Here's an American Cryptarithm:

            F I F T Y
       +  S T A T E S 
        A M E R I C A

Every letter stands for a different number. In this case, since there are ten different letters, you need to find ten different numbers that make this equation true. Remember, the first digit of each number is not allowed to be zero. Use your powers of deduction and logic to figure out what this equation stands for.

You have three options, now that you know the puzzle:

• You can stop reading here, and solve the puzzle yourself.
• You can read the summary of the steps I used to solve the puzzle, then solve the puzzle yourself, only re-reading my summary, if necessary, for ideas.
  • For obvious reasons, I hope that you pick one of the first two options.
• If you need further help, you may keep reading "The Full Treatment".
  • Read what I'm going to do
  • Try to do what I said I'm going to do, and perhaps, do it
  • Read to see what I did for that step, and do it, if you haven't already
  • Keep reading each section of my full treatment this way.
  • If you find that you can do several steps without reading, do them, then come back and read what you did.

Summary of Steps I used to Solve Puzzle

These are the steps that I went through to solve this puzzle. There may be other sequences of steps that will work. After you read one step, do that step on your own, and do any other steps you can think of, before you come back to read my next step. If you solve the puzzle this way, you will actually learn the strategy better than if you just read this article. If you don't understand my summary of a step, you can scroll further down to the "full treatment" and see a full explanation of each step.

Remember, after you figure out what each letter is equal to, write that number in place of every occurrence of that letter before you go to the next step. If you figure out that a "carry" will result, mark that "carry" above the column that it will carry to, and if you figure out that there will be no carry to the next column, put a zero there so that you know you figured it out.

1. What are the only possible numbers that "A", "S", and "M" can be equal to?

• Remember, the first letter of each word may NOT be zero, so "A" and "S" must be greater than zero.
• Remember also, all three values are different.
• Here's a shortcut to make this part easier
  • Look at the first three columns of your problem
  • If you wish, you may temporarily change the third column (F+T) to nines
  • Figure out the only three numbers (different) that A, S and M can equal
  • Remember to put the S value & the A value in both places they go
  • Mark the carry values above their columns. You will need to know the carry values later (without the carry to the hundred-thousand column, there wouldn't be a carry to the million column).

         F
     + S T
     A M E

2. Figure out what "Y" is equal to. You will figure out that there will be a carry from the one column to the ten column, so put a "1" over the ten column to remind yourself that there is a carry.

3. Will there be a carry from the thousand column to the ten-thousand column?

• If there is a carry from the hundred column, we are evaluating:  I+1+1=R
• If there is no carry from the hundred column, we are evaluating:  I+1=R
• What is the largest possible value for I?
  • S is already equal to 9, so the largest possible I is 8.
• If I=8, then either
  • There is no carry: 8+1=9
  • There is a carry: 8+1+1=10, carry the 1, R=0
  • But you should have figured out that M=0, so R cannot be zero
• Thus, we have proved that there is no carry from the thousand to the next column
• Put a zero (for "no carry") above the ten-thousand column

4. Will there be a carry from the ten column to the hundred column?

• We already know that there is a carry from the one column: T+E+1=C
• If we do not have a carry to the hundred column, we have: F+T=I
• If we do have a carry to the hundred column, we have: F+T+1=I
• Look at the ten-thousand column
  • There is no carry from the thousand column: F+T=E
• If F+T=E, then F+T cannot be equal to I
• We have just proved that there will be a carry from the ten to the hundred column
• Put a one (for "carry") over the hundred column

5. Will there be a carry when you add the hundred column?

• The hundred column can be written as: F+T+1=I
• Look at the ten-thousand column
  • There is a carry from: F+T=E to the hundred-thousand column
• If F+T has a carry, then F+T+1 will have a carry.
• This proves that we'll carry 1 from the hundred to the thousand column.
• Mark the "1" above the thousand column

6. Compare the hundred column and the ten-thousand column

• We can write an equation for each column, using 10 to show the carry to the next column
  • Hundred column: F+T+1=10+I
  • Ten-Thousand column: F+T=10+E
• Since the ten-thousand equation tells us that F+T is equal to 10+E
  • Erase the F+T in the hundred-equation and change it to 10+E
  • You now have: 10+E+1=10+I
  • Subtract 10 from both sides of the new equation
    • 10 - 10 + E + 1 = 10 - 10 + I
    • E + 1 = I
    • I = E + 1
• You have just proved that I is the next number higher than E, whatever it is.
• Remember the equation: I=E+1 (we will use it later)

7. Look at the thousand column:

• There is no carry to the ten-thousand column
• There is a carry from the hundred column.
• We can write this column as an equation:  I+A+1=R
• Earlier, you figured out that A=1
• Change A to 1: I+1+1=R
• Simplify: I+2=R
• In step 6, you figured out that I=E+1
• Substitute E+1 for the I in R's equation: E+1+2=R
• Simplify: E+3=R
• You have just proved that R is three numbers higher than E, and two numbers higher than I.
• You now have two equations to remember
  • I = E + 1
  • R = E + 3

8. Draw a mini-number line to represent the relationship of these values

E --- I --- ? --- R

9. Look at the ten-thousand column again

• F + T = E + 10 (the 10 represents the carry to the hundred-thousand column)
• We could write this as: E = F + T - 10
• Either way, this tells us that F and T are both LARGER than E.

10. Extend your mini-number line to six positions (six because the only numbers left available to us are 3, 4, 5, 6, 7, and 8). There are only two ways to extend the number line in such a way that at least two unknown values are greater than E:

E --- I --- ? --- R --- ? --- ?

and

? --- E --- I --- ? --- R --- ?

11. Since the only remaining numbers are three through eight, put a second mini-number line below each of the above number lines, with those six values shown:

E --- I --- ? --- R --- ? --- ?
3 --- 4 --- 5 --- 6 --- 7 --- 8

and

? --- E --- I --- ? --- R --- ?
3 --- 4 --- 5 --- 6 --- 7 --- 8

12. Let's look at the ten column and turn it into an equation with these number lines in mind

• The number lines tell us that either: E is equal to 3, or E is equal to 4
• The tens column has a carry from the one column (1)
• The tens column will carry a one to the hundred column (10)
• Write an equation for the ten column: T + E + 1 = C + 10
• Subtract C from both sides of the equation:
  • T + E + 1 - C = C - C + 10
  • T + E - C + 1 = 10
• Subtract E from both sides of the equation:
  • T + E - E - C + 1 = 10 - E
  • T - C + 1 = 10 - E
• Subtract 1 from both sides of the equation:
  • T - C + 1 - 1 = 10 - 1 - E
  • T - C = 9 - E
• Test this equation: (T-C=9-E) for both values of E
  • If E is equal to 3, then
    • T - C = 9 - E
    • T - C = 9 - 3
    • T - C = 6
    • The difference between T and C is 6
    • What is the difference between the first and last available numbers?
      • The first available number is 3
      • The last available number is 8
      • The difference between 3 and 8? 8-3=5
    • We have just proved that E is not equal to 3, because we cannot have two numbers, C and T, with a difference of 6.
  • E must be equal to 4
    • T - C = 9 - E
    • T - C = 9 - 4
    • T - C = 5
  • The difference between T and C is 5
  • Pick the only number line that has a distance of 5 units between two question marks
  • Now, you can figure out what T and C are equal to

13. Now that you know what E is equal to, figure out what I and R are equal to.

14. You should only have one value left and one letter left to change. Change the last remaining letter (T) to the last remaining number (whatever it is)

15. Double check the following:

• Did you use a different number for each letter?
• Did you use all ten numbers (there were ten letters to start with)?
• Did you use the same number every time a duplicate letter was used?
• Does the answer add up?

The Full Treatment
Try to do it by yourself with all the steps above
before you scroll down any further
for expanded explanations

            F I F T Y
       +  S T A T E S 
        A M E R I C A

To figure out the first three values, for A, S and M, look only at the first three columns.

         F
      +S T 
     A M E

Do we have to have a carry from the (F+T=E) column to the (+S=M) column?

Yes, S is not allowed to be equal to the same value that M is equal to, so there must be a carry from the (F+T=E) column. We can demonstrate a carry by changing both F and T to 9. (This is only temporary, so it's okay). Now, add that column. 9+9=18, change the E to 8 (temporarily) and carry the one.

