Cryptarithms or crypt-arithmetic problems are encrypted math problems, where numbers in a given mathematical expression are represented by letters or other symbols.
There are two main types of cryptarythms:
- Hindu problems.
- Alphametics.
In Hindu problems, every digit in a mathematical expression is concealed with the same symbol, usually an asterisk. Surprisingly, despite the fact that once the original digits have been replaced by asterisks and they are indistinguishable, these problems can be solved.
Alphametics are puzzles where each digit is replaced by a different symbol, usually a letter. Many alphametics “spell out” words, making them more attractive and entertaining. One of the most famous was invented by Henry Ernest Dudeney, a British puzzlist, in 1924: SEND + MORE = MONEY.
By substituting S=9, E=5, N=6, D=7, M=1, O=0, R=8, Y=2 the cryptarithm translates into: 9567+1085=10652. In the second half of the week, we’ll learn how this and other cryptarithms are solved.
A bit of history
Cryptarithms first appeared in the United States in 1864, but it is believed that they were invented much earlier, in Ancient China. These original cryptarithms were mainly of the “Hindu” type. In the early twentieth century, Simon Vatriquant, a Belgian mathematician pseudo-named ‘minos’, and mistakingly called Maurice Vatriquant, took them much more seriously. He published many cryptarithms in a mathematical journal called “Sphinx”, published during the first half of the twentieth century. Maurice Kraitchick, another well-known mathematician, was the editor of this journal. The journal was dedicated to math puzzles, a branch of mathematics that is known today as recreational mathematics. Jorge Soares, among others, has a great cryptarithm site dedicated to the Sphinx. If you are already well-versed in cryptarithms, you might want to challenge yourselves with some of the cryptarithms in his site.
Hindu problems originated in India in the middle ages and are sometimes referred to as arithmetical restorations or skeleton problems. In these problems, digits are replaced by asterisks, for example: * + * = 2.
Each asterisk represented one missing digit (from 0 and 9). Two or more asterisks in the same problem may represent different digits.
For example, for the problem * + * = 2 there are two possible solutions:
- 1 + 1 = 2 – in this solution each asterisk represents the digit 1
- 2 + 0 = 2 – in this solution one asterisk represents the digit 0 and the other asterisk represents the digit 2
Note that we are counting 2 + 0 and 0 + 2 as the same solution.
Remember the following points:
- Some puzzles might have several solutions and some might have no solution
- Two adjacent asterisks represent a 2-digit number, like 74. Three adjacent asterisks represent a 3-digit number, and so on. In these cases the leftmost asterisk cannot represent the number 0, i.e. it is illegal to write to represent the number 8 as 08.
Solve the following cryptarithm. There is only one solution!
* * * × * – * * = 1
The solution is: 100 × 1 – 99 = 1 A 3-digit number multiplied by a 1-digit number will result in either a 3 or 4-digit answer, unless the 1-digit number is 0, in which case the multiplication result is 0. This will lead to an impossible problem: 0-**=1 since negative numbers are not allowed. So, we are dealing with a multiplication that will give either a 4 or 3-digit result. It cannot be a 4-digit result, because a 4-digit number minus a 2-digit number will result in a 3 or 4-digit number and not the number 1 as required. So, we now know that the result of the multiplication is a 3-digit number which, when we subtract from it a 2-digit number, we get ‘1’. The only possible solution is: 100-99=1 so the multiplication must be 100x1 and the whole solution: 100x1-99=1.
Solve the following cryptarithm. There is only one solution!
* * – * = 99
The solution is: 99 - 0 = 99 Look at the question backwards: 99 + * = **. The only number you can add to 99 that will still give a 2-digit answer is 0. Hence, the only possible solution is 99 - 0 = 99
Solve the following cryptarithm. There is only one solution!
* * × * * = 169
The solution is:1 3 × 1 3 = 169. Here we have to use the fact that there is only one solution. Hence, both the multiplicands have to be prime numbers. Since 11 × 11 already gives a close answer (121 is close to 169), we can try the next Prime up - 13. Divide 169 by 13 to get 13 - and the problem is solved. Of course, if you happen to know that 13 squared is equal to 169 that does make things easier :-)
How many solutions are there to the following cryptarithm?
