To solve Eric's riddle, we start with the information given:
- Eric is thinking of a fraction equivalent to \( \frac{2}{5} \).
- The numerator of this fraction is \( 2 \).
- We need to determine if the denominator is greater than 18.
Since we know that equivalent fractions can be formed by multiplying both the numerator and the denominator of a fraction by the same non-zero integer, we can express the fraction Eric is thinking of as follows:
\[ \frac{2}{5} = \frac{2 \times k}{5 \times k} \]
where \( k \) is any positive integer. Here, the numerator remains 2.
Thus, any fraction equivalent to \( \frac{2}{5} \) that has a numerator of 2 will have the form:
\[ \frac{2}{d} \]
where \( d = \frac{5k}{1} \) for some integer \( k \). Knowing:
\[ d = \frac{5k}{2} \]
This means \( d \) must be multiplied by \( \frac{5}{2} \) to form the denominator keeping numerator 2. Then we check if \( d \) is less than 18:
\[ \frac{5k}{2} > 18 \] \[ 5k > 36 \implies k > \frac{36}{5} = 7.2 \]
Since \( k \) must be an integer, the smallest integer \( k \) can take is 8. Thus, when \( k = 8 \):
\[ d = \frac{5 \times 8}{2} = \frac{40}{2} = 20 \]
At this point, we know that if \( k = 8 \), then the resulting fraction is \( \frac{2}{20} \), thus the result would be \( 2/20 = 1/10 \). , confirming that the denominator 20 is not less than 18.
Hence, we realize it is possible to have:
- Expressions like \( d = \frac{5k}{2} \), so trying with lower values of \( k \):
- when \( k=1, d=5/2\) which isn't suitable because they doesn't work for integer.
Since any multiple of \( \frac{2}{5} \) with \( k>7\) yields \( d \) greater than or equal to 20 confirms it is more than less than 18 for \( d \).
In conclusion, Eric's fraction is any form of \( \frac{2}{d} \) for \( d \geq 20 \), thus the denominator is not less than 18.