Let $$X = \{n \in N : 1 \le n \le 50\}$$. If $$A = \{n \in X : n \text{ is a multiple of } 2\}$$ and $$B = \{n \in X : n \text{ is a multiple of } 7\}$$, then the number of elements in the smallest subset of X, containing both A and B, is

We begin by fixing the universal set. By definition we have $$X=\{n\in\mathbb N:1\le n\le 50\}\,.$$ Thus every natural number from $$1$$ to $$50$$ is in $$X$$.

Next we describe the two given subsets. The first is

$$A=\{n\in X:n\text{ is a multiple of }2\}\,,$$

while the second is

$$B=\{n\in X:n\text{ is a multiple of }7\}\,.$$

The problem asks for the number of elements in the smallest subset of $$X$$ that contains both $$A$$ and $$B$$. The smallest set that contains two sets is simply their union, written $$A\cup B$$. Therefore we need to calculate the cardinality (number of elements) of $$A\cup B$$.

To find $$|A\cup B|$$ we use the Principle of Inclusion-Exclusion, which states

$$|A\cup B| = |A| + |B| - |A\cap B|.$$ Here $$|A|$$ is the number of multiples of $$2$$ in $$X$$, $$|B|$$ is the number of multiples of $$7$$ in $$X$$, and $$|A\cap B|$$ is the number of numbers that are multiples of both $$2$$ and $$7$$, that is, multiples of the least common multiple $$\operatorname{lcm}(2,7)=14$$.

We evaluate each term separately.

Counting |A|: A number is in $$A$$ exactly when it is of the form $$2k$$ with $$1\le 2k\le 50$$. Dividing the inequality by $$2$$ gives $$1\le k\le 25$$, so there are $$25$$ such integers. Hence $$|A|=25$$.

Counting |B|: A number is in $$B$$ exactly when it is of the form $$7m$$ with $$1\le 7m\le 50$$. Dividing by $$7$$ yields $$1\le m\le 7$$ (since $$7\times7=49\le50$$ but $$7\times8=56>50$$). Thus $$|B|=7$$.

Counting |A∩B|: A number lies in the intersection $$A\cap B$$ precisely when it is simultaneously a multiple of $$2$$ and of $$7$$, meaning it is a multiple of $$14$$. Write such a number as $$14r$$ with $$1\le 14r\le 50$$. Dividing by $$14$$ gives $$1\le r\le 3$$ (since $$14\times3=42$$ is allowed but $$14\times4=56>50$$). Therefore $$|A\cap B|=3$$.

Now we substitute these counts into the inclusion-exclusion formula:

$$|A\cup B| = 25 + 7 - 3 = 29.$$

This number $$29$$ is the size of the smallest subset of $$X$$ that contains every element of both $$A$$ and $$B$$.

So, the answer is $$29$$.