* * * × * – * * = 4
The answers is: 4 and solutions are,
100 × 1 – 96 = 4
101 × 1 – 97 = 4
102 × 1 – 98 = 4
103 × 1 – 99 = 4
Solve the following cryptarithm:
* * × * * – * = 120
The solutions are:
11 × 11 – 1 = 120
10 × 12 – 0 = 120
The way to work this one out is to ‘add’ the single asterix to the answer to get: * * × * * = 120 + * . Now check which numbers between 120 and 129 have two 2-digit multiplicands.
Let us move on to
What are alphametics?
Alphametics are cryptarithms that spell out words. Given a mathematical expression, every digit in the expression is replaced by a letter. One of the most famous alphametics, spelling out ‘SEND MORE MONEY’ appears above. This alphametic was first published by Henry Dudeney, a British puzzlist, in 1924.
Five rules govern alphametics:
- Identical digits are replaced by the same letter. Different digits are replaced by different letters.
- After replacing all the letters with digits, the resulting arithmetic expression must be mathematically correct.
- Numbers cannot start with 0. For example, the number 0900 is illegal.
- Each problem must have exactly one solution, unless stated otherwise (unlike the “Hindu” problems where there might be no solution or multiple solutions).
- The problems will be in base 10 unless otherwise specified. This means that the letters replace some or all of the 10 digits – 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
I recommend reading this before watching the video… There are key elements to solving most alphametics.
- In many cases the result of an addition problem is one digit longer (in digit-length) than the addends - the numbers added. If there are only two addends, this implies that the extra digit is the number 1.
Let’s look at a very simple alphametic: ME+ME=BEE
The letter B must represent the digit 1, since when you add two 2-digit numbers you cannot possibly get a number larger than 198. That happens when both addends are 99. Since M and E are two different numbers, they will certainly be even smaller than 99! In any case, the hundreds digit in the sum, represented by B in our example, must be 1.
- In two addend alphametics, there may be columns that have the same letter in both the addends and the result. If such a column is the units column, that letter must be 0. Otherwise, it can either be 0 or 9 (and then there is a carry).
In the alphametic: ME+ME=BEE the column of the unit’s digits is: E+E=E There is only one digit, which has the property that when you add it to itself you get the same digit as the result – zero! Only the sum of two zeros is zero, so E must be equal to 0.
The solution to this alphametic is therefore: B=1, E=0, M=5: 50+50=100.
Here are some tips for solving more complicated alphametics.
If there are more than 2 addends, the same rules apply but need to be adjusted to accommodate other possibilities. If there are four identical letters in the units column (one of them the sum), this letter can now be: 0 or 5 (because 5+5+5=15). If there are four identical letters in a different column (one of them the sum), this letter can now be: 0 or 5 (no carry), 4 or 9 (carry 2). Four identical letters in a column other than the units column means a 1 could not have been carried over (why not?). This rule can be worked out for more than 3 addends as well…
- It is wise to turn subtraction problems into addition problems by adding the result to the smaller addend to get the larger one.
- When faced with a few options for a letter, try one out until you either get the correct answer, or find a contradiction.
Now let’s look at a slightly more advanced cryptarithm. Above video shows how to solve the alphametic: NO + GUN + NO = HUNT. Note the ‘neat’ sentence: “No gun, no hunt!”
Solve the following alphametic. It might be helpful to rewrite the alphametic as a column addition problem.
MA + PA = MOO
The solution is: 15 + 85 = 100
M=1 (hundred’s digit)
P+1 or P+2 (if there is a carry from the first column) =10+O
P can either be 9, in which case O=0, or 8, in which case O=0. In any case, O=0, hence, A must be 5 and P=8.
Solve the following alphametic. It might be helpful to rewrite the alphametic as a column addition problem.
ON + TO = OFF
The solution is: 19 + 81 = 100
Explanation: O has to be 1 because the result is 3 digits long. The first digit of the result has to be 1 because the sum of two 2-digit numbers cannot be larger than 198 at the most.
Replacing all the O’s with 1 we now have: 1N+T1=1FF
If we write the problem in columns, both the first column and the second column carry over a 1 to the next column. For the second to the first this is obvious. From the first to the second this is because the result in both columns is the same: F, yet the addends are different - N and T. This means that both N+1 and T+1+1 are larger than 9.
