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Difference between revisions of "2013 AMC 10B Problems/Problem 25"

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  <math>\textbf{5}</math> & 0 & 4 & 3 & 2 & 1 & 0 & <math>\textcolor{red}{4}</math> & 3 & 2 & 1 & 0 & 4 & 3 & 2 & 1 & 0 & 4 & <math>\textcolor{red}{3}</math> & <math>\textcolor{red}{2}</math> & 1 & 0 & 4 & 3 & 2 & 1 & 0 & 4 & 3 & 2 & <math>\textcolor{red}{1}</math> \\   
 
  <math>\textbf{5}</math> & 0 & 4 & 3 & 2 & 1 & 0 & <math>\textcolor{red}{4}</math> & 3 & 2 & 1 & 0 & 4 & 3 & 2 & 1 & 0 & 4 & <math>\textcolor{red}{3}</math> & <math>\textcolor{red}{2}</math> & 1 & 0 & 4 & 3 & 2 & 1 & 0 & 4 & 3 & 2 & <math>\textcolor{red}{1}</math> \\   
 
  <math>\textbf{6}</math> & 0 & 1 & 2 & 3 & 4 & 5 & <math>\textcolor{red}{0}</math> & 1 & 2 & 3 & 4 & 5 & 0 & 1 & 2 & 3 & 4 & <math>\textcolor{red}{5}</math> & <math>\textcolor{red}{0}</math> & 1 & 2 & 3 & 4 & 5 & 0 & 1 & 2 & 3 & 4 & <math>\textcolor{red}{5}</math> \\
 
  <math>\textbf{6}</math> & 0 & 1 & 2 & 3 & 4 & 5 & <math>\textcolor{red}{0}</math> & 1 & 2 & 3 & 4 & 5 & 0 & 1 & 2 & 3 & 4 & <math>\textcolor{red}{5}</math> & <math>\textcolor{red}{0}</math> & 1 & 2 & 3 & 4 & 5 & 0 & 1 & 2 & 3 & 4 & <math>\textcolor{red}{5}</math> \\
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  <math>\textbf{10}</math> & 0 & 4 & 8 & 2 & 6 & 0 & <math>\textcolor{red}{4}</math> & 8 & 2 & 6 & 0 & 4 & 8 & 2 & 6 & 0 & 4 & <math>\textcolor{red}{8}</math> & <math>\textcolor{red}{2}</math> & 6 & 0 & 4 & 8 & 2 & 6 & 0 & 4 & 8 & 2 & <math>\textcolor{red}{6} \\
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  <math>\textbf{10}</math> & 0 & 4 & 8 & 2 & 6 & 0 & <math>\textcolor{red}{4}</math> & 8 & 2 & 6 & 0 & 4 & 8 & 2 & 6 & 0 & 4 & <math>\textcolor{red}{8}</math> & <math>\textcolor{red}{2}</math> & 6 & 0 & 4 & 8 & 2 & 6 & 0 & 4 & 8 & 2 & <math>\textcolor{red}{6}</math> \\
 
\end{tabular}
 
\end{tabular}
 
\end{center}
 
\end{center}
  
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As we can see, there are </math>5<math> cases, including the original, that work. These are highlighted in </math>\textcolor{red}{\text{red}}<math>. So, thus, there are </math>5<math> possibilities for each case, and </math>5\cdot 5=\boxed{\textbf{(E) }25}$.
+
As we can see, there are <math>5</math> cases, including the original, that work. These are highlighted in <math>\textcolor{red}{\text{red}}</math>. So, thus, there are <math>5</math> possibilities for each case, and <math>5\cdot 5=\boxed{\textbf{(E) }25}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:37, 15 February 2016

The following problem is from both the 2013 AMC 12B #23 and 2013 AMC 10B #25, so both problems redirect to this page.

Problem

Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$. For example, if $N = 749$, Bernardo writes the numbers $10,\!444$ and $3,\!245$, and LeRoy obtains the sum $S = 13,\!689$. For how many choices of $N$ are the two rightmost digits of $S$, in order, the same as those of $2N$?

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$

Solution

First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.

