2020 AMC 10B Problems/Problem 12

Revision as of 17:59, 7 February 2020 by Pcchess (talk | contribs) (Created page with "==Problem== The decimal representation of<cmath>\dfrac{1}{20^{20}}</cmath>consists of a string of zeros after the decimal point, followed by a <math>9</math> and then several...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

The decimal representation of\[\dfrac{1}{20^{20}}\]consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?

$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$

Solution

\[\dfrac{1}{20^{20}}=\dfrac{1}{2^{40}\cdot5^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}\]

Now we do some estimation. Notice that $2^{20}=1024^2$, which means that $2^{20}$ is a little more than $1000^2=1,000,000$. Multiplying it with $10^{20}$, we get that the denominator is about $1\underbrace{00\dots0}_{26 \text{ zeros}}$. Notice that when we divide $1$ by an $n$ digit number, there are $n-1$ zeros before the first nonzero digit. This means that when we divide $1$ by the $27$ digit integer $1\underbrace{00\dots0}_{26 \text{ zeros}}$, there are $\boxed{\textbf{(D) } \text{26}}$ zeros in the initial string after the decimal point. -PCChess

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS