tough math questions?

Question

How many consecutive zeroes end 1000!, exactly?

tougher follow up question?

What is the first interger to the left of the string of zeroes in 1000! ?

Answer 1

249 zeros end 1000!

There will be a zero for every factor of 10 multiplied in 1000!. Since 10 = 2*5 there will be as many factors of 10 as there are numbers of factors of 5.
There are 200 numbers from 1 to 1000 divisible by 5. So 200 5's in these.
There is an additional 5 for every factor of 25=5*5. There are 40 of these from 1 to 1000, so add another 40.
There is an additional 5 for every factor of 125 = 5*5*5, and there are 8 of these, so 8 more.
And one more factor of 5 for 625 = 5*5*5*5.
So 200 + 40 + 8 + 1 = 249 factors of 5, thus 249 factors of 10 and 249 trailing zeros.

2 is the first digit to the left of these zeros.

Math proof too lengthy here and now. http://www.justinwhite.com/big-calc/1000.html

Pretzels: Good starting technique, however, in ignoring factors like 32 and 35 (because one has a 2 and one has a 5) you neglected to factor in the residue factors i.e., 32/2 = 16 and 35/5 = 7. These will contribute to the last non-zero digit. Also, the 3 in 30, the 4 in 40, th 6 in 60 etc. as well as the 3 in 300, the 4 in 400, the 6 in 600, etc. all contribute to this digit.
This is very complicated and will take a long boring time to type it all out. Just use a computer/calculator to figure it. It's 2.

Answer 2

I answered the first part in your 2000! question (it's 249). As far as the last nonzero digit in 1000!, You need to track the product of all the last digits with the 5s and 2s that cause the zeros taken out. Every additional 10 in the factorial multiplies by:

1*2*3*4*5*6*7*8*9*10

Taking out the factors that make zeros leaves:

3*4*6*7*8*9 = 36288

So each additional 10 muliplies the last digit by 8. There are 100 of these. So far we have 8^100.

However, this factor of 100 only accounts for 200 zeros (2 zeros were produced by the 2*5*10 that were removed. There are another 49 5s lurking in the factors and these will gobble up 49 of the 2s in the final product, converting them to zeros. We need to divide by 2^49 to remove these extra 2s. So now, our last nonzero digit matches (8^100)/(2^49) = 2^251

Now find the last digit in 2^251 and that is your answer. Look at the pattern:

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 62

It repeats every 4 times so any multiple of 4 can be subtracted from the exponent without changing the last digit so 2^251 end the same as 2^3 = 8

Your answer is 8

You can verify this technique on smaller numbers like 30! to show it is correct.

Answer 3

There is a Web site where 1000! is actually calculated out. The first non-zero digit is 2. So, whatever Scott R is doing must be right. (Sorry, Pretzels, but even the best ones miss one once in awhile...)

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OK, folks, I think I have the "easy" solution:

As you count from 1 to 999, the last NON-ZERO digit has to occur the same number of times for each digit 1, 2, 3, up to 9. Since you have 9 possible non-zero digits and 999 numbers, that means the last non-zero digit is 1 (one) 111 times, 2 (two) 111 times, etc.

The last non-zero digit of 9! is 8. Now we just have to repeat this 111 times, i.e. find the last digit of 8^111. That's easy enough, since the last digit of 8^n goes through the cycle 8, 4, 2, 6. Since the remainder of 111/4 is 3, the last digit is 8^111 is the same as 8^3.

So there is your answer: 2

Answer 4

1)
Numbers ending in 0, and numbers ending in 5, each contribute to one of trailing 0s in the product.

2)
Numbers ending in two zeros and those ending in 50, contribute one additional 0.

3)
Numbers ending in three zeros and those ending in 500, contribute one more 0.

Case 3) there are 2 (500, 1000).
Case 2) there are 10 for 00 ending, and 10 for 50 ending - total 20
Case 1) there are 100 for 0 ending and 100 for 5 ending - total 200.

The number of trailing 0s in 1000! will be 2 + 20+200 = 222.

The first integer to the left of the trailing 0s:

This should be the units digit in 8^100 (since 8 is the last such digit in 10!):
8^1 = 8
8^2 = 16
8^3=128
and so on.

So 100 being even, an even power of 8 will have 6 as the last digit.

6 is the answer.

NOTE:
If you like to generalise for 10^n!, we will have:

1)

2(1 + 10+..10^(n-1)) trailing zeros.

2)

If n is even, the last digit before the trailing zeroes is 6, otherwise 8.

Answer 5

I made a program on my TI-84+ and there are 249 zeros, and the first non-zero digit is 2, so I agree with all above except the one who said it was 8.

Answer 6

3. And 1

Answer 7

If you really can't answer this, then you need to talk to your teacher or otherwise learn it as opposed to getting answers. There are three zeroes at the end of 1000 and infinite more of them after the decimal point, but they make no difference so you can really ignore those. Integers are just positive whole numbers (not -3 or 29.34 or 4.8, etc.) so the first one to the left of the zero is 1. Let me know if you need more help.

Answer 8

the answer is 3 and 1