To solve the problem, we start by setting up the equation based on the information given. Let \( x \) be the number we are trying to find.
According to the problem, one sixth of the sum of twice the number and 1 is equal to one quarter of the difference between three times the number and 1. We can write this as:
\[ \frac{1}{6} (2x + 1) = \frac{1}{4} (3x - 1) \]
Next, we'll eliminate the fractions by multiplying both sides by 12 (the least common multiple of 6 and 4):
\[ 12 \cdot \frac{1}{6} (2x + 1) = 12 \cdot \frac{1}{4} (3x - 1) \]
This simplifies to:
\[ 2(2x + 1) = 3(3x - 1) \]
Now, distribute on both sides:
\[ 4x + 2 = 9x - 3 \]
Next, we'll get all the \( x \) terms on one side and the constant terms on the other. Subtract \( 4x \) from both sides:
\[ 2 = 5x - 3 \]
Now add 3 to both sides:
\[ 5 = 5x \]
Finally, divide both sides by 5:
\[ x = 1 \]
Thus, the number is:
\[ \boxed{1} \]
To verify, substitute \( x = 1 \) back into the original condition:
The left side:
\[ \frac{1}{6} (2(1) + 1) = \frac{1}{6} (3) = \frac{1}{2} \]
The right side:
\[ \frac{1}{4} (3(1) - 1) = \frac{1}{4} (2) = \frac{1}{2} \]
Both sides are equal, confirming that our solution is correct:
\[ \boxed{1} \]