Printable Math Puzzles - Free Printable

Let’s solve each cryptarithm one by one. In a cryptarithm, each letter stands for a different digit (0–9), and the same letter always means the same digit. We need to find which digits make the addition correct.

We’ll go step by step for each problem.

---

Problem 1:

```
R I C H
+ F O O D
---------
C R A Z Y P E O P L E
```

Wait — this doesn’t look right. The sum has 11 letters? That can’t be — adding two 4-digit numbers gives at most a 5-digit number. Let me check again.

Actually, looking closely:

It says:

```
R I C H
+ F O O D
---------
C R A Z Y
P E O P L E
```

No — that still doesn’t make sense. Maybe it’s meant to be:

Looking at the layout in the image description — actually, let’s re-read the instruction: “each letter represents a different digit... figure out what letter is replaced.”

But in Problem 1, the sum line is written as:

“CRAZY PEOPLE” — but that’s two words. Probably it’s meant to be one word: CRAZYPEOPLE? But that’s 10 letters — too long.

Wait — perhaps there’s a typo or misalignment. Let me look at the structure again.

Actually, in standard cryptarithms like this, the addends are aligned on the right, and the sum is below them.

So likely:

```
R I C H
+ F O O D
-----------
C R A Z Y
```

But then “PEOPLE” is underneath? That doesn’t fit.

Alternatively — maybe it’s:

```
R I C H
+ F O O D
-----------
C R A Z Y
P E O P L E ← this might be a separate thing?
```

This is confusing. Let me check the other problems to see the pattern.

Problem 2:

```
B E A L
+ S N A I L
----------
H O N K E Y
A N I M A L S
```

Again, two lines under the sum? That doesn’t make sense.

Wait — perhaps the second line is not part of the sum, but rather the answer is supposed to spell a phrase? Like “BEAL + SNAIL = HONKEY ANIMALS”? But that’s not math.

I think I see the issue — in the original image, the sum result is written as a phrase, but it’s actually one continuous string of letters representing the numerical sum.

For example, in Problem 1:

RICH + FOOD = CRAZYPEOPLE → but that’s 10 letters, while RICH and FOOD are 4 letters each → max sum is 9999+9999=19998 → 5 digits. So impossible.

Unless... wait — maybe it’s:

RICH
FOOD
----
CRazy
People

But that still doesn’t help.

Perhaps the puzzle is designed so that the sum spells a word or phrase, and we have to assign digits so that the arithmetic works AND the letters form those words.

But with 4-digit + 4-digit = up to 5-digit, having an 11-letter result is impossible.

Let me count the letters in the sum for Problem 1: “CRAZY PEOPLE” — if we remove space, it’s 10 letters. Still too many.

Wait — perhaps it’s a vertical alignment where the sum is written across multiple lines? No, that doesn’t make sense for addition.

Another idea: maybe “CRAZY” is the first part of the sum, and “PEOPLE” is something else? But the instruction says “the letters in the equation are replaced by digits”, implying the entire thing is one equation.

Let me look at Problem 3:

ANIMAL + STRIPE = ? — no sum given? Wait, in the image description, for Problem 3 it shows:

```
A N I M A L
+ S T R I P E
--------------
```

And nothing below? That can’t be.

Actually, reviewing the user's text:

In the initial description, for each problem, there is a "+" and then a blank or something? Let me read carefully:

User wrote:

"1
RICH
FOOD
+
CRAZY
PEOPLE"

Ah! Perhaps the "+" is between the two addends, and the sum is "CRAZY PEOPLE" meaning the result is a 10-digit number? But that’s impossible for two 4-digit numbers.

Unless... oh! I think I got it. In some cryptarithms, especially themed ones, the sum might be intended to be read as a phrase, but numerically, it must match the digit assignment.

But mathematically, 4-digit + 4-digit ≤ 19998, so only 5 digits possible.

So for Problem 1, if RICH and FOOD are both 4-digit numbers, their sum must be a 5-digit number starting with 1 (since max is 19998).

So the sum should start with '1', and have 5 letters.

But "CRAZY" is 5 letters — perhaps "CRAZY" is the sum, and "PEOPLE" is a red herring or part of the theme?

Similarly, in Problem 2: BEAL (4) + SNAIL (5) = ? Max is 9999 + 99999 = 109998 — 6 digits. "HONKEY" is 6 letters — that fits!

BEAL is 4 letters, SNAIL is 5 letters — so when adding, we align to the right:

```
B E A L
+ S N A I L
-----------
H O N K E Y
```

Yes! That makes sense. The sum is HONKEY — 6 letters, which matches the maximum possible sum (less than 110000).

Similarly, for Problem 1: RICH (4) + FOOD (4) = ? Should be 5 digits. "CRAZY" is 5 letters — so probably "CRAZY" is the sum, and "PEOPLE" is either a mistake or part of the puzzle name.

Looking back at user's input:

"1
RICH
FOOD
+
CRAZY
PEOPLE"

Perhaps "PEOPLE" is not part of the sum, but the title or something. Or maybe it's "CRAZY" is the sum, and "PEOPLE" is the next problem? No.

Another possibility: in some puzzles, the sum is written as a phrase, but we only care about the numerical value. For Problem 1, if we assume the sum is "CRAZY", then it's 5 letters, which works.

Similarly, for Problem 4: SKIRT + SHIRTS = CLOTHES — SKIRT is 5, SHIRTS is 6, sum CLOTHES is 7 letters — 5-digit + 6-digit = up to 99999 + 999999 = 1099998 — 7 digits, yes.

SKIRT is 5 letters, SHIRTS is 6 letters — so:

```
S K I R T
+ S H I R T S
-------------
C L O T H E S
```

Aligning to the right:

The second number has 6 digits, first has 5, so:

```
S K I R T
+ S H I R T S
---------------
C L O T H E S
```

Sum is 7 letters — perfect.

For Problem 1, RICH and FOOD are both 4 letters, so sum should be 5 letters. "CRAZY" is 5 letters, so likely "CRAZY" is the sum, and "PEOPLE" is either a distractor or part of the puzzle's theme (like "rich food makes crazy people" or something). But for solving, we'll take the sum as "CRAZY".

Similarly, for Problem 2: BEAL (4) + SNAIL (5) = HONKEY (6) — good.

Problem 3: ANIMAL (6) + STRIPE (6) = ? In user's input, it shows:

"3
ANIMAL
STRIPE
+"

No sum specified? But in the image description, it might be implied. Looking back:

User wrote for Problem 3:

"3
ANIMAL
STRIPE
+"

And then for Problem 4:

"4
SKIRT
SHIRTS
+
CLOTHES"

So for Problem 3, no sum is given? That can't be.

Perhaps in the original image, the sum is written, but in the text description, it's omitted. Let me check the very first line:

"Intermediate Cryptarithms" and then "DIRECTIONS: ..."

Then for each problem, it lists the addends and the sum.

In user's message:

After "1", it has RICH, FOOD, +, CRAZY, PEOPLE — so perhaps for Problem 1, the sum is "CRAZY PEOPLE" but that's 10 characters, which is impossible.

Unless "CRAZY PEOPLE" is meant to be the sum, but we have to consider it as a single string, and the addition is set up with proper alignment.

But RICH and FOOD are both 4-letter, so when added, the sum can have at most 5 digits. "CRAZY PEOPLE" has 10 letters if combined, or 5 and 5.

Another idea: perhaps the "+" is not between RICH and FOOD, but RICH is one number, FOOD is another, and their sum is CRAZY, and PEOPLE is the next problem? But the numbering is 1,2,3, etc.

Let's list what user provided:

Problem 1:
RICH
FOOD
+
CRAZY
PEOPLE

Problem 2:
BEAL
SNAIL
+
HONKEY
ANIMALS

Problem 3:
ANIMAL
STRIPE
+

Problem 4:
SKIRT
SHIRTS
+
CLOTHES

Problem 5:
SNAKE
FROG
+
TURTLE

Problem 6:
EASTER
EGGS
+
SPRING

For Problem 3, no sum is given — that must be a mistake. Similarly, for Problem 5 and 6, the sum is given after "+", but for Problem 3, it's missing.

Perhaps in Problem 3, the sum is implied to be something, but it's not stated.

Maybe for Problem 3, "ANIMAL + STRIPE = " and the sum is not specified, but that doesn't make sense.

Another possibility: in some cryptarithms, the sum is the concatenation or something, but that's unlikely.

Let's look at Problem 5: SNAKE + FROG = TURTLE — SNAKE is 5, FROG is 4, TURTLE is 6 — 5-digit + 4-digit = up to 99999 + 9999 = 109998 — 6 digits, yes.

Similarly, Problem 6: EASTER (6) + EGGS (4) = SPRING (6) — 6-digit + 4-digit = up to 999999 + 9999 = 1009998 — 7 digits, but SPRING is 6 letters — contradiction.

EASTER is 6 letters, EGGS is 4 letters, so when adding:

```
E A S T E R
+ E G G S
-------------
S P R I N G
```

The sum should be 6 or 7 digits. Since EASTER is at least 100000, EGGS at least 1000, sum at least 101000, at most 999999 + 9999 = 1009998, so could be 6 or 7 digits.

SPRING is 6 letters, so if the sum is 6 digits, it must be less than 1000000, which is possible if EASTER < 990001, etc.

But typically, if the first digit of the sum is S, and EASTER starts with E, and EGGS is small, S might be E or E+1.

But in any case, for Problem 6, SPRING is 6 letters, so we'll assume the sum is 6 digits.

For Problem 3, since no sum is given, perhaps it's a typo, and in the original image, the sum is provided. Maybe "ANIMAL + STRIPE = " and the sum is "CREATURE" or something, but it's not specified.

Perhaps in Problem 3, the sum is "ANIMALS" or "STRIPES", but that's guesswork.

Another thought: in the user's initial description, for Problem 3, it might be that the sum is not listed because it's obvious, but that doesn't help.

Let's skip Problem 3 for now and do the others where the sum is clear.

Also, for Problem 1, if we take sum as "CRAZY", 5 letters, it works.

For Problem 2, sum is "HONKEY", 6 letters, and BEAL (4) + SNAIL (5) = HONKEY (6) — good.

For Problem 4, SKIRT (5) + SHIRTS (6) = CLOTHES (7) — good.

For Problem 5, SNAKE (5) + FROG (4) = TURTLE (6) — good.

For Problem 6, EASTER (6) + EGGS (4) = SPRING (6) — possible if no carry to 7th digit.

For Problem 3, perhaps the sum is "ANIMALS" or "STRIPES", but let's see the context. In the image description, it might be that for Problem 3, the sum is "CREATURE" or something, but since it's not specified, I'll assume it's a mistake, and perhaps the sum is "ANIMALS" or we can infer.

To move forward, I'll solve the problems where the sum is clearly given, and for Problem 3, I'll note that the sum is missing, but perhaps in standard versions, it's known.

Upon second thought, in many such puzzles, for Problem 3, "ANIMAL + STRIPE = CREATURE" or "ANIMAL + STRIPE = ZEBRA" but that doesn't fit.

Another idea: perhaps "ANIMAL + STRIPE = " and the sum is the combination, but let's calculate the number of letters.

ANIMAL is 6, STRIPE is 6, sum could be 6 or 7 digits. If sum is 7 digits, it must start with 1.

But in the user's input, for Problem 3, after "+", there is nothing, so perhaps it's incomplete.

To avoid delay, I'll solve Problems 1,2,4,5,6, and for Problem 3, I'll skip or assume a common version.

But let's try to find a consistent approach.

Perhaps for Problem 1, "CRAZY PEOPLE" is meant to be the sum, but we have to consider that "PEOPLE" is not part of the numerical sum, but the puzzle is to make RICH + FOOD = CRAZY, and PEOPLE is the answer to what it represents, but that doesn't help for solving.

I recall that in some cryptarithms, the sum is a phrase, and we solve for the digits so that the arithmetic holds, and the letters form the words.

For Problem 1, if RICH + FOOD = CRAZY, then it's 4+4=5, good.

Similarly, for Problem 2, BEAL + SNAIL = HONKEY, 4+5=6, good.

For Problem 3, perhaps ANIMAL + STRIPE = ANIMALS or something, but ANIMALS is 7 letters, while ANIMAL and STRIPE are 6 each, sum could be 7 digits.

Assume for Problem 3, the sum is "ANIMALS" — 7 letters.

Similarly, for Problem 5, SNAKE + FROG = TURTLE, 5+4=6, good.

For Problem 6, EASTER + EGGS = SPRING, 6+4=6, which requires that the sum is less than 1000000, so EASTER < 990001, etc.

For Problem 4, SKIRT + SHIRTS = CLOTHES, 5+6=7, good.

So let's define the problems as:

1. RICH + FOOD = CRAZY (4+4=5)
2. BEAL + SNAIL = HONKEY (4+5=6)
3. ANIMAL + STRIPE = ANIMALS (6+6=7) — assuming "ANIMALS" is the sum, as it's a common extension.
4. SKIRT + SHIRTS = CLOTHES (5+6=7)
5. SNAKE + FROG = TURTLE (5+4=6)
6. EASTER + EGGS = SPRING (6+4=6)

For Problem 3, if sum is ANIMALS, then it's 7 letters, and ANIMAL and STRIPE are 6 letters, so when adding, we have:

```
A N I M A L
+ S T R I P E
-------------
A N I M A L S
```

But then the sum is ANIMALS, which is almost the same as ANIMAL, with an S at the end. That might work.

Similarly, for others.

I think this is reasonable. So I'll proceed with these assumptions.

Now, let's solve each one.

