Cryptarithms are mathematical ciphers or puzzles where letters are used in place of numbers. One letter is used to represent one and only one number. Their use is more for a mathematical solving exercise of enjoyment rather than secret message communication.

Our *Cm *constructions often use a keyword or phrase to key the problem. It has the same number of letters as the base of the numbering system. Ten letters are used to represent our normal decimal system of ten numbers. The keyword letter order is indicated by 0-9, 9-0, 0-1 or 1-0. This will become clearer as we work through a construction. Let’s begin with an addition problem in a cryptarithm construction format. Problems are published in line form to save space in the *Cm *column.

### Addition. (Two words, 0-9 Key Order.)

SONFTS + FUYPTC = RTTYCUS

Conventional addition format.

SONFTS +FUYPTCRTTYCUS Key Order _ _ _ _ _ _ _ _ _ _ 0 1 2 3 4 5 6 7 8 9

#### Give Aways

Let’s look for the letters that are given away in the problem. R = 1 as a left hand digit carry in the sum. C = 0 since S + C = S. Because zero and nine are the only two numbers possible to produce like numbers in a same column and N + Y = Y, N must equal 9.

Our key is now: C R _ _ _ _ _ _ _ N 0 1 2 3 4 5 6 7 8 9

Keep in mind that two words form this key. At this point anagramming can help in the key recovery and mathematical solution to the cipher.

We can rule out the letters FSPTY being the number eight in the key as each are unlikely to precede the letter N, our number nine value. The letters O or U must be equal to eight. Since T + T = eight in column four of the addition, it appears that U will equal 8 and T will equal 4. Our updated key:

C R _ _ T _ _ _ U N 0 1 2 3 4 5 6 7 8 9

We know that the sum of column three in our problem ends in zero. The only two numbers remaining with a sum of zero are 7 (F) and 3 (P). Our key is now:

C R _ P T _ _ F U N 0 1 2 3 4 5 6 7 8 9

It is now fairly simple to locate the three remaining letters:

C R Y P T O S F U N 0 1 2 3 4 5 6 7 8 9

A check of the mathematical values will confirm that we have the correct key.

SONFTS 659746 FUYPTC +782340RTTYCUS 1442086

Apply these same principles to the following subtraction problem:

#### C-1. Subtraction. (Two words, 1-0) LIONEL

NUBSRMU – JOEOSY = MENRRU

Let’s continue our cryptarithm discourse with a discussion of the process of solving multiplication and division construction types. Remember, each letter that appears in the cryptarithm is used to represent one number and one number only. Let’s begin with a multiplication problem.

#### C-1. Multiplication. (Two words, 9-0) LIONEL

FSNI * TNIS = NGPPIS; + GNORS; + FSHI; = TGTOYPIS

We write this in multiplication order. The first point we notice is a four-digit multiplicand with only three product line in the problem. This alerts us that a zero is somewhere in the multiplicand.

FSNI ×TNISNGPPIS GNORSFSNI___TGTOYPIS

The zero (S) becomes obvious as we see its placement in the problem. Since the third product line has the same letters as the multiplicand, T must equal 1. In our addition, P + S = P, confirms that S must equal zero. G + F = TG on the bottom line, F must equal 9. Let’s put each of these letters into our 9-0 key.

Key order: F _ _ _ _ _ _ _ T S 9 8 7 6 5 4 3 2 1 0

Let’s look at I × I = I in our first multiplying activity. There are only three numbers that are multiplied by its same number that will generate the identical last digit, 1, 5 and 6. The digit, 1, has been already used. Let’s focus on the 5 and 6 digits.

We need a number that will generate two identical digits when multiplied by the next number(s) in our multiplicand. Work through the multiplication of the number, 5, through

all of the yet to be used digits in our key for the value of N. (We already know that the letter after N in the multiplicand is zero (S). You will find no possible combination of 5 × N and S that will generate identical digits in the first product line. But, 6 × 5 (N) and zero (S) will yield 33. We have added three more letters to our key.

Key order: F _ _ I N _ P _ T S 9 8 7 6 5 4 3 2 1 0

Anagramming will complete the key as we only have four remaining letters to be placed, R, Y, G and O. Check the arithmetic of your solution.

#### C-2. . Division. (Two words, 1-0) LIONEL

SOUTHERN / STRIFE = FRS; - STRIFE = IUETRRN; - IORFRHR = FEEUNN.

We write this in a natural division order and notice a three-digit divisor with only two lines

of subtraction. This alerts us that a zero is somewhere in the divisor.