       1
         9
      +S 9 
     A M 8

Stop here, figure out what S, M, and A must equal, then scroll down to continue my analysis. Remember, they must be three different values, because S, A and M are three different letters (different from each other, but not necessarily different from our "temporary" numbers).

Look at the second column. What is the only value for S that will let us carry 1 to the first column?

1 + 7 = 8
1 + 8 = 9
1 + 9 = 10

Obviously, 9 is the only one-digit number that results in a two-digit number when you add the carried 1 to it.

Change S, A and M to the values that you figured out.

Only S=9 allows us to carry the 1 over to the millions column (A), and that makes M=0.

Let's fill all of this in to our original equation. Don't forget to change the other occurrences of A and S.

          1
            F I F T Y
       +  9 T 1 T E 9 
        1 0 E R I C 1

Now, look at the one's column and figure out what "Y" must be equal to.

          1
            F I F T Y
       +  9 T 1 T E 9 
        1 0 E R I C 1

Y + 9 = 1

Obviously, something plus nine cannot be equal to one, so there must be a carry, which means our answer is 11 (the ten is our carry to the next column over to the left..

Y + 9 = 11

Only one number makes this statement true:

2 + 9 = 11

"Y" must be equal to 2. Change the Y to 2 in your addition problem. Don't forget to mark your "carry".

          1       1
            F I F T 2
       +  9 T 1 T E 9 
        1 0 E R I C 1

Let's look at the thousand column.

          1       1
            F I F T 2
       +  9 T 1 T E 9 
        1 0 E R I C 1

Will there be a carry from the thousand column (I+A=R) to the next column over, the ten-thousand column?

Let's assume that there is a carry from the prior column: I + A + 1 = R

We already know that A is equal to 1: I + 1 + 1 = R

The largest possible value for I is 8, because S is already 9: 8 + 1 + 1 = 10

This would make R equal to zero, with a carry of 1.

BUT ... M is already equal to zero, so R cannot be equal to zero.

We have proved that there is NO carry from the thousand column to the ten thousand column. It may not seem important now, but this little detail will be important shortly.

Put a zero, for "no carry," above the ten-thousand column to show that we've done this.

          1 0     1
            F I F T 2
       +  9 T 1 T E 9 
        1 0 E R I C 1

Let's look at the hundred column, ignoring for the moment, any possible carry to the thousand column:

          1 0     1
            F I F T 2
       +  9 T 1 T E 9 
        1 0 E R I C 1

F + T = I

Is there a carry from the ten column to this column? In other words, we must choose between:

F + T = I (no carry)
or
F + T + 1 = I (with carry)

To answer this question, let's look at the ten-thousand column (again, ignoring the carry to the next higher column):

          1 0     1
            F I F T 2
       +  9 T 1 T E 9 
        1 0 E R I C 1

F + T = E (no carry)

Since we already know that there is NO CARRY for: F + T = E,
we must conclude that there IS A CARRY for: F + T = I
because, but, by the rules of puzzles like this, I must NOT be equal to E.

F + T + 1 = I

Put a 1 over that column to show that we figured this out.

          1 0   1 1
            F I F T 2
       +  9 T 1 T E 9 
        1 0 E R I C 1

Will there be a carry from the hundred column to the thousand column?

F + T + 1 = I

Look at the ten thousand column:

F + T = E + 10 (the carry to the hundred thousand column)

If F+T is equal to E+10, then F+T+1 must be equal to E+11. (Get it???)

In other words, since F+T had a carry, F+T+1 must have a carry.

Put a 1 above the thousand column.

          1 0 1 1 1
            F I F T 2
       +  9 T 1 T E 9 
        1 0 E R I C 1

Now, let's find some Algebraic expressions to represent the four columns that have letters in them.

Let's look at the two similar columns first. In fact, we just looked at them up above.

F + T + 1 = I + 10
and
F + T = E + 10

If you know a little bit of Algebra, you know that we can manipulate equations to solve for variables.

Since F+T is equal to E+10, according to the second equation, substitute E+10 for F+T in the first equation.