Start with Problem 1: RICH + FOOD = CRAZY

Letters: R,I,C,H,F,O,D,A,Z,Y — 10 distinct letters, so digits 0-9 all used.

Since it's 4-digit + 4-digit = 5-digit, the sum must be at least 10000, so C = 1 (because the carry from thousands place will make it 1xxxx).

So C = 1.

Now, the addition:

```
R I C H
+ F O O D
-----------
C R A Z Y
```

With C=1, so:

```
R I 1 H
+ F O O D
-----------
1 R A Z Y
```

Now, look at the units column: H + D = Y or 10 + Y, etc.

Tens column: C + O + carry = Z or 10 + Z, but C=1, so 1 + O + carry1 = Z or 10 + Z.

Hundreds column: I + O + carry2 = A or 10 + A.

Thousands column: R + F + carry3 = R or 10 + R, but the sum has R in thousands place, and C=1 in ten-thousands.

Let's write it with columns:

Let me denote the columns from right to left as col0 (units), col1 (tens), col2 (hundreds), col3 (thousands), col4 (ten-thousands).

Col0: H + D = Y + 10*c0 (c0 is carry to col1, 0 or 1 or 2)

Col1: C + O + c0 = Z + 10*c1 but C=1, so 1 + O + c0 = Z + 10*c1

Col2: I + O + c1 = A + 10*c2

Col3: R + F + c2 = R + 10*c3 ? But the sum has R in col3, and c3 is carry to col4.

The sum is C R A Z Y, so col4: c3 = C = 1

Col3: R + F + c2 = R + 10*c3 ? That would imply F + c2 = 10*c3

But c3 = 1, so F + c2 = 10

Since F is digit 1-9 (can't be 0 as leading digit), and c2 is 0,1,2, so F + c2 = 10, so F = 10 - c2

c2 is carry from col2, which is at most 2 (since I+O+c1 ≤ 9+8+1=18, so c2≤1? Wait, max I+O+c1 = 9+8+1=18, so c2=1 if sum>=10, or 0 if <10, but 18/10=1.8, so c2 can be 1, or if sum=19, c2=1, max sum is 9+8+1=18, so c2 ≤ 1.

I and O are digits, c1 is carry from col1, which is at most 1 (since 1+O+c0 ≤ 1+9+2=12, so c1≤1).

So col2: I + O + c1 ≤ 9+8+1=18, so c2 = floor((I+O+c1)/10) ≤ 1.

So c2 is 0 or 1.

Then F + c2 = 10, so if c2=0, F=10, impossible; if c2=1, F=9.

So F=9, c2=1.

Good.

So F=9.

Now, col3: R + F + c2 = R + 9 + 1 = R + 10, and this equals R + 10*c3, and c3=1, so R+10 = R + 10*1, yes, holds for any R, but the sum digit is R, so R + 10 = 10 + R, so the digit is R, and carry 1, perfect.

So no restriction on R yet, except it can't be 0 or 1 (C=1), and not F=9.

Now, col4: c3=1=C, good.

Now, col2: I + O + c1 = A + 10*c2 = A + 10*1 = A + 10

So I + O + c1 = A + 10

c1 is carry from col1, 0 or 1.

I and O are digits, A is digit.

Also, all letters distinct.

Now, col1: 1 + O + c0 = Z + 10*c1

c0 is carry from col0, 0,1,2.

Col0: H + D = Y + 10*c0

Now, we have F=9, C=1.

Letters used: C=1, F=9.

Available digits: 0,2,3,4,5,6,7,8

Letters: R,I,H,O,D,A,Z,Y — 8 letters, plus R, so 9 letters total? List: R,I,C,H,F,O,D,A,Z,Y — that's 10 letters, C and F assigned, so 8 left for 8 digits.

Digits available: 0,2,3,4,5,6,7,8

Now, from col2: I + O + c1 = A + 10

Since I and O are at least 0, but probably not 0 if leading, but I is in RICH, which is not leading, so can be 0? R is leading, so R≠0, I can be 0.

Similarly, O in FOOD, not leading, can be 0.

A in CRAZY, not leading, can be 0.

But in sum, C=1, R is thousands digit, so R≠0.

Now, I + O + c1 = A + 10

Min I+O+c1 = 0+2+0=2 (since digits distinct, min two different from available), but A+10 ≥10, so I+O+c1 ≥10.

Max I+O+c1 = 8+7+1=16, so A+10 ≤16, A≤6.

A is digit 0-6.

Also, since I+O+c1 = A+10, and A≥0, so I+O+c1≥10.

Now, col1: 1 + O + c0 = Z + 10*c1

Z is digit, so 1+O+c0 = Z + 10*c1

c1 is 0 or 1.

If c1=0, then 1+O+c0 = Z ≤9, so O+c0 ≤8

If c1=1, then 1+O+c0 = Z +10, so O+c0 = Z+9 ≥9, since Z≥0.

Now, also from col2, I+O+c1 = A+10

Let me try possible values.

Also, in FOOD, O is repeated, so O is the same digit.

Now, let's consider c1.

Suppose c1=1.

Then from col2: I + O +1 = A +10, so I + O = A +9

From col1: 1 + O + c0 = Z +10*1 = Z+10, so O + c0 = Z +9

c0 is 0,1,2.

Z is digit 0-8 (since 1,9 used).

O + c0 = Z +9

Since Z≥0, O+c0≥9

O≤8, c0≤2, so O+c0≤10, so Z+9≤10, Z≤1, but C=1, so Z≠1, so Z=0, then O+c0=9

Or if Z=0, O+c0=9

Z cannot be 1, so only possibility Z=0, O+c0=9

O is digit, c0=0,1,2, so O=9-c0

But F=9, so O≠9, so if c0=0, O=9, conflict; c0=1, O=8; c0=2, O=7

So possible: c0=1, O=8 or c0=2, O=7

Now, also I + O = A +9

And A is digit, I is digit.

Also, from col0: H + D = Y + 10*c0

Now, let's case on c0.

First, suppose c0=1, then O=8

Then I + O = I +8 = A +9, so I = A +1

Also, H + D = Y + 10*1 = Y +10

Now, digits used: C=1, F=9, O=8, Z=0

Available: 2,3,4,5,6,7

Letters left: R,I,H,D,A,Y

And I = A +1

Possible pairs (A,I): A=2,I=3; A=3,I=4; A=4,I=5; A=5,I=6; A=6,I=7

Now, H + D = Y +10

H,D,Y distinct digits from available, and not equal to others.

Also, R is left, and must be from available.

Now, H + D = Y +10, so H and D are large, since sum at least 10+0=10, but Y≥0, so H+D≥10.

Min H+D=2+3=5<10, but we need ≥10, so possible pairs for (H,D) that sum to at least 10.

Available digits: 2,3,4,5,6,7

Possible pairs summing to 10 or more: 3+7=10, 4+6=10, 4+7=11, 5+6=11, 5+7=12, 6+7=13

Y = H+D -10

Y must be in available digits, and not equal to H,D,A,I,R.

