Question

The telephone numbers in a small town have two digits. They run from 00 to 99.

Answer 1

Answer:

$Since there are 10 cases (0-9), add up the possible telephone numbers for each case:\displaystyle \sum_{n=1}^{10}n=1+2+3+4+5+6+7+8+9+10=\boxed{55}n=1∑10n=1+2+3+4+5+6+7+8+9+10=55Alternatively, recall that the sum of this series can be found using \frac{n(n+1)}{2}2n(n+1) , where nn is the number of values in the set. In this case, n=10n=10 , and we have:1+2+3+4+5+6+7+8+9+10=\frac{10(11)}{2}=\frac{110}{2}=\boxed{55}1+2+3+4+5+6+7+8+9+10=210(11)=2110=55$

Answer 2

Answer:

$55$

Step-by-step explanation:

We'll be using case-work to solve this problem. Let's call the ones digit of the telephone number $B$ and the tens digit $A$. Each telephone number can be represented as $AB$.

Since the question states that numbers that are smaller when their digits are reversed are not used, we have the following inequality:

$B\geq A$

This is because if $A>B$, the number would become smaller when $A$ and $B$ are switched in $AB$. However, if $A=B$ or $B>A$, the number will not become smaller.

Let's work our way up starting with $A=0$. If $A=0$, there are 10 other numbers (0-9) that we can choose for $B$ that adhere to the condition $B\geq A$:

$0B,\\01, 02, 03,...$

Therefore, there are 10 possible telephone numbers when the tens digit is 0.

Repeat the process, now assigning $A=1$. Now, we only have the digits 1-9 to choose from for $B$, since $B$ 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 $A$ by 1), the number of possible telephone numbers decreases by 1, since there becomes one less option for $B$.

The last case would be $A=9$ in which case there would only be one option for $B$ and that would be 9.

Since there are 10 cases (0-9), add up the possible telephone numbers for each case:

$\displaystyle \sum_{n=1}^{10}n=1+2+3+4+5+6+7+8+9+10=\boxed{55}$

Alternatively, recall that the sum of this series can be found using $\frac{n(n+1)}{2}$, where $n$ is the number of values in the set. In this case, $n=10$, and we have:

$1+2+3+4+5+6+7+8+9+10=\frac{10(11)}{2}=\frac{110}{2}=\boxed{55}$