To find the smallest integer written down by Pearl, let’s denote the seven consecutive integers starting with \( n \). Therefore, the integers can be represented as:
\[ n, n+1, n+2, n+3, n+4, n+5, n+6 \]
The sum \( S \) of these integers can be calculated as follows:
\[ S = n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6) \] \[ S = 7n + (0 + 1 + 2 + 3 + 4 + 5 + 6) \] \[ S = 7n + 21 \]
The largest of these seven integers is \( n+6 \). According to the problem, the sum of the integers is equal to \( \frac{56}{11} \) times the largest integer:
\[ 7n + 21 = \frac{56}{11}(n + 6) \]
To eliminate the fraction from the equation, we can multiply both sides by \( 11 \):
\[ 11(7n + 21) = 56(n + 6) \] \[ 77n + 231 = 56n + 336 \]
Now, we rearrange the equation to isolate \( n \):
\[ 77n - 56n = 336 - 231 \] \[ 21n = 105 \] \[ n = 5 \]
Now that we have determined \( n \), which is the smallest integer Pearl wrote down, let’s summarize:
The integers are \( 5, 6, 7, 8, 9, 10, 11 \).
Finally, we double-check the sum and the condition given in the problem. The sum of these integers is:
\[ 5 + 6 + 7 + 8 + 9 + 10 + 11 = 56 \]
And the largest integer is:
\[ n + 6 = 5 + 6 = 11 \]
Now, we check if \( 56 \) is indeed \( \frac{56}{11} \) times the largest integer:
\[ \frac{56}{11} \times 11 = 56 \]
Thus, the condition holds true.
Therefore, the smallest integer that Pearl wrote down is
\[ \boxed{5} \]