STORK + KOI + BOAS + TROUT + TOADS + RATS = BIRDS
Step-by-step solution
Six animal words add up to BIRDS. Find the ten letter values, then read 7059 6812354 as letters. This is a harder cryptarithmetic problem: we reduce the possibilities with carries and the sum of all ten digits, then explicitly check the remaining branches.
Each letter represents a digit from 0 to 9. Matching letters keep the same value; different letters cannot share a digit. The leading letters S, K, B, T, and R cannot be zero.
Jump to the answer1. Write one equation for each column
Call the carries leaving the units, tens, hundreds, and thousands c1, c2, c3, and c4. Carries are ordinary numbers, not letter assignments: they may equal a letter value or each other. There is no carry beyond the five-digit result.
The units contain K + I + S + T + S + S and end in S, so cancel one S. In the tens, D appears on both sides and cancels completely. This does not make D arbitrary: all ten different letters must use all ten digits exactly once.
| Column | Equation |
|---|---|
| Units | K + I + T + 2S = 10c1 |
| Tens | R + O + A + U + T + c1 = 10c2 |
| Hundreds | 3O + K + 2A + c2 = R + 10c3 |
| Thousands | T + B + 2R + O + c3 = I + 10c4 |
| Ten-thousands | S + 2T + c4 = B |
2. Bound the leftmost letters and the carries
Suppose c4 = 0. Substituting B = S + 2T into the thousands gives I = S + 3T + 2R + O + c3. The distinct nonzero digits S, T, and R make S + 3T + 2R at least 10: even the smallest arrangement gives 3 + 3 × 1 + 2 × 2 = 10. That cannot fit the single digit I. Therefore c4 is at least 1.
Now S + 2T + c4 = B ≤ 9 gives T ≤ 3 and S ≤ 6. The units equation is positive and at most 9 + 9 + 3 + 2 × 6 = 33, so c1 is 1, 2, or 3. In the tens, the four distinct digits R, O, A, U total at most 9 + 8 + 7 + 6 = 30; adding T and c1 gives at most 36. Thus c2 is also 1, 2, or 3.
The hundreds can carry at most 5. The thousands then total at most 3 + 9 + 2 × 9 + 9 + 5 = 44, so c4 is at most 4. Combining this with S + 2T + c4 ≤ 9 gives 2 ≤ T + c4 ≤ 6.
1 ≤ S ≤ 6; 1 ≤ T ≤ 3; 1 ≤ c1, c2 ≤ 3; 1 ≤ c4 ≤ 4
3. Use 0 + 1 + … + 9 = 45 to recover D
Add the nine letter values other than D. Use the tens equation to replace R + O + A + U + T, giving S + K + I + B + 10c2 − c1. The units equation replaces K + I, and B = S + 2T + c4 then reduces this to T + c4 + 9c1 + 10c2.
Consequently D = 45 − T − c4 − 9c1 − 10c2. Since D is between 0 and 9 and T + c4 is between 2 and 6, the value 9c1 + 10c2 must be between 30 and 43. For c1 and c2 in 1, 2, 3, only 39, 38, and 37 fit: the carry pairs are (1, 3), (2, 2), and (3, 1).
The pair (3, 1) would require K + I + T + 2S = 30. Because K + I ≤ 17 and S ≤ 8 − 2T, this forces T = 1 and S = 6. The leftmost column then forces c4 = 1 and B = 9, while K + I = 17 needs the digits 8 and 9. That reuses B, so discard this pair.
D = 45 − T − c4 − 9c1 − 10c2
4. Rule out c1 = 1, c2 = 3
For this pair, D = 6 − T − c4 and K + I = 10 − T − 2S. Check S from 1 to 6, T from 1 to 3, and c4 from 1 to 4; calculate B and D, then split K + I into unused digits with K nonzero. Reject values outside 0–9 or repeated letters. Only the two rows below remain; the tuples follow the header order.
