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Difference between revisions of "2016 USAJMO Problems/Problem 2"

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We divide the positive numbers into <math>999999</math> groups.
 
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==See also==
 
==See also==
 
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{{USAJMO newbox|year=2016|num-b=1|num-a=3}}

Revision as of 13:40, 29 June 2020

Problem

Prove that there exists a positive integer $n < 10^6$ such that $5^n$ has six consecutive zeros in its decimal representation.

Solution

Let digit $1$ of a number be the units digit, digit $2$ be the tens digit, and so on. Let the 6 consecutive zeroes be at digits $k-5$ through digit $k$. The criterion is then obviously equivalent to

\[5^n \mod 10^k < 10^{k-6}\]

We will prove that $n = 20+2^{19}, k = 20$ satisfies this, thus proving the problem statement (since $n = 20+2^{19} < 10^6$).

We want

\[5^{20+2^{19}} \mod 10^{20}\]

\[5^{20} \left(5^{2^{19}} \mod 2^{20}\right)\]

\[5^{20} \left(5^{\varphi \left(2^{20}\right)} \mod 2^{20}\right)\]

($\varphi$ is the Euler Totient Function.) By Euler's Theorem, since gcd$\left(5, 2^{20}\right)$ = 1,

\[5^{\varphi \left(2^{20}\right)} \mod 2^{20} = 1\]

so

\[5^{20} \left(5^{\varphi \left(2^{20}\right)} \mod 2^{20}\right) = 5^{20}\]

Since $2^{20} > 10^6, 5^{20} < 10^{14} = 10^{20-6}$, so

\[5^n \mod 10^k < 10^{k-6}\]

for $n = 20+2^{19}$ and $k = 20$, and thus the problem statement has been proven. $\blacksquare$

Motivation for Solution

Modifying our necessary and sufficient inequality, we get:

\[5^n \mod 10^k < 10^{k-6}\]

\[5^k \left(5^{n-k} \mod 2^k\right) < 10^{k-6}\]

\[5^{n-k} \mod 2^k < \frac{2^{k-6}}{5^6}\]

Since gcd$\left(5^{n-k}, 2^k\right) = 1$ if $k \neq 0$ (which is obviously true) and $n \neq k$ which is also true given that $5^k < 10^k$, we need the RHS to be greater than $1$:

\[\frac{2^{k-6}}{5^6} > 1\]

\[2^{k-6} > 5^6\]

\[2^k > 10^6\]

The first $k$ that satisfies this inequality is $k = 20$, so we let $k = 20$:

\[5^{n-20} \mod 2^{20} < \frac{2^{14}}{5^6}\]

\[5^{n-20} \mod 2^{20} \leq 1\]

\[5^{n-20} \mod 2^{20} = 1\]

From this, Euler's Theorem comes to mind and we see that if $n-20 = \varphi\left(2^{20}\right)$, the equality is satisfied. Thus, we get that $n = 20 + 2^{19}$, which is less than $10^6$, and we should be done. However, this requires slightly more formalization, and can be proven directly more easily if $n = 20+2^{19}$ is known or suspected.

Solution 2

We divide the positive numbers into $999999$ groups.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2016 USAJMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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