Science:Math Exam Resources/Courses/MATH312/December 2005/Question 05
Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.
• Q1 • Q2 • Q3 • Q4 • Q5 • Q6 • Q7 • Q8 • Q9 • Q10 • Q11 •
Question 05 |
---|
Multiple Choice The number of zeros at the end of the decimal representation of 153! is
|
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? |
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. |
Hint 1 |
---|
To count the number of zeroes in this number, it suffices to count the number of factors of 10 the number has. |
Hint 2 |
---|
To accomplish hint 1, notice that the number of factors of 2 inside must exceed the number of factors of 5 so it suffices to count the number of factors of 5 inside . |
Hint 3 |
---|
The number of 5s occurring in the expansion of is given by
This is true since you get a factor of 5 every 5 integers. You get another factor of 5 every 25 integers and so on. |
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
|
Solution |
---|
Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. The correct answer is c = 37. Following the hints, we seek to count the number of factors of 10 in the number . To do this we count the number of 5s occuring in its expansion. The number of 5s occurring in the expansion of is given by
This is true since you get a factor of 5 every 5 integers. You get another factor of 5 every 25 integers and so on. Counting this, we see that . |