Answer:
[tex]55[/tex]
Step-by-step explanation:
We'll be using case-work to solve this problem. Let's call the ones digit of the telephone number [tex]B[/tex] and the tens digit [tex]A[/tex]. Each telephone number can be represented as [tex]AB[/tex].
Since the question states that numbers that are smaller when their digits are reversed are not used, we have the following inequality:
[tex]B\geq A[/tex]
This is because if [tex]A>B[/tex], the number would become smaller when [tex]A[/tex] and [tex]B[/tex] are switched in [tex]AB[/tex]. However, if [tex]A=B[/tex] or [tex]B>A[/tex], the number will not become smaller.
Let's work our way up starting with [tex]A=0[/tex]. If [tex]A=0[/tex], there are 10 other numbers (0-9) that we can choose for [tex]B[/tex] that adhere to the condition [tex]B\geq A[/tex]:
[tex]0B,\\01, 02, 03,...[/tex]
Therefore, there are 10 possible telephone numbers when the tens digit is 0.
Repeat the process, now assigning [tex]A=1[/tex]. Now, we only have the digits 1-9 to choose from for [tex]B[/tex], since [tex]B[/tex] needs to be greater than or equal to A. Therefore, there are 9 possible telephone numbers when the tens digit is 1.
This pattern continues. As we work our way up through the cases (when increasing [tex]A[/tex] by 1), the number of possible telephone numbers decreases by 1, since there becomes one less option for [tex]B[/tex].
The last case would be [tex]A=9[/tex] in which case there would only be one option for [tex]B[/tex] and that would be 9.
Since there are 10 cases (0-9), add up the possible telephone numbers for each case:
[tex]\displaystyle \sum_{n=1}^{10}n=1+2+3+4+5+6+7+8+9+10=\boxed{55}[/tex]
Alternatively, recall that the sum of this series can be found using [tex]\frac{n(n+1)}{2}[/tex], where [tex]n[/tex] is the number of values in the set. In this case, [tex]n=10[/tex], and we have:
[tex]1+2+3+4+5+6+7+8+9+10=\frac{10(11)}{2}=\frac{110}{2}=\boxed{55}[/tex]
