lydtid. The
inverse of that is lydtid/h = illad. In other words, if
the product is divided by the single digit multiplier
that produced it, the divisor of the problem which
served as the multiplicand should emerge.
Before doing that, there is more useful information in
the problem that may result in the elimination of some
of the possibilities. Notice that two of the products
have r as their highest order digit Since r = 1, the
digit multipliers that produced them must be smaller
than the other two letters that also produced 6-digit
products. So l and d < h and b. In the first group h (9)
is always greater l (4, 8) and d(3). On the other hand,
if l = 4, b = 7 or e. If l = 8, b can only be E.
In the second set d > h. But d should be < h. Hence the
second set cannot be correct. So we will confine our
attention to the first set. Back to the proposed
division: h = 9; lydtid is l03ti3. l can be 4 or 8. One
of those must be correct. It's a 50/50 chance to guess
correctly. Let's start with l = 4.
********* Remember this is base 12. 4 x 12 + 0 is 48.
9 / 403ti3. 9 goes into 48, 5 times, giving 9 * 5 = 45.
39 base 10 or 39 base 12 (3 * 12 + 9 = 45).
*** Hence i = 5. Subtracting 9 from 0 or 12 =
33 3. 39 divided by 9 = 4. 4 x 9 = 36 or 30
30 base 12. Hence l = 4. Since we are looking
** for illad, we hope the next quotient will
3? be 4 also. Though we don't know t's
** equivalent, we do know we can subtract 30
again! 403ti3 has become 403t53. illad is
544a3. We move to the end of this
division. 9 * 3 = 27 or 23 base 12. Hence
the previous subtraction must have produce
a 2 in its units place.
Since the units digit in the dividend is 5, 9 * a = 3
mod 12. The multiplicative structure and the
multiplication table both show that the only possible
multipliers of 9 that fit are 7 and E. 7 * 9 = 53 base
12, making t = 5, not possible since i = 5. E * 9 = 83,
making t = 8 and a = E. The completed partial product
is: 9 * 544E3 = 403853. The equivalent table now
0 E X 9 8 7 6 5 4 3 2 1
y a h t i l d r. Letters without values are b,
c, u and e.
"b" is the units value of the quotient. b * 544E3 =
31ucu9. 31 base 12 or 37 base 10 divided by 5 = 7; thus
b = 7. If you carry out the multiplication, you will
discover the values of u and c. That would leave only
one place for e. If you can do a little anagraming, you
can read the key without those last computations.
We were fortunate. Had we chosen l = 8, we would soon
have run into contradictions leading to the discarding
of that possibility.
We have identified all the lower case letter equivalents
and not yet one single upper case equivalent. Now that's
just a matter of solving five subtraction problems. That
shouldn't prove too difficult and will be fine base 12
Errata for Lecture 14
a) In the explanatory paragraph below the duodecimal
Example 4. Division, after writing down the 32 we must
first subtract it from 48, making 16 the difference, and
then bring down the next digit of the dividend, t.
b) The final product of the duodecimal multiplication at
the end of the lecture, before the Appendix, is 7C8e8t8
not 7C8et8 as written. Unfortunately this typo was not
spotted in time before publication.
LECTURE 17 ANSWERS
17-1 Headline Puzzle
Paul Derthick's HEADLINE PUZZLE . by Larry Gray
The following are all headlines from a recent daily
newspaper. Each of the five is a different mono -
alphabetic substitution, and all five are derived from
the same mixed alphabet at different settings against
1. PXYWFXKLJE DFYMJYV VGHKJ `DFYM-US' GF ZYFGJVG
2. JUBHFGO EUHKEOF HR WEUDBGO, FHSJF DKD RO ZGI YRE
3. NEZZY AEZYVKU AEVP NFUVLKY LR ALVVKU JLBPV ECKU
4. ZEHCGOL LZCCOMMSS WEMSAQ MZALD AFB AZFMS MZ DZBZA
5. PTQQU WQRKWCQBSD WQEKLLQUBX BZOKWEQ MKW ENJWSQX JUB
Setting = ANOLE Key = GECKO Hat= CHAMELEON
17-2 Playfair. While Rome Burns. BARRISTER: ON44:CE17
Tip= "ers are"
OCMAF ZDAPZ BYPGY BOKYT BYVMT AVIBY PVGPP RBCFH
XEAPI VTCPV VBKGV MEWCB IEGMQ PPBOL ENRHZ MRFSC
DRNAI ZEITN SUNA.
TWO HINTS: The title is significant and does not follow
LANAKI's Red Herring rule and look for naturals such as
PO = QP or OPQ. A Natural is a cipher digraph not in
the keyword whose letters because of the standard
alphabetical relationships stay in the natural
alphabetical order in the cipher square.
H U C K L
E B R Y F
I N A D G
M O P Q S
T V W X Z
Message: Pupil's answers are que(x)er, to wit: Nero was
a cruel tyrant who would torture his poor subjects by
playing the violin.
17-3 Foursquare 'anasonly' ZEMBIE
UB XB MS SF SQ MS TH DE UB HM GL NL BW GB LW NQ NF UB FM
QH EM BW BI GT LD UQ IG WM CF TQ ET CT NF IP LS UQ FK UH
IZ UQ YF TN XP NS FF UV HV NF HI CE NQ UO UQ GK ET HT ND
PV BI BE ND BD YM DE LX UB GA CX ET XT DE PE NL BF PY IQ
NG QW IS NC CK XB TF GK ED LA EL LE RW MI EX SF MS UP XQ
NF EV FF BI KK NA MX.
Answer in complex Caesar: (ISUPV OMPAY - UGBSK NGKPN)
ZKXJO GGKGN SFXPC DYJKP MRPPJ
hints: run down 10 letters ; Look for thinks 2x, germs
2x, if all fails - square to = i/juxtaposition, square 4
17-4 Short Bifid. Clue - DIAMONDS is there somewhere and
the text talks about them being HIDDEN. Period = 7.
ETIALIG LDMNITV NFEMISI EEIDGEI
HPCEDUT PINOFLW INDLEEK
ANS: The diamonds are hidden in the side pocket of me
LECTURE 18 PROBLEMS
18-1 Unidecimal square root. (Three words 0-E) MARSHEN
LO'SE gives root it; - KF = EKSE; - ERRE = EWH
18-2 Duodecimal division. (Two words, 0-E) CODEX
BRIDGE / CLUBS = CC; - DUHRE = BRHEE; - DUHRE = BOLO
REFERENCES AND CRYPTOGRAPHIC RESOURCES
The CDB (Crypto Drop Box) was updated last week with
140,000 bytes of references. I will update them again
after Lecture 19 is complete.
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