# Cryptarithms

A cryptarithm is just a math puzzle or a math riddle. We can turn any regular addition, subtraction, multiplication, or division problem into a cryptarithm by replacing the numbers with letters.

Your job now is find the numerical values of these letters.

In ths lesson, I will show you, using my own techniques and thinking ability, how to solve cryptarithms.

Example #1: Solve the cryptarithm below using only the numbers 0, 1, 2, 3, 6, 7, and 9. No letter can represent two different digits. For instance n cannot be 2 and 6 at the same time.

sun
+  fun
______
SWIM

Guess and test is the strategy that we are going to use. However, just guess and test is not enough. You may need to make some good observations or use some basic math facts to solve the puzzle in a timely fashion. Otherwise, we may end up guessing all day.

Solution:

s    u    n
+      f    u    n
_______________
s   w    i    m

Notice that n cannot be 0. When adding n and n, the result is m and m is different than n. However, 0 + 0 = 0, so you are not getting a different number. By the same token, u cannot be 0.

Notice also that n cannot be 2, 7, or 9.

2 + 2 = 4                     7 + 7 = 14                            9 + 9 = 18

When adding, you will have to write down 4, or 8. However, 4 and 8 are not listed among the numbers we can use.

By the same token u cannot be 2, 7 or 9 and n or u can only be 1, 3, or 6

Notice that the following additions are not possible.

s     6    1

+  f     6    1                            _______________

s  w     2    2

2 appears twice

s     6    3

+    f     6    3                          _______________

s  w    2    6

6 appears twice

s      1    6

+   f      1    6                            _______________

s  w     2    2

2 appears twice

For the one in the middle, you could try to swap 3 and 6 and see what happens.

s    3    6
+      f    3    6
_______________
s  w    7    2

We have 3 numbers left to use  0, 1, and 9. If you going to get a number with 4 digits as an answer, then 9 must be either s or f

9    3    6
+      1    3    6
_______________
1  0    7    2

1    3    6
+      9    3    6
_______________
1   0    7    2

The one on the right is of course our answer. The other two additions below are more possibilities that would not have worked if we had tried them.

9    3    1
+      7    3    1
_______________
1  6     6    2

9    1    3
+      7    1    3
_______________
1  6     2    6

Example #2: Find A, B, C, and D. The digits that we can pick from are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9

ABCD
x          4
____________
DCBA

Solution:

A    B     C    D
x                        4
_________________
D   C      B    A

Important observation: The answer has 4 digits

If the digit in red ( A ) is bigger than 2, the answer will have 5 digits instead 4.
Therefore A cannot be 3, 4, 5, 6,7, 8, or 9. A can be either 1 or 2.

Suppose that A is 1 as shown below. Then, 4 times D = 1

1    B     C    D
x                       4
_________________
D   C      B     1

4 times 1 = 4,   4 times  2 = 8,    4 times 3 = 12,   4 times  4 = 16,  4 times 5 = 20,
4 times 6 = 24 ,   4 times 7 = 28,   4 times 8 = 32,     4 times  9 = 36

There are no numbers multiplied by 4 that will give you a 1 in the ones place. A cannot be 1 and since A cannot be 1, A must be 2.

2    B     C    D

x                     4
_________________
D   C      B     2

Only 4 times 3 or 4 times 8 will give you a 2 in the ones place, so D is 3 or 8

2    B     C    3

x                     4
_________________
3   C      B     2

2    B     C    8

x                     4
_________________
8   C      B     2

At this point, I feel more inclined to use the one on the right because 4 times 2 gives me the 8 that I want.

2    B     C    8

x                     4
_________________
8   C      B     2

My strategy now is to pick the right number for B. As I try to do this I notice that B cannot be bigger than 3. Anything bigger than 3 will have a carry that will force the 8 in the answer to be bigger. 2 is not a choice either since 2 cannot be repeated. My only choice for B then is 1.

2    1     C    8

x                    4
_________________
8   C      1     2

When I do  4 times 8, I have a carry of 3.

4 times  C + 3 must give a number with 1 in the ones place. My two choices for C are 2 and 7. 2 cannot be chosen since the 2 will be repeated. My only choice for C is 7.

2    1     7    8

x                     4
_________________
8   7      1      2

Some cryptarithms will be tougher than that to solve. However, you can eventualy solve them by making some good observations.

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