Also, I = A+1

Let's try A=2, I=3

Then available for H,D,Y,R: 4,5,6,7

H+D=Y+10

Possible: if H=4,D=6, sum=10, Y=0, but Z=0 already, conflict.

H=4,D=7,sum=11,Y=1, but C=1, conflict.

H=5,D=6,sum=11,Y=1, conflict.

H=5,D=7,sum=12,Y=2, but A=2, conflict.

H=6,D=7,sum=13,Y=3, but I=3, conflict.

No good.

Next, A=3, I=4

Available: 2,5,6,7 (since A=3,I=4 used)

H+D=Y+10

Possible pairs: 5+6=11,Y=1 conflict; 5+7=12,Y=2; 6+7=13,Y=3 but A=3 conflict.

So H=5,D=7,Y=2 or H=7,D=5,Y=2

Y=2, which is available.

So possible.

Say H=5,D=7,Y=2 or vice versa.

Now, letters left: R, and we have used C=1,F=9,O=8,Z=0,A=3,I=4,H and D are 5 and 7, Y=2

So used: 0,1,2,3,4,5,7,8,9 — missing 6, and R must be 6.

Check if all distinct: R=6, I=4, C=1, H=5 or 7, F=9, O=8, D=7 or 5, A=3, Z=0, Y=2

All distinct, good.

Now, verify the addition.

First, suppose H=5, D=7

Then RICH = R I C H = 6 4 1 5

FOOD = F O O D = 9 8 8 7

Sum = 6415 + 9887 = let's calculate: 6415 + 9887 = 16302

CRAZY = C R A Z Y = 1 6 3 0 2 — yes! 16302.

Perfect.

If H=7, D=5, then RICH=6417, FOOD=9885, sum=6417+9885=16302, same sum, CRAZY=16302, so Z=0,Y=2, same.

But in this case, D=5, H=7, but in FOOD, D is units, in RICH, H is units.

In the sum, it's the same.

But we have two possibilities? But in the letters, for RICH, H is last digit, for FOOD, D is last digit.

In first case, H=5,D=7; second H=7,D=5.

But in both cases, the sum is 16302, so CRAZY is fixed.

But the assignment of H and D is swapped.

However, in the puzzle, all letters must represent different digits, which they do in both cases, but we need to see if there are constraints.

In FOOD, O is repeated, but we have O=8, fine.

Is there any other constraint? The problem is to find what letter is replaced, but probably we need to output the digit for each, or perhaps just verify.

But in this case, for Problem 1, we have a solution.

Specifically, with R=6,I=4,C=1,H=5,F=9,O=8,D=7,A=3,Z=0,Y=2 or H=7,D=5.

But if H=7,D=5, then RICH=6417, FOOD=9885, sum=16302, same.

But in CRAZY, Y=2, which is fine.

However, in the sum, it's the same, so both work numerically, but perhaps the puzzle expects specific assignment.

Notice that in FOOD, the two O's are the same, which is satisfied.

But in the first solution, with H=5,D=7, or H=7,D=5.

But let's see if there is a reason to prefer one.

In the tens column: col1: C + O + c0 =1 +8 + c0

c0 is carry from units.

In first case, H+D=5+7=12, so c0=1, Y=2

Then col1: 1+8+1=10, so Z=0, c1=1, good.

In second case, H+D=7+5=12, same c0=1, same everything.

So both give same sum and same digits for other letters.

The only difference is swapping H and D.

But in the puzzle, since H and D are different letters, and in one case H=5,D=7, other H=7,D=5, but both satisfy the equation.

However, typically in such puzzles, we might have additional constraints, but here both work.

Perhaps we can choose one.

But let's see the other problems; perhaps for consistency.

Maybe the puzzle has a unique solution, so I need to check.

Another thing: in RICH, if H=5 or 7, both ok.

But let's list the digits.

In both cases, the mapping is almost the same, only H and D swapped.

But for the purpose, since the sum is the same, and the question might be to find the sum or something, but the instruction is to "figure out what letter is replaced", probably meaning to find the digit for each letter.

But since there are two solutions, perhaps I made a mistake.

Earlier I assumed c1=1, and got this.

But there was another possibility: c0=2, O=7

Let me check that case to see if there are other solutions.

Earlier, when c1=1, we had c0=1,O=8 or c0=2,O=7

We did c0=1,O=8, got solution.

Now c0=2, O=7

Then from earlier, Z=0 (since O+c0=Z+9, 7+2=9=Z+9, so Z=0)

Then from col2: I + O + c1 = I +7 +1 = I+8 = A+10, so I = A+2

Digits used: C=1,F=9,O=7,Z=0

Available: 2,3,4,5,6,8

I = A+2

Possible (A,I): A=2,I=4; A=3,I=5; A=4,I=6; A=5,I=7 but O=7 conflict; A=6,I=8

So A=2,I=4; A=3,I=5; A=4,I=6; A=6,I=8

Now col0: H + D = Y + 10*c0 = Y + 20

But H and D are single digits, max 8+6=14<20, impossible.

H+D ≤8+6=14<20, cannot be Y+20≥20.

So impossible.

Thus, only c0=1,O=8 case works, with two subcases for H and D.

But in both subcases, the sum is 16302, and CRAZY is 16302.

For the puzzle, perhaps it's acceptable, or maybe we need to specify.

But let's move to other problems and come back.

Perhaps for Problem 1, the "PEOPLE" is irrelevant, and we have the solution.

So for Problem 1, one possible assignment is:

R=6, I=4, C=1, H=5, F=9, O=8, D=7, A=3, Z=0, Y=2

Or H=7,D=5.

But in the sum, it's the same.

To choose one, let's take H=5,D=7 for now.

So RICH=6415, FOOD=9887, sum=16302=CRAZY.

Good.

Now Problem 2: BEAL + SNAIL = HONKEY

Letters: B,E,A,L,S,N,I,H,O,K,Y — 11 letters? But only 10 digits, so must have overlap or something.

List: B,E,A,L from BEAL; S,N,A,I,L from SNAIL — A and L repeated; H,O,N,K,E,Y from HONKEY — E and N repeated.

So unique letters: B,E,A,L,S,N,I,H,O,K,Y — that's 11 letters, but only 10 digits, impossible.

Mistake.

BEAL: B,E,A,L

SNAIL: S,N,A,I,L — so A and L are in both.

HONKEY: H,O,N,K,E,Y — N and E are in previous.

So all letters: B,E,A,L,S,N,I,H,O,K,Y — still 11 distinct letters? Let's list: B,E,A,L,S,N,I,H,O,K,Y — yes 11.

But only 10 digits, so impossible unless some letter is the same, but the puzzle says "each letter represents a different digit", so must be 10 or fewer letters.

Perhaps I miscounted.

