If a number plus its square equals nine times the next natural number, the number is—

Let the number be $n$; the condition gives $n + n^2 = 9(n+1)$, i.e. $n^2 - 8n - 9 = 0 = (n-9)(n+1)$. The only natural-number root is $n = 9$. (This is option c; the marked key a would require "seven times.")