Notice that:
N+1 = 10+F and T+1+1 = 10+F
So N = T + 1
The only possibilities left for N and T are: 5,6; 6,7; 7,8 or 8,9 which is in fact, the correct answer. N=9 and T=8. Substituting these back into the problem yields the value of F - 0.
Solve the following alphametic. It might be helpful to rewrite the alphametic as a column addition problem.
BAD + DAD = DORA
The solution is: 921 + 121 = 1042
The solution is: 921 + 121 = 1042
Explanation: D must be 1 since it is the first digit in the 4-digit result of a two 3-digit addend sum. Replacing all D’s with 1, we get in the first column 1+1=A. So, A must be 2. Replacing the A’s with 2 - we get in the second column 2+2=R, so R is 4. In the third column, B+1 = 10, so B=9.
Solve the following alphametic. It might be helpful to rewrite the alphametic as a column addition problem.
WOW + WOW + WOW + WOW + WOW = SNOW
The solution is: 575 + 575 + 575 + 575 + 575 = 2875
Explanation: Rewrite the problem as WOWx5=SNOW
It is obvious from this that W=5 since the last digit in the result of multiplication of any number by 5 is always 5. Replacing all W’s with 5, we get: 5O5x5=SNO5.
Since 5x5=25, Ox5+2 must equal O. The only options for O are: 2 or 7. Trial and error give us the solution.
It is more difficult to find multiplication cryptarithms that also spell out words. Still, alphametic multiplication (and division) problems are still fun! Try this one. There is no need to write it down as column multiplication: AB x AB = ABB
he solution is: 10 x 10 = 100
For BxB to equal B, B can only be 0,1,5 or 6.
If we look at the problem as AB squared = ABB, we know that ABB must be a 3-digit square number. That rules out B=1, B=5 and B=6, so B=0.
To give a 3 digit result, A can be either 1,2 or 3.
However for A0xA0=A00, the only possible answer is A=1.
Some multiplication rules can really help us when we get stuck on a cryptarithm. Here are some rules you should know before taking the ‘plunge’ into the more difficult problems…
- A number is divisible by 2 if it’s unit’s digit is an even number.
- A number is divisible by 3 if the sum of its digits is divisible by 3.
- A number is divisible by 5 if the number ends with a 0 or a 5.
- A number is divisible by 6 if it’s an even number and the sum of its digits is divisible by 3.
- A number is divisible by 9 if the sum of its digits is divisible by 9.
- A number is divisible by 11 when the difference between the sum of the digits in the even positions (looking at the number from left to right) and the sum of the digits in the odd positions is divisible by 11.
Examples
The number 43242543 is divisible by 9 and by 3 because the digit-sum of the number: 4+3+2+4+2+5+4+3=27 is divisible both by 3 and by 9.
Is the number 1352467928 divisible by 11?
The sum of the digits in the even spots: 3+2+6+9+8=28
The sum of the digits in the odd spots: 1+5+4+7+2=19
Since 28-19=9 and 9 is not divisible by 11 the whole number is not divisible by 11.
We can ‘fix’ the number to be divisible by 11 if we can ‘fix’ the difference to be, i.e. 11. This can be done by adding 2 to one of the digits in the even spots. Let’s change the number 3 to 5. We get: 1552467928. This number is divisible by 11:
The sum of the digits in the even spots: 5+2+6+9+8=30
The sum of the digits in the odd spots: 1+5+4+7+2=19
Since 30-19=11 and 11 is divisible by 11 the whole number, 1552467928, is divisible by 11.
Another fun genre is when the cryptarithms spell special words. Here is an example that Truman Collins has on his website:http://www.tkcs-collins.com/truman/alphamet/alphamet.shtml
SATURN + URANUS = PLANETS.
There are three possible solutions for this cryptarithm.
Truman has a lot of similar examples, and you can even generate your own ‘related words’ cryptarithms with his online cryptarithm generator. http://www.cadaeic.net/alphas.htmOther literary cryptarithms can be found on Mike Keith’s excellent website, http://www.tkcs-collins.com/truman/alphamet/alpha_gen.shtml
Try and solve the following cryptarithm: I x KINGJO = JOKING The hint is: I=4