Say that $N \equiv a \pmod{6}$

also that $N \equiv b \pmod{5}$

Substituting these equations into the question and setting the units digits of 2N and S equal to each other, it can be seen that $a=b$, and $b < 5$, (otherwise $a$ and $b$ always have different parities) so $N \equiv a \pmod{6}$, $N \equiv a \pmod{5}$, $\implies N=a \pmod{30}$, $0 \le a \le 4$

Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$

Keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$, ; Also, we have already found which digits of $y$ will add up into the units digits of $2N$.

Now, examine the tens digit, $x$ by using $\mod{25}$ and $\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work) Then we see that $N=30x+y$ and take it $\mod{25}$ and $\mod{36}$ to find the last two digits in the base $5$ and $6$ representation. \[N \equiv 30x \pmod{36}\] \[N \equiv 30x \equiv 5x \pmod{25}\] Both of those must add up to \[2N\equiv60x \pmod{100}\]

($33 \ge x \ge 4$)

Now, since $y=0,1,2,3,4$ will always work if $x$ works, then we can treat $x$ as a units digit instead of a tens digit in the respective bases and decrease the mods so that $x$ is now the units digit. \[N \equiv 6x \equiv x \pmod{5}\] \[N \equiv 5x \pmod{6}\] \[2N\equiv 6x \pmod{10}\]

Say that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x) Then

\[N \equiv 5m+n \pmod{5}\] \[N \equiv 25m+5n \pmod{6}\] \[2N\equiv30m + 6n \pmod{10}\]

and this simplifies to

\[N \equiv n \pmod{5}\] \[N \equiv m+5n \pmod{6}\] \[2N\equiv 6n \pmod{10}\]

From inspection, when

$n=0, m=6$

$n=1, m=6$

$n=2, m=2$

$n=3, m=2$

$n=4, m=4$

This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \boxed{\textbf{(E) }25}$

Shortcut

Notice that there are exactly $1000-100=900=5^2\cdot 6^2$ possible values of $n$. This means, from $100\le n\le 999$, every possible combination of $2$ digits will happen exactly once. We know that $n=900,901,902,903,904$ works because $900\cong\dots00_5\cong\dots00_6$.

We know for sure that the units digit will add perfectly every $30$ added or subtracted, because $\text{lcm }5,6=30$. So we only have to care about cases of $n$ every $30$ subtracted. In each case, $2n$ subtracts $6$/adds $4$, $n_5$ subtracts $1$ and $n_6$ adds $1$.

\begin{center} \begin{tabular}{c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c } 

$\textbf{5}$ & 0 & 4 & 3 & 2 & 1 & 0 & $\textcolor{red}{4}$ & 3 & 2 & 1 & 0 & 4 & 3 & 2 & 1 & 0 & 4 & $\textcolor{red}{3}$ & $\textcolor{red}{2}$ & 1 & 0 & 4 & 3 & 2 & 1 & 0 & 4 & 3 & 2 & $\textcolor{red}{1}$ \\  
$\textbf{6}$ & 0 & 1 & 2 & 3 & 4 & 5 & $\textcolor{red}{0}$ & 1 & 2 & 3 & 4 & 5 & 0 & 1 & 2 & 3 & 4 & $\textcolor{red}{5}$ & $\textcolor{red}{0}$ & 1 & 2 & 3 & 4 & 5 & 0 & 1 & 2 & 3 & 4 & $\textcolor{red}{5}$ \\
$\textbf{10}$ & 0 & 4 & 8 & 2 & 6 & 0 & $\textcolor{red}{4}$ & 8 & 2 & 6 & 0 & 4 & 8 & 2 & 6 & 0 & 4 & $\textcolor{red}{8}$ & $\textcolor{red}{2}$ & 6 & 0 & 4 & 8 & 2 & 6 & 0 & 4 & 8 & 2 & $\textcolor{red}{6}$ \\

\end{tabular} \end{center}

As we can see, there are $5$ cases, including the original, that work. These are highlighted in $\textcolor{red}{\text{red}}$. So, thus, there are $5$ possibilities for each case, and $5\cdot 5=\boxed{\textbf{(E) }25}$.

See also

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 12 Problems and Solutions
2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
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

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