BEAL: positions: B,E,A,L — 4 letters

SNAIL: S,N,A,I,L — 5 letters, but A and L are shared, so new letters S,N,I

HONKEY: H,O,N,K,E,Y — 6 letters, but N and E are already in, so new H,O,K,Y

So total unique letters: from BEAL: B,E,A,L

From SNAIL: S,N,I (since A,L already)

From HONKEY: H,O,K,Y (since N,E already)

So list: B,E,A,L,S,N,I,H,O,K,Y — that's 11 letters.

But 11 > 10, impossible for different digits.

Unless "HONKEY" is not all new, but still, the set has 11 distinct letters.

Perhaps in "HONKEY", it's H,O,N,K,E,Y, and N and E are already used, but the letter 'N' is the same, so it's the same digit, but the count of distinct letters is still 11 if all are different symbols.

The letters are: B,E,A,L,S,N,I,H,O,K,Y — 11 different characters, so 11 different letters, but only 10 digits, so impossible.

This suggests that my assumption is wrong.

Perhaps for Problem 2, the sum is "HONKEY ANIMALS" but that's even worse.

Another possibility: in the original puzzle, "BEAL + SNAIL = HONKEY" and "ANIMALS" is not part of it, but in user's input, it's listed as "HONKEY" and then "ANIMALS" on the next line, but perhaps "ANIMALS" is the answer or something.

But in the text, for Problem 2, it's:

"2
BEAL
SNAIL
+
HONKEY
ANIMALS"

So perhaps "ANIMALS" is the sum for another, but it's under the same number.

Perhaps it's BEAL + SNAIL = HONKEY, and ANIMALS is a separate thing, but that doesn't make sense.

Another idea: perhaps "HONKEY ANIMALS" is the sum, but that's 11 letters, while BEAL+SNAIL is 4+5=9 digits max sum 9999+99999=109998, 6 digits, so "HONKEY" is 6 letters, "ANIMALS" is 7, too big.

Unless it's not the numerical sum, but the puzzle is to make the addition correct with the letters forming the words, but still, the numerical value must match.

Perhaps for Problem 2, the sum is "HONKEY", and "ANIMALS" is a distractor or the theme.

But then we have 11 letters for 10 digits, impossible.

Unless some letters are the same, but the puzzle says "different digit" for each letter, so each letter symbol represents a different digit, so if there are 11 letter symbols, it's impossible.

So likely, in "HONKEY", it's 6 letters, but some are repeated in the addends, but the distinct count is 11.

Let's list all letter occurrences.

In BEAL: B,E,A,L

In SNAIL: S,N,A,I,L — so A and L are repeated, but same letters, so not new.

In HONKEY: H,O,N,K,E,Y — N and E are already in SNAIL and BEAL, so not new letters.

So the distinct letters are: B,E,A,L,S,N,I,H,O,K,Y — still 11.

B from BEAL, E from BEAL and HONKEY, A from BEAL and SNAIL, L from BEAL and SNAIL, S from SNAIL, N from SNAIL and HONKEY, I from SNAIL, H from HONKEY, O from HONKEY, K from HONKEY, Y from HONKEY.

So the set is {B,E,A,L,S,N,I,H,O,K,Y} — 11 elements.

Yes.

This is a problem.

Perhaps "ANIMALS" is not part of Problem 2, but the beginning of Problem 3.

In user's input:

"2
BEAL
SNAIL
+
HONKEY
ANIMALS
3
ANIMAL
STRIPE
+"

So perhaps "ANIMALS" is the sum for Problem 2, but then for Problem 3, "ANIMAL" is given, which is similar.

If for Problem 2, sum is "ANIMALS", then BEAL + SNAIL = ANIMALS

BEAL 4 letters, SNAIL 5 letters, ANIMALS 7 letters — 4+5=9 digits max sum 109998, 6 digits, but ANIMALS is 7 letters, impossible.

Max sum is 9999 + 99999 = 109998, which is 6 digits, so sum must be 6 digits or less.

"HONKEY" is 6 letters, so likely sum is "HONKEY", and "ANIMALS" is either a mistake or for Problem 3.

For Problem 3, "ANIMAL + STRIPE = " and perhaps sum is "ANIMALS" or "CREATURE", but "ANIMALS" is 7 letters, ANIMAL and STRIPE are 6 each, sum could be 7 digits.

So perhaps for Problem 2, sum is "HONKEY", and "ANIMALS" is the start of Problem 3's sum or something.

To resolve, I'll assume for Problem 2, sum is "HONKEY", and ignore "ANIMALS" for now, but then we have 11 letters, which is impossible.

Unless in "HONKEY", the 'E' and 'N' are the same as in addends, but still, the letter 'B' is only in BEAL, 'S' only in SNAIL, etc., so all are distinct symbols.

Perhaps the puzzle allows that, but mathematically impossible.

Another possibility: "BEAL" might be "BEAR" or something, but in user's input, it's "BEAL".

Perhaps "HONKEY" is "MONKEY" or something, but still 6 letters.

Let's count the letters in the sum for Problem 2: "HONKEY" has 6 letters, addends have 4 and 5, so 9 digits involved, but with overlaps, the number of distinct letters may be less than 11 if some letters are the same, but in this case, all seem different.

List the letters with their occurrences:

- B: only in BEAL

- E: in BEAL and HONKEY

- A: in BEAL and SNAIL

- L: in BEAL and SNAIL

- S: only in SNAIL

- N: in SNAIL and HONKEY

- I: only in SNAIL

- H: only in HONKEY

- O: only in HONKEY

- K: only in HONKEY

- Y: only in HONKEY

So letters that appear only once: B,S,I,H,O,K,Y — 7 letters

Letters that appear twice: E,A,L,N — 4 letters

Total distinct letters: 7 + 4 = 11, yes.

But since each letter represents a digit, and there are 11 letters, but only 10 digits, it's impossible to assign different digits to each.

This suggests that my initial assumption is incorrect, and for Problem 2, the sum is not "HONKEY" alone, but "HONKEY ANIMALS" or something, but that's larger.

Perhaps "ANIMALS" is the sum, and "HONKEY" is not part of it, but in the text, it's listed as "HONKEY" then "ANIMALS", so perhaps the sum is "ANIMALS", and "HONKEY" is a typo or something.

Let's try that.

Suppose for Problem 2, BEAL + SNAIL = ANIMALS

Then BEAL 4 letters, SNAIL 5 letters, ANIMALS 7 letters.

Max sum 9999 + 99999 = 109998, which is 6 digits, but ANIMALS is 7 letters, so impossible unless the sum is at least 1000000, but 9999+99999=109998<1000000, so cannot be 7 digits.

So impossible.

Therefore, the only logical conclusion is that for Problem 2, the sum is "HONKEY", and "ANIMALS" is either a mistake or belongs to Problem 3.

For Problem 3, "ANIMAL + STRIPE = ANIMALS" or " = CREATURE" etc.

And for Problem 2, perhaps "ANIMALS" is not there, or it's " the answer is ANIMALS" but not part of the equation.

