Set $$A$$ has $$m$$ elements and set $$B$$ has $$n$$ elements. If the total number of subsets of $$A$$ is 112 more than the total number of subsets of $$B$$, then the value of $$m \cdot n$$ is___.

We have two finite sets, set $$A$$ with $$m$$ elements and set $$B$$ with $$n$$ elements.

First, we state the basic formula: for any set containing $$r$$ elements, the total number of its subsets is $$2^{\,r}$$. This is because each element can be either “chosen” or “not chosen”, giving two possibilities per element and hence $$2 \times 2 \times \dots \times 2 = 2^{\,r}$$ possibilities in all.

Applying this formula to the two given sets, the number of subsets of $$A$$ is $$2^{\,m}$$ and the number of subsets of $$B$$ is $$2^{\,n}$$.

We are told that

$$2^{\,m} = 2^{\,n} + 112.$$

To handle the difference of two powers of two, we isolate the common factor $$2^{\,n}$$ on the right hand side. Subtracting $$2^{\,n}$$ from both sides and then factoring gives

$$2^{\,m} - 2^{\,n} = 112 \quad\Longrightarrow\quad 2^{\,n}\bigl(2^{\,m-n} - 1\bigr) = 112.$$

Let us introduce a new positive integer $$k$$ defined by $$k = m - n$$. Because $$m > n$$ gives a positive difference, we have $$k \ge 1$$. Substituting $$k$$ into the previous expression, we get

$$2^{\,n}\bigl(2^{\,k} - 1\bigr) = 112.$$

The integer $$112$$ can be written in its prime-factor form:

$$112 = 16 \times 7 = 2^{4} \times 7.$$

This factorisation tells us that any power of two dividing $$112$$ must be at most $$2^{4}$$, so we must have $$2^{\,n} \le 2^{4}$$ and therefore $$n \le 4$$ (since $$n$$ is a non-negative integer).

Now we test each possible value of $$n$$ from $$0$$ to $$4$$, computing the corresponding value of $$2^{\,k} - 1$$ and checking whether it is an integer power of two.

• If $$n = 0$$, then $$2^{\,n} = 1$$ and

$$1\bigl(2^{\,k} - 1\bigr) = 112 \;\Longrightarrow\; 2^{\,k} - 1 = 112 \;\Longrightarrow\; 2^{\,k} = 113,$$ which is not a power of two. So $$n = 0$$ is impossible.

• If $$n = 1$$, then $$2^{\,n} = 2$$ and

$$2\bigl(2^{\,k} - 1\bigr) = 112 \;\Longrightarrow\; 2^{\,k} - 1 = 56 \;\Longrightarrow\; 2^{\,k} = 57,$$ and $$57$$ is not a power of two. So $$n = 1$$ is impossible.

• If $$n = 2$$, then $$2^{\,n} = 4$$ and

$$4\bigl(2^{\,k} - 1\bigr) = 112 \;\Longrightarrow\; 2^{\,k} - 1 = 28 \;\Longrightarrow\; 2^{\,k} = 29,$$ but $$29$$ is not a power of two. So $$n = 2$$ is impossible.

• If $$n = 3$$, then $$2^{\,n} = 8$$ and

$$8\bigl(2^{\,k} - 1\bigr) = 112 \;\Longrightarrow\; 2^{\,k} - 1 = 14 \;\Longrightarrow\; 2^{\,k} = 15,$$ and $$15$$ is not a power of two. So $$n = 3$$ is impossible.

• If $$n = 4$$, then $$2^{\,n} = 16$$ and

$$16\bigl(2^{\,k} - 1\bigr) = 112 \;\Longrightarrow\; 2^{\,k} - 1 = 7 \;\Longrightarrow\; 2^{\,k} = 8.$$

Because $$8 = 2^{3}$$ is indeed a power of two, this case works with $$k = 3$$. Therefore

$$m - n = k = 3 \;\Longrightarrow\; m = n + 3 = 4 + 3 = 7.$$

Now that we have $$m = 7$$ and $$n = 4$$, we compute the required product:

$$m \cdot n = 7 \times 4 = 28.$$

So, the answer is $$28$$.