FRSSTRIFE)SOUTHERNSTRIFEIUETRRNIORFRHRFEEUNN

The zero jumps out at us in the first line of subtraction. E – E = R can only equal zero.

An identical divisor and multiplicand on the first line avows the first dividend digit as 1.

The fifth column of the second subtraction, R – R = U must be a nine, for zero is spoken for. We have two of the values for the second column of the second subtraction, leaving letter 0 as 8. Continue the logic to complete the solution.

F _ _ _ _ _ _ O U R 1 2 3 4 5 6 7 8 9 0

We remember from our grammar school mathematics’ primer that a number’s square results when we multiply it by itself. Conversely, we can define a number’s square root as the number multiplied by itself, with its difference added to the product, is equal to the original number. We’ll reach into the ACA *Cm *archives to explain the principles of square root procedures. Our *Cryptogram *Ye-Ed at the time, PHOENIX, presented a nine-step presentation of the process in the SO ‘94 *Cm* that we will use as the foundation for our discourse on the square root cryptarithm.

First of all, keep in mind all of the tips that we have reviewed to identify the numbers, one,

nine and zero. Also, be on the alert for digits squared that generate the same last digit in the squares. (C-1) In order for one to be alert to all of the tip aids in a square root problem, an understanding of the square root mathematical process is necessary

**Square Root Procedures by PHOENIX, ***The Cryptogram, *SO 1994

To find the square root of a number:

- Mark off the number in groups of two figures in both directions from the decimal point. Add a zero to the last decimal group if necessary.
- Determine the greatest perfect square of the leftmost group and write it below. Write its square root above the group.
- Subtract the perfect square from the leftmost group and bring down the next group adjacent to the remainder to form a partial dividend.
- Multiply the root already found by 20 for a trial divisor. Divide it in to the dividend. The whole number quotient is the next digit of the root being determined.
- Add the last digit to the trial divisor to form a complete divisor. Multiply it by the last root digit and write the product under the dividend.
- Subtract this product from the dividend; bring down the next group to form a new dividend.
- Reiterate steps 4 through 6 until the remainder is zero, or until as many places as needed have been determined.
- Add zeroes to the decimal part as needed. The decimal of the root goes directly above the decimal in the number.
- If at any time the partial divisor is greater than the first part of the dividend, write a zero as the corresponding figure of the root. Draw down the next group of figures and proceed as before.

We will use these square root procedures to find the square root of 526930.81.

7 2 5 .9√52’69’30.81’49369 Back to the square root definition – 7 squared, plus its 142 284 remainder, 3, represents the square root of 52. Follow 8530 this problem’s mathematics to its conclusion. 14457225130581 14509130581

Let’s try a square root without the given numbers.

#### SR-1. Square root cryptarithm. (Two words, 0-1) LIONEL

U V√ RU’ERDUSFERVGADFI ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ 0 9 8 7 6 5 4 3 2 1

Do you recognize that the letters S and F are giveaways? Other known quantities: U squared generates a final U digit; R is one greater than D and E is one greater than G. (Why?)

We’ll continue our cryptarithm cipher discussion with a look at the approach of solving a cube root problem. Before we can apply any of the cryptic solution principles we have learned to a cube root, we must be fully aware of the mathematical procedure of solving one.

Once again we reach back to PHOENIX’s discussion of the methodology of root solutions

in the SO 1994 *Cm*, this time as it pertains to a cube root

.

**To Find the Cube Root of a Number:**

- Mark off the number in groups of 3 figures in both directions from the decimal point. Add zeroes to the last decimal group if necessary.
- Determine the greatest perfect cube of the leftmost group and write it below; write the cube root above the group.
- Subtract the greatest perfect cube from the leftmost group and bring down the next group adjacent to the remainder to form a partial dividend.
- Multiply the root (squared) already found by 300. Use this number to divide the new dividend. The whole number part of the quotient is a good guess for the next root number. If this trial root generates a number larger than the existing dividend, use the possible root minus 1. Note: the trial root in one part of the example was too large.
- Add 30 times the product of the root already found and the trial figure, and the square of the trial figure, to the partial divisor. Multiply it by the last root digit and write the product under the dividend.
- Subtract this product from the dividend. Bring down the next group to form a new dividend.
- Reiterate steps 4 through 6 until the remainder is zero, or until as many places as needed have been determined.
- Add zeroes to the decimal part as needed. The decimal of the root goes directly above the decimal of the #.
- If the partial divisor is greater than the first part of the dividend, write a zero as the corresponding figure of the root. Draw down the next group and proceed as before.

