A cryptogram is a mathematical puzzle, where various symbols are used to represent digits, and a given system has to be true. In the previous page, we have discussed various techniques when tackling a cryptogram. However, it is not immediately clear how to solve cryptogram that involves various techniques. For example, for the cryptogram above, for the uninitiated, we could start by considering the case . It would be unwise to begin with the trial and error approach because there is close to ways to begin with it. Thus it is definitely not an ideal way to approach it by hand. Hence, it will be beneficial to know which technique is best applicable.
In this wiki, we will work through various examples, and extensively apply this table to identify the conditions and techniques to use.
|Technique||Type of problem||Conditions to apply|
|1) Converting cryptogram to equation||All||Few symbols, which occur 1-2 times each|
|2) Rearrangement of columns||At most 1-2 symbols in each column|
|3) Carry Over||(Most useful technique) to identify additional constraints|
|4) Divisibility Rules||A term has a factor with a simple rule of divisibility|
|5) Factoring of terms||If we know the entire term and can easily factorize it|
|6) Modular Arithmetic||If we have a lot of information modulo a value (like 2, 3, 4, 5, 7)|
|7) Chinese Remainder Theorem||Last few digits of 2 or more numbers are the same|
|8) Trial and error||All||Few variables, but they occur many times|
|9) Programming approach||All||Any cryptogram will do|
We should note that all cryptograms that have the subtractions can be converted to a strictly addition problem. For example, can be converted to .
For the above cryptogram, solve for the unknown letters and find the possible value(s) of the 5-digit number .
We first convert this cryptogram to an addition problem.
Since this is a simple addition problem, it might be beneficial to apply Converting cryptogram to equation, Rearrangement of column and/or Carry over. However, because there are so many unknowns from the start, we will not accomplish much. Since there are 's on both sides in the first column, it's best to start to apply (3) Carry over:
For the first column, we have the last digit of as , so must equal 0.
Since we are taking the sum of two 3 digit numbers, the carry over for the third column is non-zero and must equal 1, implying . This simplifies the cryptogram to:
After the value of is known, we can rearrange the letters/digits in their respective columnns such that there are at most 2 symbols in each column. Thus we apply (2) Rearrangement of columns:
Because all the values in the first column is known, and there is no carry over. We can eliminate the first column. Therefore we apply (3) Carry over:
Since there are a few symbols left, we can apply (1) Converting cryptogram to equation:
Solving for gives . Finally, we can convert the cryptogram to an equation: .
For the above cryptarithm, solve for the unknown letters and find the possible value(s) of the 5-digit number .
Since there are many unknown letters to consider, it is very likely to consider trial and error as a possible approach. However we should always reduce the number of possible cases to check. Thus we should begin with applying (8) Trial and error then follow up with (3) Carry over:
From the first column, we have only. Now, because , the third column implies that there was a carry in the second column and thus and .
From the fourth column, there may or may not be carry over, so . Also, the fifth column gives
Since from OSAKA we know that the leading digit is not 0, i.e from we have only. Then since from above, .
Notice how we have reduced the possibility of to ! So we are left to check less cases. Again, we apply (8) Trial and error & (3) Carry over for the values of :
Let's start with the extreme values first (the largest and smallest), that is, for or . We start with them first because they usually tend to tighten the bounds and quickly force out a solution or contradiction. However, keep in mind that this method does not always guarantee results.
Then . Because we know from that there is a carry from the second column, then . But from we have which is impossible. So is not a possible solution. Notice that we don't need to solve the entire cryptogram to determine that .
Then , then from , we have . With , from the second column, we have . But and must be distinct. This means that is also impossible.
Finally, we are left to check for or only. Amazing isn't it? After bounding the possible values of the values above, we just need to check for two cases!
Then , and from , we have only. But since we know that the second column has a carry, then can't be satisfied. So .
So we are left with . But of course, we should check whether each of these letters are distinct. With , , then .
So far, we have confirmed that . What's left to solve is the values of and . Once more, we apply (8) Trial and error & (3) Carry over for the values of :
Suppose , then . Also, from , we have . This means that from the fourth column, which means that is at least a two digit integer, which is impossible. So only.
Now to solve for and , from the second column, . Since we know from that , then there is no carry over in the fourth column, that is .
Putting them all together shows that is indeed true.
From the examples shown above, we can note that the two techniques: Carry Over and Trial and error are often used together. This is because Carry Over gives us an additional insight to the restriction to the unknowns and Trial and error is used to pinpoint the exact values of these unknowns.
The above shows an incomplete long multiplication. We can obviously multiply out both numbers to find the unknown non-zero single digit integers and However, that can be a tedious task. Without performing the long multiplication, but instead using the techniques we have learned above, show that and .
Because there are only a few symbols to consider, it is a viable approach to apply (1) Converting cryptogram to equation:
We can represent this product as for and as the 8-digit number.
Because with a leading digit of the leftmost column to be 1, this forces to be less than or equals to only. Thus .
Because we can gain the insights for modulo a certain value, apply (4) Divisibility Rules:
It's simplest to consider the divisibility of 9 as we are only taking the sum of its digits. We have , with and . Thus , which means that when the sum of the digits of is divided by 9, it yields a remainder of 8. So
For the above cryptogram, each letter represents a distinct single digit non-negative integer, with and being non-zero. Find the value of .
Since this is a multiplication between two numbers, factoring of terms and Chinese Remainder Theorem are possible viable approaches . However, since the last two digits of these three numbers are the same, we can Apply (7) Chinese Remainder Theorem:
By considering the last two digits, .
So we have . Like the earlier example, we have .
This means that only.
So what's left to check? Well, it's just the possible values of . Because knowing value of can solve the entire long multiplication, the most sensible approach is to apply (8) Trial and error:
Suppose . We apply trial and error for the possible values of , which are . By inspection, none of them yields a product in the form of .
Thus . Using the same approach in the earlier paragraph, we get . Hence
For the letters used in the cryptogram above, find the value of .
Since we have additions of 4 rows of multiplications, we can obtain the additional information by applying (1) Converting the cryptogram to equations:
From the very top, the third row shows that
And the fifth row shows that
By considering the last digit of the last equation, we have
If we reconstruct the last cryptogram again with the knowledge of , we have
As highlighted above, the brown colored symbols show an addition. Since there is only at most 2 unknowns in the column, we can apply (1) Converting the cryptogram to equations:
Refer to the and rows highlighted in blue below. Because there is a sum of two numbers and to produce the number we can apply (3) Carry over:
Looking at the leftmost column, we see that the carry over must be equal to 1 because . Thus
Because is a 7-digit number, we start by bounding the equation. That is, apply (1) Converting cryptogram to an equation:
Since is an 8-digit number, then . With , then or only. Hence connecting and gives
Now we are left with 3 cases to check only! Thus we apply (8) Trial and error. By inspection, we have
Completing the entire cryptogram yields the values of as respectively. Thus,
From the examples shown above, we can note that the two techniques: Divisibility Rules and Modular Arithmetic are often used together. This is because Divisibility Rules is used in conjunction with Modular Arithmetic as the former is a further generalization of the latter.