22. Consider numbers of the form $10 n+1$, where $n$ is a positive integer. We shall call such a number 'grime' if it cannot be expressed as the product of two smaller numbers, possibly equal, both of which are of the form $10 k+1$, where $k$ is a positive integer.
How many 'grime numbers' are there in the sequence $11,21,31,41, \ldots, 981,991$ ?
A 0
B 8
C 87
D 92
E 99
Solution
C Instead of checking each of the 99 numbers in the sequence $11,21, \ldots, 981,991$ in turn to see whether or not they are 'grime numbers', it is easier to count the numbers in this sequence that are not 'grime numbers'. A number is not a 'grime number' if, and only if, it is a product, say $r s$, where both $r$ and $s$ are numbers of the form $10 k+1$, where $k$ is a positive integer. We can assume that $r \leq s$, and since we are only interested in numbers $r s$ such that $r s \leq 991$, we can also assume that $r \leq 31$, because $32^{2}>991$.
We see that the only products of this form which are not greater than 991 are the twelve numbers
$$
\begin{array}{c}
11 \times 11, \quad 11 \times 21, \quad 11 \times 31, \quad 11 \times 41, \quad 11 \times 51, \quad 11 \times 61, \\
11 \times 71, \quad 11 \times 81, \quad 21 \times 21, \quad 21 \times 31, \quad 21 \times 41 \quad \text { and } \quad 31 \times 31 .
\end{array}
$$

It can be seen, without calculating their values, that all these products are different as they have different prime factorisations.

So there are 12 numbers in the sequence $11,21, \ldots, 981,991$ that are not 'grime numbers'. Hence there are $99-12=87$ 'grime numbers' in this sequence.