Question

Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 56/11 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?

Answer 1

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

\[
\boxed{5}
\]