How many zeroes are at the end of (45^8)(88^5)

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The discussion focuses on determining the number of trailing zeroes in the expression (45^8)(88^5) without using a calculator. The breakdown of the numbers into their prime factors reveals that there are 8 factors of 10, which consist of pairs of 2 and 5. Despite initial confusion due to a calculator's scientific notation output, it is clarified that the full number contains eight trailing zeroes. The calculator's display can truncate significant digits, leading to misunderstandings about the total count of zeroes. Ultimately, the conclusion is that the expression indeed ends with eight zeroes.
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Homework Statement



How many zeroes are at the end of (45^8)(88^5), don't use a calculator.

Homework Equations



Using the unique factorization of integers theorem, you can break any integer down into the product of prime integers.

The Attempt at a Solution



So I broke it down

(45^8) = (3 * 3 * 5) ^ 8
(88^5) = (2 * 2 * 2 * 11) ^ 5

If you put it back together as separate factors you get something like this

(3^16) * (5^8) * (2^15) * (11^5)

now my thinking is that you can find the number of zeroes by figuring out how many factors of 10 (which equals 2 * 5) you can make.

You can make 8 factors of 10 so it looks like
(3 ^ 16) * (2 ^ 7) * (11 ^ 5) * (10 ^ 8)

And from this I assume that there would be 8 zeroes at the end, however if you check it with a calculator you get a different answer..

(45^8) * (88^5) = 8.87387835 × 10^22

Anyone care to explain where my thinking is wrong..?
 
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You are right, but checking it with a calculator doesn't work because the screen of the calculator can't display all the digits of the product, if you use the calculator on your computer it should display the full number.
 
Animuo said:

Homework Statement



How many zeroes are at the %nd of (45^8)(88^5), don't use a calculator.

Homework Equations



Using the unique factorization of integers theorem, you can break any integer down into the product of prime integers.

The Attempt at a Solution



So I broke it down

(45^8) = (3 * 3 * 5) ^ 8
(88^5) = (2 * 2 * 2 * 11) ^ 5

If you put it back together as separate factors you get something like this

(3^16) * (5^8) * (2^15) * (11^5)

now my thinking is that you can find the number of zeroes by figuring out how many factors of 10 (which equals 2 * 5) you can make.

You can make 8 factors of 10 so it looks like
(3 ^ 16) * (2 ^ 7) * (11 ^ 5) * (10 ^ 8)

And from this I assume that there would be 8 zeroes at the end, however if you check it with a calculator you get a different answer..

(45^8) * (88^5) = 8.87387835 × 10^22

Anyone care to explain where my thinking is wrong..?

Your answer is right. The calculator is giving you a truncated answer in scientific notation. Even though the last shown figure is "5", there are still many nonzero figures to the right of that 5 when you write the whole number out. There will still be eight trailing zeroes.
 
Wow... didn't think of that, I used the google calculator and now I feel like a dumbass -.-. Thanks guys, feel better now. Here's another one I'm having a little difficulty with, and I don't feel like spamming these forums.

Moderator note: I made a separate thread for the new problem.[/color]
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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