Digit Detective: A Three-Digit Card With One Condition Too Many?
The 'Digit Detective' round at a school olympiad hands Team Asha a sealed card. The card states five conditions about a hidden number: it is three-digit; its three digits are all different and none of them is zero; the digits add up to 12; the number exceeds its own reversal by 297; and the hundreds digit is even. The reversal is formed by writing the same three digits in the opposite order, so the reversal of 100a + 10b + c is 100c + 10b + a. Team Asha is told that these conditions pin down exactly one number, and that they should not find the answer by listing 100 to 999 one by one. A rival team claims the last condition - hundreds digit even - is redundant, because the other four conditions already leave only one possibility. The judge asks Asha's team to settle that claim as well as to find the number.
1. Find the number using algebra, not exhaustive listing.
Read the worked answer
Write the number as 100a + 10b + c, where a, b, c are the hundreds, tens and units digits.
Step 1 - use the reversal condition. (100a + 10b + c) − (100c + 10b + a) = 99(a − c) = 297, so a − c = 3.
Step 2 - use the digit sum. a + b + c = 12, so b = 12 − a − c = 12 − a − (a − 3) = 15 − 2a.
Step 3 - bound a. c = a − 3 ≥ 1 gives a ≥ 4, and b ≥ 1 gives 15 − 2a ≥ 1, so a ≤ 7. Thus a ∈ {4, 5, 6, 7}.
Step 4 - test the four cases for distinct digits.
- a = 4: b = 7, c = 1 → 471, all digits distinct ✔
- a = 5: b = 5, c = 2 → b = a, rejected
- a = 6: b = 3, c = 3 → b = c, rejected
- a = 7: b = 1, c = 4 → 714, digits distinct ✔
Two survivors so far: 471 and 714. The final condition, hundreds digit even, keeps only 471. Check: 471 − 174 = 297 and 4 + 7 + 1 = 12.
2. Is the rival team right that the 'hundreds digit is even' condition is redundant? Justify your answer.
Read the worked answer
No - the rival team is wrong, and Step 4 above shows exactly why.
Without the parity condition the other four conditions admit two numbers, 471 and 714. Both are three-digit, both use distinct non-zero digits, both have digit sum 12, and both exceed their reversals by 297 (714 − 417 = 297).
So the parity condition is doing real work: it removes 714 and leaves a unique answer. If the card had omitted it, the honest answer would have been '471 or 714', not a single number.
This is the key judgement the question tests: a condition is only redundant if the remaining conditions already force a unique solution. 'It did not feel necessary' is not an argument - you must check what happens without it.
3. Suppose the 297 is replaced by 198 while the hundreds-digit-even rule stays. Would the answer still be unique?
Read the worked answer
Redo Step 1: 99(a − c) = 198, so a − c = 2.
Then b = 12 − a − c = 12 − a − (a − 2) = 14 − 2a. Constraints: c = a − 2 ≥ 1 gives a ≥ 3, and b ≥ 1 gives a ≤ 6, so a ∈ {3, 4, 5, 6}.
- a = 3: b = 8, c = 1 → 381
- a = 4: b = 6, c = 2 → 462
- a = 5: b = 4, c = 3 → 543
- a = 6: b = 2, c = 4 → 624
All four have distinct non-zero digits and digit sum 12. Imposing the even-hundreds rule keeps only 462 and 624, so the answer is not unique.
The general fact behind all three parts: for any three-digit number, the difference between the number and its reversal is always 99(a − c), a multiple of 99, which is why these puzzles reduce to comparing the outer two digits.