+658 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 999,999,999,9982, count how many (2 * 5)s it has.
+343 Aren't those for terminating zeros? I don't think that is what the question is asking though.
+658 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.
+658 Good idea!
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So by the (a - b)2 = (a + b)(a - b), can we work this further? I am struggling as much as you right now.
+658 1,000,000,000,002 * 999,999,999,996
How will we know how many zeros that has?
+422 Wait a sec didnt you say its 999,999,999? If it is that then it should be (100,000,000-1)^2 right?
+658 But you said... wait now I'm super confused and I lost my train of thought
+658 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 \(\boxed{11}\)?
+129852 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
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999,999,999,996,000,000,000,004
11 zeros
Just as AELN found....good job, AELN !!!!!