F + T + 1 = I + 10

E + 10 + 1 = I + 10

Subtract 10 from both sides of our new equation:

E + 10 - 10 + 1 = I + 10 - 10

E + 1 = I

That's our first equation to remember: I = E + 1

Now, look at the thousand column. I + A + 1 = R

We know that A is equal to 1: I + 1 + 1 = R  or  R = I + 2

Substitute E+1 for I (first equation):  R = E+1 + 2  or  R = E + 3

Now, we have two equations to remember.

• I = E + 1
• R = E + 3
• If we were to show these values on a number line, we would see

----- E ----- I ----- ? ----- R -----

Let's look at the ten thousand column again:

F + T = E + 10 (10 is the carry to the next column over)

What does this tell us about the individual values for F and T?

F must be greater than E
E must be greater than E

Why? If one of them was less than E, the other would be greater than ten.

What does that do to our number line that we sketched above? Our number line will end up with exactly six values on it, because our unused numbers right now are 3, 4, 5, 6, 7, and 8.

The mini-number-line has four positions on it right now, one of which is an unknown value greater than E. Our two possible number lines, with six positions total, are:

<----- E ----- I ----- ? ----- R ----- ? ----- ? ----->
or
<----- ? ----- E ----- I ----- ? ----- R ----- ? ----->

As I said above, there are only six remaining values that we can use. Make a number line for those values underneath each of the above number lines.

<----- E ----- I ----- ? ----- R ----- ? ----- ? ----->
<----- 3 ----- 4 ----- 5 ----- 6 ----- 7 ----- 8 ----->

or

<----- ? ----- E ----- I ----- ? ----- R ----- ? ----->
<----- 3 ----- 4 ----- 5 ----- 6 ----- 7 ----- 8 ----->

What does this tell you?

E is either equal to 3, or it is equal to 4. It cannot be equal to any other value.

Let's use a little bit of trial and error to see if E is equal to the first value.

Let E = 3

Look at the ten column of our puzzle problem:

T + E + 1 = C + 10

The "+1" is the carry from the one column to the ten column.
The "+10" is the carry from the ten column to the hundred column.

Substitute E=3 and simplify:

T + 3 + 1 = C + 10

T + 4 = C + 10

Subtract 4 from both sides of this equation:

T + 4 - 4 = C + 10 - 4

T = C + 6

This means that our unknown value for T is six higher than our unknown value for C.

But, our only remaining values, the only numbers available for us to choose from are 3, 4, 5, 6, 7 and 8. None of these choices will allow us to have two numbers six apart.

This proves that E is NOT EQUAL TO 3.

Which means, tah dah (drum roll), E is equal to 4. Earlier we proved that E was either equal to 3 or 4. 4 is the only other number left.

Look at your first two Algebraic equations? (Remember, I told you to remember them?)

• I = E + 1

• R = E + 3

Or, look at your number line, the line that has E=4.

<----- ? ----- E ----- I ----- ? ----- R ----- ? ----->
<----- 3 ----- 4 ----- 5 ----- 6 ----- 7 ----- 8 ----->

Substitute E=4, I=5, and R=7 into your problem.

          1 0 1 1 1
            F 5 F T 2
       +  9 T 1 T 4 9 
        1 0 4 7 5 C 1

Now, I'm going to let you finish the work. Remember what I did when I tested to see if E=3?

I want you to do the same thing, with E=4.

Solve this equation to see how apart C and T must be?

F + E + 1 = C + 10

Here's what you need to do:

• Substitute 4 for the E
• Simplify
• Subtract 5 from both sides of the equation
• See what unknown values on your number line will make your new statement true
• Fill those values into your addition problem

After you do this, you should only have one value left to fill into your puzzle.

Once you think you have the answer, you need to double-check your answer:

• Check the addition.
  Does the sum of FIFTY + STATES equal your number for AMERICA?
• Did you use ten different numbers?
  After all, you did start with ten different letters.
• The letters A and T showed up three places each in the problem.
  Did you use the same number all three times for A, and all three times for T?
• The letters F, I, S and E showed up two times each.
  Did you use the same number both times for each letter?

*****

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