In the first row, the unused digits for R, O, A, U are 4, 6, 7, 9. The thousands total T + B + 2R + O + c3 is at least 3 + 8 + 2 × 4 + 6 = 25, but I + 10c4 is only 10. In the second row, the unused digits are 4, 7, 8, 9: the total is at least 1 + 6 + 2 × 4 + 7 = 22, but must be 20. Both rows fail.
| S, T, c4 | B, D, K, I | Minimum > target |
|---|---|---|
| 1, 3, 1 | 8, 2, 5, 0 | 25 > 10 |
| 2, 1, 2 | 6, 3, 5, 0 | 22 > 20 |
c1 = 2; c2 = 2; D = 7 − T − c4; K + I = 20 − T − 2S
5. Check the remaining branches systematically
Repeat the same bounded check with c1 = c2 = 2. Calculate B = S + 2T + c4, D = 7 − T − c4, and K + I = 20 − T − 2S. The table lists all 31 ordered choices of K and I grouped by S, T, c4. Each of the six assigned letters S, T, B, D, K, I must be different.
For each choice, try distinct O and A from the four unused digits. Calculate X = 3O + K + 2A + 2. Its last digit must be an unused nonzero R, and its tens digit is c3. Check T + B + 2R + O + c3 = I + 10c4. Give U the remaining digit and check the tens equation as well. The final column counts complete assignments that pass these checks, not guesses about which branch looks promising.
Only the first row has a valid completion, with K = 7 and I = 9. This determines S = 1, T = 2, c4 = 1, B = 6, and D = 4. The next step follows that surviving branch in detail.
| S, T, c4 | (K, I) | Valid completions |
|---|---|---|
| 1, 2, 1 | (7, 9); (9, 7) | 1 |
| 1, 3, 2 | (7, 8); (8, 7) | 0 |
| 2, 1, 2 | (7, 8); (8, 7) | 0 |
| 2, 1, 3 | (6, 9); (9, 6) | 0 |
| 3, 1, 1 | (4, 9); (9, 4) | 0 |
| 3, 1, 2 | (5, 8); (8, 5) | 0 |
| 3, 1, 4 | (5, 8); (6, 7); (7, 6); (8, 5) | 0 |
| 3, 2, 1 | (5, 7); (7, 5) | 0 |
| 4, 1, 1 | (2, 9); (3, 8); (8, 3); (9, 2) | 0 |
| 4, 1, 3 | (5, 6); (6, 5) | 0 |
| 5, 1, 2 | (2, 7); (3, 6); (6, 3); (7, 2) | 0 |
| 6, 1, 1 | (3, 4); (4, 3); (7, 0) | 0 |
6. Finish with R, O, A, and U
The unused digits are 0, 3, 5, 8. The thousands equation becomes 2R + O + c3 = 11. R cannot be zero, and R = 8 is already too large, leaving 3 or 5.
If R = 3, then O + c3 = 5. O = 8 is too large. O = 0 would need c3 = 5, but the hundreds total 7 + 2A + 2 is at most 25. O = 5 would need c3 = 0, but the hundreds total 15 + 7 + 2A + 2 is at least 24. So R cannot be 3.
Therefore R = 5 and O + c3 = 1. Only O = 0 is available, giving c3 = 1. The hundreds equation is now 7 + 2A + 2 = 15, so A = 3. The last unused digit is U = 8. The tens check is 5 + 0 + 3 + 8 + 2 + 2 = 20, as required.
R = 5; O = 0; A = 3; U = 8
7. Decode 7059 6812354
Reverse the letter-to-digit mapping. The first group reads 7 → K, 0 → O, 5 → R, 9 → I. The second reads 6 → B, 8 → U, 1 → S, 2 → T, 3 → A, 5 → R, 4 → D.
The hidden animal name is KORI BUSTARD. We obtain it from the verified arithmetic, rather than using a guessed animal name to force the letter values.
7059 6812354 → KORI BUSTARD
Answer and verification
- S
- = 1
- T
- = 2
- O
- = 0
- R
- = 5
- K
- = 7
- I
- = 9
- B
- = 6
- A
- = 3
- U
- = 8
- D
- = 4
12057 + 709 + 6031 + 25082 + 20341 + 5321 = 69541
All letters have distinct digit values, no leading digit is zero, and the equation checks out.
Takeaway
A letter that cancels from a column is not necessarily free. Here, using all ten digits recovers D and sharply restricts the carries. Some branching remains: list it explicitly, reject cases against the column equations, and verify the final sum before decoding the message.