My house has got a number. If it is a multiple of 3, then it is in between 50…

A conditional clue of the form ‘If P, then Q’ only constrains the house number when P is actually true for it; when P is false, the clue is automatically satisfied no matter what Q says. So each option must be checked against all three clues, using its own divisibility facts to decide whether a clue’s trigger applies — a value is valid only if no triggered clue is violated.

1. Test the range 50–59: any number here that is not a multiple of 4 would force the second clue to place it between 60 and 69 instead — a contradiction — so it would have to be a multiple of 4. For the same reason under the third clue, it would also have to be a multiple of 6. A number that is a multiple of both 4 and 6 must be a multiple of 12, and no multiple of 12 falls between 50 and 59 — so no valid house number lies in this range.

2. Test the range 60–69: any number here would have to NOT be a multiple of 3 (else the first clue would place it between 50 and 59). It would also have to BE a multiple of 6 (else the third clue would place it between 70 and 79). But every multiple of 6 is automatically a multiple of 3 — a direct contradiction. So no valid house number lies in this range either.

3. That leaves the range 70–79. Here the number must NOT be a multiple of 3 (else the first clue misplaces it in the fifties) and MUST be a multiple of 4 (else the second clue misplaces it in the sixties). Among 70–79, the multiples of 4 are 72 and 76; of these, 72 is a multiple of 3 (fails) while 76 is not — so 76 is the only number in this range that satisfies every clue.

Cross-check directly on 76: it is not a multiple of 3, so the first clue’s trigger never fires and it is automatically satisfied; it is a multiple of 4, so the second clue’s trigger never fires either; it is not a multiple of 6, so the third clue does apply — and 76 indeed lies between 70 and 79, satisfying it. All three clues hold together only for this value.

So the house number is 76.