To proceed, I'll assume that for Problem 2, the sum is "HONKEY", and we have to live with 11 letters, but that's impossible, so perhaps in "HONKEY", the 'E' and 'N' are the same as in addends, but still, the letter 'B' is unique, etc.

Unless the puzzle has a letter that is not used, but all are used.

Perhaps "BEAL" is "REAL" or something, but in user's input, it's "BEAL".

Another idea: perhaps "HONKEY" is "DONKEY" or "MONKEY", but still 6 letters.

Let's look online or recall if this is a standard puzzle.

Since this is taking too long, and for the sake of time, I'll solve the problems that are clear.

For Problem 1, we have a solution.

For Problem 4: SKIRT + SHIRTS = CLOTHES

SKIRT: S,K,I,R,T — 5 letters

SHIRTS: S,H,I,R,T,S — 6 letters, but S,I,R,T repeated, so new H

CLOTHES: C,L,O,T,H,E,S — 7 letters, T,H,S repeated, so new C,L,O,E

Distinct letters: from SKIRT: S,K,I,R,T

From SHIRTS: H (since S,I,R,T already)

From CLOTHES: C,L,O,E (since T,H,S already)

So list: S,K,I,R,T,H,C,L,O,E — 10 letters, good.

Digits 0-9.

Addition:

```
S K I R T
+ S H I R T S
-------------
C L O T H E S
```

Align to right:

So:

Col0 (units): T + S = S + 10*c0 ? Sum digit is S, so T + S = S + 10*c0, which implies T = 10*c0

T is digit 0-9, so c0=0, T=0 or c0=1, T=10 impossible, so T=0, c0=0

So T=0, and no carry.

Col1 (tens): R + T + c0 = E + 10*c1

T=0, c0=0, so R + 0 + 0 = E + 10*c1, so R = E + 10*c1

R and E digits, so c1=0, R=E, but letters must be different, contradiction.

R = E + 10*c1, if c1=0, R=E, not allowed; if c1=1, R=E+10, impossible for digits.

So contradiction.

What's wrong?

Col1: the addends are SKIRT and SHIRTS.

SKIRT: digits S,K,I,R,T so positions: 10000s:S, 1000s:K, 100s:I, 10s:R, 1s:T

SHIRTS: S,H,I,R,T,S so 100000s:S, 10000s:H, 1000s:I, 100s:R, 10s:T, 1s:S

When adding, align to right:

So:

S K I R T -> this is 5-digit, so in 6-digit grid, it's 0 S K I R T or something? No.

Standard way: when adding numbers of different lengths, align to the right.

So SHIRTS is 6-digit, SKIRT is 5-digit, so:

```
S K I R T
+ S H I R T S
---------------
C L O T H E S
```

So the first number is padded with a leading zero for alignment, but since it's a number, the leading digit can't be zero, but in addition, we can think of it as:

Position: 100000s | 10000s | 1000s | 100s | 10s | 1s

SKIRT: 0 | S | K | I | R | T

SHIRTS: S | H | I | R | T | S

Sum: C | L | O | T | H | E | S ? Sum is CLOTHES, 7 letters, so 7 digits.

CLOTHES: C,L,O,T,H,E,S — so 7 digits: 1000000s:C, 100000s:L, 10000s:O, 1000s:T, 100s:H, 10s:E, 1s:S

So addition:

Col0 (1s): T + S = S + 10*c0 => T = 10*c0, so T=0, c0=0 (since T digit)

Col1 (10s): R + T + c0 = E + 10*c1

T=0, c0=0, so R +0+0 = E +10*c1, so R = E +10*c1

As before, c1=0 implies R=E, conflict; c1=1 implies R=E+10>9, impossible.

So contradiction.

This suggests that my assumption for the sum is wrong.

Perhaps for Problem 4, the sum is "CLOTHES", but the addition is set up differently.

Another possibility: "SKIRT + SHIRTS = CLOTHES" but SHIRTS has 6 letters, SKIRT 5, sum 7, but in the addition, the first number is SKIRT, which is 5-digit, second is SHIRTS 6-digit, sum 7-digit, so the carry must make it 7-digit.

In col0: T + S = S + 10*c0, so T=10*c0, so T=0, c0=0.

Then col1: R + T + c0 = R +0+0 = E +10*c1, so R = E +10*c1, same problem.

Unless the sum's units digit is not S, but in CLOTHES, the last letter is S, so units digit is S.

Perhaps "CLOTHES" is not the sum, but something else.

I think I need to give up and provide the answer for Problem 1, and for others, skip or assume.

For the sake of completing, let's take Problem 5: SNAKE + FROG = TURTLE

SNAKE: S,N,A,K,E — 5 letters

FROG: F,R,O,G — 4 letters

TURTLE: T,U,R,T,L,E — 6 letters, but T repeated.

Distinct letters: S,N,A,K,E,F,R,O,G,T,U,L — 12 letters? List: from SNAKE: S,N,A,K,E

From FROG: F,R,O,G (all new)

From TURTLE: T,U,R,T,L,E — R and E already, T new, U new, L new, but T repeated, so new T,U,L

So total: S,N,A,K,E,F,R,O,G,T,U,L — 12 letters, impossible.

TURTLE has T twice, but same letter, so distinct letters are S,N,A,K,E,F,R,O,G,T,U,L — 12, yes.

Same problem.

For Problem 6: EASTER + EGGS = SPRING

EASTER: E,A,S,T,E,R — 6 letters, E repeated

EGGS: E,G,G,S — 4 letters, G repeated, E and S may be shared

SPRING: S,P,R,I,N,G — 6 letters

Distinct letters: from EASTER: E,A,S,T,R (E repeated)

From EGGS: G (E and S already)

From SPRING: P,I,N (S,R,G already)

So list: E,A,S,T,R,G,P,I,N — 9 letters, good.

Addition:

```
E A S T E R
+ E G G S
-------------
S P R I N G
```

Align to right:

So:

Col0 (1s): R + S = G + 10*c0

Col1 (10s): E + G + c0 = N + 10*c1

Col2 (100s): T + G + c1 = I + 10*c2

Col3 (1000s): S + E + c2 = R + 10*c3

Col4 (10000s): A + 0 + c3 = P + 10*c4 (since EGGS has no 10000s digit, so 0)

Col5 (100000s): E + 0 + c4 = S + 10*c5 (EGGS has no 100000s, so 0)

Col6 (1000000s): c5 = 0 or 1, but sum is 6 digits, so c5=0, and E + c4 = S, since no higher digit.

Sum is SPRING, 6 digits, so no 7th digit, so c5=0, and from col5: E + c4 = S

Also, E is leading digit of EASTER, so E≠0.

S is leading digit of SPRING, so S≠0.

From col5: E + c4 = S

c4 is carry from col4, 0 or 1 or 2.

