Find a four-digit number that is a multiple of 12 and whose digits have a product greater than 25 but less than 30. Give any one such number.

Step-by-step solution

1. 1

   Narrow the digit product to the possible values. The product must be $26$, $27$, $28$ or $29$. Test each for being a product of four digits from $1$ to $9$ (a zero digit would force the product to $0$):

   $26=2\times 13\ \text{— needs a factor }13,\ \text{impossible}$

   $29\ \text{is prime and}>9,\ \text{impossible}$

   $27=3^3=1\times 3\times 3\times 3\ \text{— possible in principle}$

   $28=2^2\times 7=1\times 2\times 2\times 7\ \text{or}\ 1\times 1\times 4\times 7\ \text{— possible}$

2. 2

   Recall what divisibility by 12 requires. Since $12=4\times 3$ with $\gcd(4,3)=1$, the number must be divisible by both:

   $\text{last two digits form a multiple of }4,\qquad \text{digit sum is a multiple of }3$

3. 3

   Rule out the digit product 27. Its only digit multiset is $\{1,3,3,3\}$, whose sum is $10$ — not a multiple of $3$, so no arrangement is divisible by $3$, let alone by $12$. That leaves product $28$.

4. 4

   Handle the digit multisets that give 28. For $\{1,2,2,7\}$ the digit sum is $12$, a multiple of $3$ $\checkmark$, so only the divisibility-by-$4$ condition remains. For $\{1,1,4,7\}$ the sum is $13$, not a multiple of $3$, so that multiset is out. Arranging $1,2,2,7$ so the last two digits form a multiple of $4$ (namely $12$ or $72$) gives the complete list:

   $1272,\quad 2172,\quad 2712,\quad 7212$

5. 5

   Verify one answer fully. Take $N=1272$:

   $1272\div 12=106\quad\checkmark$

   $1\times 2\times 7\times 2=28,\qquad 25<28<30\quad\checkmark$

   An exhaustive search over all four-digit multiples of $12$ confirms these four numbers are the only solutions, so any one of them may be given as the answer.