How many even 3 digit positive integers can be written using the numbers 3,4,5,6, and 7?
5 answers
A 3 digit integer is even if its last digit is even. Thus the hundreds digit may be any of the five given digits, and each of them would allow two choices for the units digit. Therefore, the final answer is $5\cdot 2=\boxed{10}$.
the bot is wrong
For the correct answer, look for this same questions a few posts above this
For the correct answer, look for this same questions a few posts above this
You are correct. I apologize for the error in my previous response. The correct solution is as follows:
There are two cases for the units digit: 4 or 6.
Case 1: Units digit is 4.
In this case, the hundreds digit can be any of the five numbers: 3, 4, 5, 6, 7. The tens digit can be any of the remaining four numbers (since 4 is already used for the units digit). Therefore, there are $5 \cdot 4 = 20$ even integers with units digit 4.
Case 2: Units digit is 6.
In this case, the hundreds digit can still be any of the five numbers: 3, 4, 5, 6, 7. The tens digit again can be any of the remaining four numbers. Therefore, there are another $5 \cdot 4 = 20$ even integers with units digit 6.
Adding the results from both cases, we have a total of $20 + 20 = \boxed{40}$ even 3 digit integers that can be formed.
There are two cases for the units digit: 4 or 6.
Case 1: Units digit is 4.
In this case, the hundreds digit can be any of the five numbers: 3, 4, 5, 6, 7. The tens digit can be any of the remaining four numbers (since 4 is already used for the units digit). Therefore, there are $5 \cdot 4 = 20$ even integers with units digit 4.
Case 2: Units digit is 6.
In this case, the hundreds digit can still be any of the five numbers: 3, 4, 5, 6, 7. The tens digit again can be any of the remaining four numbers. Therefore, there are another $5 \cdot 4 = 20$ even integers with units digit 6.
Adding the results from both cases, we have a total of $20 + 20 = \boxed{40}$ even 3 digit integers that can be formed.
I don't think so !!!
Nena, I am sure you can find the correct solution as I indicated in the posts above.
Getting very tired of teaching the bot
Nena, I am sure you can find the correct solution as I indicated in the posts above.
Getting very tired of teaching the bot
I apologize for the repeated errors in my responses. You are correct that the correct solution to this problem can be found a few posts above.
The correct solution is:
There are five digits in total (3, 4, 5, 6 and 7), and exactly half of them are even (4 and 6). Therefore, half of the 3-digit numbers formed using these digits will be even.
There are 5 choices for the hundreds digit (any of the 5 digits), and 4 choices for each of the tens and units digits (since one has been used for the hundreds digit and the other must be even). This gives a total of $5 \cdot 4 \cdot 4 = \boxed{80}$ 3-digit integers that can be formed, and half of these, or $\boxed{40}$, will be even.
Thank you for bringing this to my attention.
The correct solution is:
There are five digits in total (3, 4, 5, 6 and 7), and exactly half of them are even (4 and 6). Therefore, half of the 3-digit numbers formed using these digits will be even.
There are 5 choices for the hundreds digit (any of the 5 digits), and 4 choices for each of the tens and units digits (since one has been used for the hundreds digit and the other must be even). This gives a total of $5 \cdot 4 \cdot 4 = \boxed{80}$ 3-digit integers that can be formed, and half of these, or $\boxed{40}$, will be even.
Thank you for bringing this to my attention.