E and S digits, S = E + c4

Since S > E (because c4≥0, and if c4=0, S=E, but letters different, so c4≥1, S = E + c4 ≥ E+1

c4 ≤2, so S = E+1 or E+2

Now, col4: A + 0 + c3 = P + 10*c4

A and P digits.

Col3: S + E + c2 = R + 10*c3

Col2: T + G + c1 = I + 10*c2

Col1: E + G + c0 = N + 10*c1

Col0: R + S = G + 10*c0

Also, in EGGS, G is repeated, so same digit.

In EASTER, E is repeated, same digit.

Now, let's try to solve.

From col0: R + S = G + 10*c0

G is digit, so R+S ≥10 if c0=1, etc.

From col5: S = E + c4, with c4=1 or 2.

Assume c4=1, then S = E+1

Or c4=2, S=E+2

Now, also, from col4: A + c3 = P + 10*c4

A and P digits 0-9, c3 carry from col3, 0,1,2.

If c4=1, then A + c3 = P + 10

So A + c3 ≥10, P = A + c3 -10

If c4=2, A + c3 = P + 20, so A+c3≥20, but A≤9, c3≤2, max 11<20, impossible.

So c4=1, S = E+1

Then A + c3 = P + 10

P = A + c3 -10

P must be digit 0-9, so A + c3 ≥10, and P = A+c3-10

c3 is 0,1,2, so A ≥10 - c3 ≥8 (since c3≤2)

So A=8 or 9

If A=8, c3≥2, so c3=2, P=8+2-10=0

If A=9, c3≥1, P=9+c3-10=c3-1, so if c3=1, P=0; c3=2, P=1

Now, S = E+1

E≠0, S≠0, and S=E+1, so E from 1 to 8, S from 2 to 9.

Now, col3: S + E + c2 = R + 10*c3

S + E = (E+1) + E = 2E+1

So 2E+1 + c2 = R + 10*c3

R is digit, so 2E+1+c2 = R + 10*c3

c3 is 1 or 2 (from above, since A=8 or 9, c3=2 or 1 or 2)

2E+1+c2 ≤ 2*8+1+2=19, so if c3=2, R +20 ≤19, impossible, so c3 cannot be 2.

From above, if A=8, c3=2; if A=9, c3=1 or 2.

But if c3=2, then 2E+1+c2 = R +20, left side ≤2*8+1+2=19<20, impossible.

So c3 cannot be 2.

Therefore, A cannot be 8 (which requires c3=2), so A=9, and c3=1 or 2, but c3=2 impossible, so c3=1, then P = c3 -1 =1-1=0

So P=0

A=9, c3=1, P=0

S = E+1

2E+1 + c2 = R + 10*1 = R +10

So 2E+1 + c2 = R +10

R = 2E+1 + c2 -10

R must be digit 0-9, so 2E+1+c2 ≥10, and R=2E+1+c2-10

E≥1, but S=E+1≤9, so E≤8

2E+1+c2 ≥10, c2≥0, so 2E≥9-c2≥7, so E≥4 (since 2*3=6<7, 2*4=8≥7)

E≥4

R = 2E+1+c2-10

c2 is carry from col2, 0,1,2

R must be digit 0-9, and not equal to others.

Also, from col0: R + S = G + 10*c0

S=E+1, so R + (E+1) = G + 10*c0

But R = 2E+1+c2-10, so substitute:

(2E+1+c2-10) + (E+1) = G + 10*c0

3E + c2 -8 = G + 10*c0

G is digit, so 3E + c2 -8 = G + 10*c0

Left side, E≥4, c2≥0, so min 3*4+0-8=4, max 3*8+2-8=24-8=16

So G +10*c0 between 4 and 16, so c0=0 or 1

If c0=0, G=3E+c2-8

If c0=1, G=3E+c2-18

G≥0, so if c0=1, 3E+c2-18≥0, 3E+c2≥18, E≥6 (3*6=18)

Now, also, we have other equations.

Let's list what we have:

A=9, P=0, c3=1, S=E+1, c4=1

R = 2E+1+c2-10

G from above.

Also, col2: T + G + c1 = I + 10*c2

Col1: E + G + c0 = N + 10*c1

Col0: R + S = G + 10*c0

And letters: E,A=9,S,T,R,G,P=0,I,N,U,L — and from SPRING, U and L are also there, but in the sum, SPRING has S,P,R,I,N,G, so U and L are not in the sum? No, SPRING is S,P,R,I,N,G — 6 letters, but in the addition, the sum is SPRING, so digits for S,P,R,I,N,G.

But in the addends, we have E,A,S,T,E,R for EASTER, so T is there, and for EGGS, G, and for SPRING, I,N,U,L? No, SPRING is S,P,R,I,N,G — so letters S,P,R,I,N,G.

But in the distinct list, we have E,A,S,T,R,G,P,I,N — and U and L are not mentioned? In SPRING, it's S,P,R,I,N,G — no U or L.

In user's input for Problem 6: "EASTER + EGGS = SPRING" , and SPRING is 6 letters: S,P,R,I,N,G.

So distinct letters: from EASTER: E,A,S,T,R (E repeated)

From EGGS: G (E and S already)

From SPRING: P,I,N (S,R,G already)

So letters: E,A,S,T,R,G,P,I,N — 9 letters, as I had earlier.

No U or L; I misread SPRING as having U, but it's S-P-R-I-N-G, so no U.

Good.

So letters: E,A,S,T,R,G,P,I,N

With A=9, P=0, S=E+1, R=2E+1+c2-10, etc.

Now, let's try E=4, then S=5

R = 2*4+1+c2-10 = 8+1+c2-10 = c2 -1

R must be digit ≥0, so c2-1≥0, c2≥1

R = c2 -1

c2=1, R=0, but P=0, conflict.

c2=2, R=1

So R=1

Then from col0: R + S =1+5=6 = G +10*c0

So G+10*c0=6, so c0=0, G=6

Now, check if digits distinct: A=9, P=0, E=4, S=5, R=1, G=6

Used: 0,1,4,5,6,9

Available: 2,3,7,8

Letters left: T,I,N

Now, col1: E + G + c0 =4+6+0=10 = N +10*c1

So 10 = N +10*c1, so c1=1, N=0, but P=0, conflict.

N=0, but P=0 already.

So not good.

Next, E=5, S=6

R = 2*5+1+c2-10 =10+1+c2-10=1+c2

R=1+c2

c2=0, R=1; c2=1, R=2; c2=2, R=3

Col0: R + S = R+6 = G +10*c0

Also from earlier, 3E + c2 -8 = G +10*c0

E=5, so 3*5 + c2 -8 =15+c2-8=7+c2 = G +10*c0

So G +10*c0 =7+c2

c2=0,1,2

If c2=0, G+10*c0=7, so c0=0, G=7

R=1+0=1

Then col0: R+S=1+6=7=G, good, c0=0

Digits: A=9,P=0,E=5,S=6,R=1,G=7

Used: 0,1,5,6,7,9

Available: 2,3,4,8

Letters left: T,I,N

Col1: E + G + c0 =5+7+0=12 = N +10*c1

So 12 = N +10*c1, so c1=1, N=2

Good, N=2

Col2: T + G + c1 = T +7 +1 = T+8 = I +10*c2

c2=0 (assumed), so T+8 = I +0, so I = T+8

T and I digits, available 3,4,8 (since 2 used for N)

T+8 = I, so if T=3, I=11 invalid; T=4, I=12 invalid; T=8, I=16 invalid. Impossible.

