THIS × IS = BASIC
Step-by-step solution
How do you solve THIS × IS = BASIC? The answer is 3027 × 27 = 81729. Here is how the repeated I and S restrict the last two columns, and why the five-digit result rules out every other assignment.
Each letter represents a digit from 0 to 9. Matching letters keep the same value; different letters cannot share a digit. The leading letters T, I, and B cannot be zero.
Jump to the answer1. Start with S × S in the units column
The units digit of the product comes from S × S. Its last digit is C; call the carry into the tens c1. For example, S = 7 would give C = 9 and c1 = 4, because 7 × 7 = 49.
S cannot be 0, 1, 5, or 6: each square ends in the same digit, which would make C = S. The only possible S values are 2, 3, 4, 7, 8, and 9. This is a list of candidates, not yet the answer.
S × S = C + 10c1
2. Use the repeated I to restrict the tens
In the tens column, I × S and S × I both contribute. Add c1; the total must end in I. Equivalently, I × (2S − 1) + c1 must be a multiple of 10.
Try each surviving S with I from 1 to 9, keeping I, S, and C different. The table contains every surviving pair. Here c2 is the carry into the hundreds. Multiplication carries can exceed 9: they are not letter values and do not follow the different-letter rule.
| I | S | C | Carry c2 |
|---|---|---|---|
| 2 | 3 | 9 | 1 |
| 2 | 7 | 9 | 3 |
| 4 | 3 | 9 | 2 |
| 6 | 3 | 9 | 3 |
| 6 | 9 | 1 | 11 |
| 7 | 4 | 6 | 5 |
| 8 | 3 | 9 | 4 |
2 × I × S + c1 = I + 10c2
3. Find H and use the five-digit limit
The hundreds column combines H × S, I × I, and c2; it must end in S. For each pair above, test H from 0 to 9 without reusing I, S, or C. The eight surviving rows are listed below, together with their carry c3.
Next, T × S + H × I + c3 gives digit A and carry c4. The leftmost column is T × I + c4 = B. Since BASIC has exactly five digits, B must be at most 9.
The last table column is a lower bound for B: temporarily set T = 1, even if that digit is already used. Increasing T cannot decrease either T × I or c4. Any row whose minimum B exceeds 9 is therefore impossible. Four rows remain.
| I | S | H | c3 | Minimum B |
|---|---|---|---|---|
| 2 | 3 | 6 | 2 | 3 |
| 2 | 7 | 0 | 0 | 2 |
| 4 | 3 | 5 | 3 | 6 |
| 6 | 3 | 8 | 6 | 11 |
| 6 | 9 | 8 | 11 | 12 |
| 7 | 4 | 0 | 5 | 7 |
| 7 | 4 | 5 | 7 | 11 |
| 8 | 3 | 5 | 8 | 13 |
H × S + I × I + c2 = S + 10c3; T × S + H × I + c3 = A + 10c4; T × I + c4 = B
4. Eliminate three of the remaining rows
For I = 2, S = 3, H = 6: T = 1 gives B = 3, reusing S. The next unused nonzero T is 4, which already gives B = 10. Larger T values also fail.
For I = 4, S = 3, H = 5: T = 1 makes the thousands total 26, so A = 6 and c4 = 2. But B = 4 + 2 = 6 as well. T = 2 already gives B = 10, so this row cannot work.
For I = 7, S = 4, H = 0: T = 1 gives B = 7, reusing I. Any T of at least 2 makes T × I alone at least 14, too large for B.
I = 2; S = 7; H = 0; C = 9; c3 = 0
5. Finish with T, A, and B
In the surviving row, the thousands total is 7T. T = 1 gives c4 = 0 and B = 2, which repeats I. T = 2 is already used by I. T = 4 would give c4 = 2 and B = 10, and every larger T also exceeds the five-digit limit. Only T = 3 remains.
Now 7 × 3 = 21 gives A = 1 and c4 = 2. The leading digit is B = 3 × 2 + 2 = 8. All seven letters have different values, including H = 0, which is allowed because H is not a leading letter.
Check the multiplication in two parts: 3027 × 7 = 21189 and 3027 × 20 = 60540. Adding these partial products gives 81729, exactly BASIC.
3027 × 27 = 21189 + 60540 = 81729
Answer and verification
- T
- = 3
- H
- = 0
- I
- = 2
- S
- = 7
- B
- = 8
- A
- = 1
- C
- = 9
3027 × 27 = 81729
All letters have distinct digit values, no leading digit is zero, and the equation checks out.
Takeaway
Repeated letters connect the multiplication columns, while the length of the product bounds the leading digits. Track carries separately from letter values, list the surviving cases, and reject both repeated digits and an unwanted sixth digit. Together, these checks leave exactly one solution.