2020 AMC 10B Problems/Problem 12
Contents
Problem
The decimal representation of consists of a string of zeros after the decimal point, followed by a
and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
Solution 1
Now we do some estimation. Notice that , which means that
is a little more than
. Multiplying it with
, we get that the denominator is about
. Notice that when we divide
by an
digit number as long as n is not a power of 10, there are
zeros before the first nonzero digit. This means that when we divide
by the
digit integer
, there are
zeros in the initial string after the decimal point. -PCChess
Solution 2
First rewrite as
. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in
.
and memming
(alternatively use the fact that
),
digits.
Our answer is .
Solution 3 (Brute Force)
Just as in Solution we rewrite
as
We then calculate
entirely by hand, first doing
then multiplying that product by itself, resulting in
Because this is
digits, after dividing this number by
fourteen times, the decimal point is before the
Dividing the number again by
twenty-six more times allows a string of
zeroes to be formed. -OreoChocolate
Solution 4 (Alternate Brute Force)
Just as in Solutions and
we rewrite
as
We can then look at the number of digits in powers of
.
,
,
,
,
,
,
and so on. We notice after a few iterations that every power of five with an exponent of
, the number of digits doesn't increase. This means
should have
digits since there are
numbers which are
from
to
, or
digits total. This means our expression can be written as
, where
is in the range
. Canceling gives
, or
zeroes before the
since the number
should start on where the one would be in
. ~aop2014
Solution 5 (Logarithms Without Words)
~phoenixfire
Video Solution
~IceMatrix
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.