Let's perform the following operation: between any two adjacent numbers, insert the number that results from subtracting the left number from the right one. We will perform the same operation on the new row, and so on. Find the sum of the numbers in the row that results after one hundred such operations.

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Let's see how the sum of the numbers in a row changes after one operation. Let $a_{1}, a_{2}, \ldots, a_{\mathrm{n}}$ be the row to which the operation is applied. Then the new row has the form $a_{1}, a_{2}-a_{1}, a_{2}, a_{3}-a_{2}, \ldots, a_{\mathrm{n}-1}, a_{\mathrm{n}}-a_{\mathrm{n}-1}, a_{\mathrm{n}}$. The sum of the numbers in the new row is $a_{1}+\ldots+a_{\mathrm{n}}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\ldots+\left(a_{\mathrm{n}}-a_{\mathrm{n}-1}\right)=s+a_{\mathrm{n}}-a_{1}$, where $s$ is the sum of the numbers in the original row. Notice now that for any row obtained from the row $1,9,8,8$ by the operation described in the problem, $a_{\mathrm{n}}=8, a_{1}=1$. Therefore, after each such operation, the sum of the numbers increases by seven. The sum of the numbers in the original row is 26. Therefore, the sum of the numbers in the row that will result after one hundred such operations is $26+7 \cdot 100=726$.

## Answer

726