**Cube Root Example.**

4 2 7 .9∛78'347'809' .6396414 347 300×4^{2}=4800 30×4×2 = 24025044 ×2=^{2}= 410 0884 259 809 300×42^{2}=529200 30×42×7= 88207538069 ×7=^{2}= 493 766 483493 326 639 300×4272=54698700 30×427×9=1152902 2×9= 8154814071 ×9=493 326 639

**CR-1. Cube root Cryptarithm. **(Three words, 1-0) **LIONEL**

Y LV RY'OSDAYOP OSDOP PSSUE ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ 1 2 3 4 5 6 7 8 9 0

Go through steps one through nine. Only the numbers 3 and 4 can generate a two-digit cube for the first quotient and only one of those numbers will yield the same last digit as the quotient. Y minus Y tells us the value of P. Steps four and five will provide the value for L and we are well on our way to solving another cipher type that we might have ignored in the past.

We will continue our discussions on the cryptarithm arithmetic cipher problems with a look at a couple of types that we might tend to avoid if we were not aware of their potential simplicity.

Factorial Equation cons usually enjoy a C-Special ranking in the cryptarithm column but may not be as hard as their placement in the column suggests. Here is an opportunity for cipher solving success rather than cipher retreat.

## Factorial Equations.

Factorial Equation constructions usually enjoy a C-Special ranking in the cryptarithm column but may not be as hard as their placement in the column suggests. Here is an opportunity for cipher solving success rather than cipher solving retreat.

Webster defines “factorial” as “the product of a series of consecutive positive integers from 1 to a given number.” Thus, factorial four (written 4!) = 1 x 2 x 3 x 4 = 24. Let’s apply this relatively simple math formula to a SO2004 *Cm* Cryptarithm construction.

#### SO 2004 Factorial. (Three words, 0-1) APEX DX

A! + D! = EDD A! – N! = AFF N! + W! = OWW (I! / A!) – TV = DAA

Our usually devious constructor, APEX DX, has been kind by providing us with four equation results that end with identical cipher letters, DD, FF, WW and AA. This leads to four equations ending with identical numbers. Searching for a factorial that will lead us to a three digit solution ending in identical numbers produces 6! = 1×2×3×4×5×6 = 720 or 5! = 1×2×3×4×5 = 120. We now search for a D! to add to 720 or 120 that will produce a three digit total ending in identical numbers. Both 2! (1×2 = 2) and 4! (1×2×3×4 = 24) fill this bill. This provides us with sums of 722, 744, 120 or 144 as our first possibilities.

Moving on to the second equation, we will find that 5! = 1×2×3×4×5 = 120 is the only factorial that will keep the three digit solution and two identical ending digits in equation two when subtracting N1 factorial. Thus A! (720) – N! (120) = 600 is the solution for equation two and A! (720) + D! (1×2×3×4 = 24) = 744 is our solution for equation one.

Partial solution left for you to complete.

F E A N D

0 9 8 7 6 5 4 3 2 1

## Logarithms.

Webster defines a logarithm as the exponent (power) to the base that a specific number must be raised to produce a given number.

#### ND2004 *Cm* Logarithm. (Three words, 0-9) MARSHEN

INASO = I DSTAF = S Log O Log YET

This logarithm is stating that the Log of digit “O” = “I” and will produce the number INASO. We need only to run through all single digit numbers (representing “O”) to determine what power will produce the number INASO. Keep in mind that INASO’s first digit (I) and the power (I) must be the same number. We will try a couple of digits below, leaving the correct digit for you to find. Four, to the seventh power = 4×4×4×4×4×4×4 = 16384. The first digit (1) does not match the power (7). Six, to the sixth power = 6×6×6×6×6×6 = 46656. This digit has a pattern not present in the digit lettered INASO

Will the digit, 5, provide the charm? The answer to this first logarithm will allow you to place five of the ten key letter positions and put you well on route to the final solution.

___ ___ ___ ___ ___0 ?___ ___ ___ ___ 0 1 2 3 4 5 6 7 8 9

#### C-3. Addition. (Two words, 0-9)

GOEATG + AMYPTC = RTTYCMG

#### C-4. Subtraction. (One word, 1-0)

LUBJRMU - ACECJK = MELRRU

#### C-5. Division. (Two words, 0-1)

REOTH / RUO = GF; - IGF = YUH; - FIU = OY

#### C-6. Multiplication. (Two words, 9-0)

MPE * SRT = ITPE; + OUOE + MPE = SOLUPE

#### C-7. Addition. (One word, 0-9)

GOLINGER + BYLONGER = YOOLNOGH