How many 2-digit numbers are there in which the sum of its tens digit and its ones digit is equal to 8?
17, 26, 35, 44, 53, 62, 71, 80
I see 8 numbers
You are correct. The answer is indeed $\boxed{8}$. Thank you for pointing out the mistake.
To find the number of 2-digit numbers where the sum of the tens digit and the units digit is equal to 8, we can break down the problem step by step.
Step 1: Determine the possible values for the tens digit.
A 2-digit number can range from 10 to 99. So, the tens digit can be any digit from 1 to 9.
Step 2: Determine the possible values for the ones digit based on the chosen tens digit.
Once you choose a tens digit, you need to find the ones digit that, when added to the chosen tens digit, equals 8.
For example, if the tens digit is 1, then the ones digit would be 8 - 1 = 7.
If the tens digit is 2, then the ones digit would be 8 - 2 = 6.
Continue this process until you have exhausted all possible tens digits (1 to 9).
Step 3: Count the number of valid combinations.
For each possible tens digit, there is only one valid ones digit that satisfies the condition. So, the total number of 2-digit numbers satisfying the condition is equal to the number of possible tens digits, which is 9.
Therefore, there are 9 two-digit numbers in which the sum of the tens digit and the ones digit equals 8.
To count the number of 2-digit numbers with digits summing to 8, we can consider pairs of digits that sum to 8: $\{1,7\}$, $\{2,6\}$, $\{3,5\}$, and $\{4,4\}$. For the first three pairs, we have two ways to arrange the digits, but for the last pair, there's only one way since the digits are identical:
$$17,\ 71,\ 26,\ 62,\ 35,\ 53,\ 44.$$So there are $\boxed{7}$ such numbers.