So c2=0 not work.

Next, c2=1, then R=1+1=2

G +10*c0 =7+1=8, so c0=0, G=8 or c0=1, G= -2 invalid, so G=8, c0=0

Col0: R+S=2+6=8=G, good.

Digits: A=9,P=0,E=5,S=6,R=2,G=8

Used: 0,2,5,6,8,9

Available: 1,3,4,7

Letters left: T,I,N

Col1: E+G+c0=5+8+0=13 = N +10*c1

So 13 = N +10*c1, so c1=1, N=3

Good.

Col2: T + G + c1 = T +8 +1 = T+9 = I +10*c2 = I +10*1 = I+10

So T+9 = I+10, thus I = T -1

T and I digits, available 1,4,7 (3 used for N)

I = T-1, so possible T=4, I=3 but N=3 conflict; T=7, I=6 but S=6 conflict; T=1, I=0 but P=0 conflict. No good.

Last, c2=2, then R=1+2=3

G +10*c0 =7+2=9, so c0=0, G=9 but A=9 conflict; or c0=1, G= -1 invalid. So not possible.

So E=5 not work.

Next, E=6, S=7

R = 2*6+1+c2-10 =12+1+c2-10=3+c2

R=3+c2

c2=0, R=3; c2=1, R=4; c2=2, R=5

From 3E+c2-8=18+c2-8=10+c2 = G +10*c0

So G +10*c0 =10+c2

c2=0, G+10*c0=10, so c0=1, G=0 but P=0 conflict; or c0=0, G=10 invalid.

c2=1, G+10*c0=11, so c0=1, G=1; or c0=0, G=11 invalid.

c2=2, G+10*c0=12, c0=1, G=2; or c0=0, G=12 invalid.

So possible:

Case c2=1, c0=1, G=1

R=3+1=4

Col0: R+S=4+7=11 = G +10*c0 =1+10*1=11, good.

Digits: A=9,P=0,E=6,S=7,R=4,G=1

Used: 0,1,4,6,7,9

Available: 2,3,5,8

Letters left: T,I,N

Col1: E+G+c0=6+1+1=8 = N +10*c1

So 8 = N +10*c1, so c1=0, N=8

Good.

Col2: T + G + c1 = T +1 +0 = T+1 = I +10*c2 = I +10*1 = I+10

So T+1 = I+10, thus I = T -9

T and I digits, I = T-9, so T=9, I=0 but A=9,P=0 conflict; T=8, I= -1 invalid. No good.

Other subcase for c2=2, c0=1, G=2

R=3+2=5

Col0: R+S=5+7=12 = G +10*c0 =2+10*1=12, good.

Digits: A=9,P=0,E=6,S=7,R=5,G=2

Used: 0,2,5,6,7,9

Available: 1,3,4,8

Letters left: T,I,N

Col1: E+G+c0=6+2+1=9 = N +10*c1

So 9 = N +10*c1, so c1=0, N=9 but A=9 conflict. Not good.

So E=6 not work.

Next, E=7, S=8

R = 2*7+1+c2-10 =14+1+c2-10=5+c2

R=5+c2

3E+c2-8=21+c2-8=13+c2 = G +10*c0

So G+10*c0=13+c2

c2=0, G+10*c0=13, c0=1, G=3; c0=0, G=13 invalid.

c2=1, G+10*c0=14, c0=1, G=4

c2=2, G+10*c0=15, c0=1, G=5

Now, try c2=0, c0=1, G=3

R=5+0=5

Col0: R+S=5+8=13 = G +10*c0 =3+10*1=13, good.

Digits: A=9,P=0,E=7,S=8,R=5,G=3

Used: 0,3,5,7,8,9

Available: 1,2,4,6

Letters left: T,I,N

Col1: E+G+c0=7+3+1=11 = N +10*c1

So 11 = N +10*c1, so c1=1, N=1

Good.

Col2: T + G + c1 = T +3 +1 = T+4 = I +10*c2 = I +0 = I (c2=0)

So T+4 = I

T and I digits, available 2,4,6 (1 used for N)

I = T+4, so T=2, I=6; T=4, I=8 but S=8 conflict; T=6, I=10 invalid.

So T=2, I=6

Good.

Now, check if all good.

Letters: E=7, A=9, S=8, T=2, R=5, G=3, P=0, I=6, N=1

All distinct: 7,9,8,2,5,3,0,6,1 — yes, all different.

Now verify the addition.

EASTER = E,A,S,T,E,R = 7,9,8,2,7,5 so 798275

EGGS = E,G,G,S = 7,3,3,8 so 7338

Sum = 798275 + 7338 = let's calculate: 798275 + 7000 = 805275, +338 = 805613

SPRING = S,P,R,I,N,G = 8,0,5,6,1,3 so 805613 — yes! Perfect.

So for Problem 6, we have a solution.

Similarly, for other problems, we can solve, but for the sake of time, since the user may expect answers, and for Problem 1 we have, for Problem 6 we have.

For Problem 1, we had RICH + FOOD = CRAZY with R=6,I=4,C=1,H=5,F=9,O=8,D=7,A=3,Z=0,Y=2 or H=7,D=5.

In the sum, it's the same, so perhaps both are valid, but usually puzzles have unique solution, so maybe there is a constraint I missed.

In "PEOPLE", if it's part of the puzzle, but in this case, with the digits, P,E,O,P,L,E — but O=8, E is not assigned yet, in our assignment, for Problem 1, we have letters R,I,C,H,F,O,D,A,Z,Y — 10 letters, all assigned, no E or P or L, so "PEOPLE" is not part of the numerical puzzle, probably just the theme.

So for Problem 1, we can take H=5,D=7 or vice versa.

Let's take H=5,D=7 for definiteness.

So for Problem 1: R=6,I=4,C=1,H=5,F=9,O=8,D=7,A=3,Z=0,Y=2

Sum CRAZY=16302

For Problem 6: E=7,A=9,S=8,T=2,R=5,G=3,P=0,I=6,N=1

Sum SPRING=805613

Now for the other problems, since time is short, and the user may want the answers, I'll provide for the ones I have.

But the user asked to solve the problem, and there are 6 problems.

Perhaps for Problem 2, if we assume sum is "HONKEY", and ignore the 11 letters, but it's impossible, so maybe in some versions, it's different.

Another thought: in Problem 2, "BEAL" might be "BEAR", but in user's input, it's "BEAL".

Perhaps "HONKEY" is "DONKEY",