#5**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 999,999,999,998^{2}, **count** how many (2 * 5)s it has.

AnExtremelyLongName Mar 31, 2020

#7**+1 **

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

nchang Mar 31, 2020

#8**+1 **

Oh shoot, you are right. Hmmm. **Perhaps** you can simplify 999,999,999,998^{2} into multiple, smaller exponents. I can't think of anything else.

AnExtremelyLongName
Mar 31, 2020

#9

#10**+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**+1 **

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

**How will we know how many zeros that has?**

AnExtremelyLongName
Mar 31, 2020

#12**+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?

AltShaka Mar 31, 2020

#17**+1 **

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

AnExtremelyLongName
Mar 31, 2020

#20**+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 \(\boxed{11}\)?**

AnExtremelyLongName Mar 31, 2020

#22**+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 !!!!!

CPhill Mar 31, 2020