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How many zeros are in the expansion of \(999,\!999,\!999,\!998^2\)?

 Mar 31, 2020
edited by Guest  Mar 31, 2020
edited by nchang  Mar 31, 2020
 #1
avatar+416 
+1

Whats the number?

 Mar 31, 2020
 #3
avatar+416 
0

Cal why are u answering homework??

 Mar 31, 2020
 #4
avatar+336 
+1

that has a 9 on the end, not an 8

 Mar 31, 2020
 #5
avatar+632 
0

Hello nchang! I like your name, it probably reveals your last name in real life! >:D

 

Know:

A number only has a 0 in it if when factored, has 2 * 5 in it.

 

Solve:

When you factor 999,999,999,9982count how many (2 * 5)s it has.

 Mar 31, 2020
 #6
avatar+1956 
+1

nchang, it will still help you.

 Mar 31, 2020
 #7
avatar+336 
+1

Aren't those for terminating zeros? I don't think that is what the question is asking though.

 Mar 31, 2020
 #8
avatar+632 
+1

Oh shoot, you are right. Hmmm. Perhaps you can simplify 999,999,999,9982 into multiple, smaller exponents. I can't think of anything else.

AnExtremelyLongName  Mar 31, 2020
 #9
avatar+336 
+1

maybe (1000000000000-2)^2?

 Mar 31, 2020
 #10
avatar+632 
+1

Good idea!

_________

So by the (a - b)2 = (a + b)(a - b), can we work this further? I am struggling as much as you right now.

AnExtremelyLongName  Mar 31, 2020
 #13
avatar+632 
+1

1,000,000,000,002 * 999,999,999,996

How will we know how many zeros that has?

AnExtremelyLongName  Mar 31, 2020
 #11
avatar+336 
0

maybe something like 1000000000000^2-1000000000000*2*2+4?

 Mar 31, 2020
 #12
avatar+416 
+1

Wait a sec didnt you say its 999,999,999? If it is that then it should be (100,000,000-1)^2 right?

 Mar 31, 2020
 #15
avatar+336 
+1

Please read the question

nchang  Mar 31, 2020
 #16
avatar+416 
0

But you said that it ended with a 9 a few minutes ago.....

AltShaka  Mar 31, 2020
 #17
avatar+632 
+1

But you said... wait now I'm super confused and I lost my train of thought

AnExtremelyLongName  Mar 31, 2020
 #18
avatar+336 
+1

I never said that!

nchang  Mar 31, 2020
 #19
avatar+416 
0

It lit says above.... Oh sorry the end is an 9 not an 8

AltShaka  Mar 31, 2020
 #14
avatar+336 
+2

I am just wondering how to do this without a calculator.

 Mar 31, 2020
 #20
avatar+632 
+2

Here is hunch:

An expression like x * y keeps the number of zeroes as long as x and y do not contain (2 * 5) or any factor of ten like (25 * 4) etc.

Ex: one zero (10 * 11 = 110) four zeroes (100 * 100 = 10,000)

 - In the current problem, the numbers obviously do not have any factor of ten. So:

1,000,000,000,002 * 999,999,999,996​ we count the zeroes, getting an answer of \(\boxed{11}\)?

 Mar 31, 2020
edited by AnExtremelyLongName  Mar 31, 2020
 #21
avatar+336 
+2

That is so clever! Thank you so much, I understand now.

 Mar 31, 2020
 #23
avatar+111321 
0

Agree  !!!!

 

I'm going to have to look at  AELN's  method.....very  exquisite    !!!!

 

cool cool cool

CPhill  Mar 31, 2020
 #22
avatar+111321 
+3

Note  that  we  can  write

 

(1,000,000,000 ,000  - 2)^2

 

By  binomial expansion we have

 

( 1, 000,000,000,000)^2  -  4  ( 1,000,000,000,000)  +  4  =

 

(10^12)^2  -  4 ( 10^12)  +  4  = 

 

(10)^24  - 4 (10^12)  + 4  =

 

   1, 000,000,000,000,000,000,000,004

-                              4,000,000,000,000

 _________________________________

      999,999,999,996,000,000,000,004

 

11  zeros

 

Just as AELN  found....good job,  AELN    !!!!!

 

 

cool cool cool

 Mar 31, 2020
 #24
avatar+632 
+1

I like your method better! It insures 100% accuracy!

AnExtremelyLongName  Mar 31, 2020
 #25
avatar+336 
+2

That is kinda like my idea! Thank you CPill! (btw, who is Allen)?

 Mar 31, 2020
 #26
avatar+632 
+1

My abbreviation lol

AnExtremelyLongName  Mar 31, 2020
 #27
avatar+111321 
0

AELN  =  shorthand  for AnExtremelyLong Name.....LOL!!!!

 

 

cool cool cool

CPhill  Mar 31, 2020
 #28
avatar+336 
+2

LOL! :)

 Mar 31, 2020

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