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

Revision as of 08:42, 5 February 2017

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, in $100\le n\le 999$ , every possible combination of $2$ digits will happen exactly once. We know that $n=900,901,902,903,904$ works because $900\equiv\dots00_5\equiv\dots00_6$ .

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 $1$ for the $10$ 's digit.

$\textbf{5 }\textcolor{red}{\text{ 0}}\text{ 4 3 2 1 0 }\textcolor{red}{\text{4}}\text{ 3 2 1 0 4 3 2 1 0 4 }\textcolor{red}{\text{3 2}}\text{ 1 0 4 3 2 1 0 4 3 2 }\textcolor{red}{\text{1}}$

$\textbf{6 }\textcolor{red}{\text{ 0}}\text{ 1 2 3 4 5 }\textcolor{red}{\text{0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5 0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5}}$

$\textbf{10}\textcolor{red}{\text{ 0}}\text{ 4 8 2 6 0 }\textcolor{red}{\text{4}}\text{ 8 2 6 0 4 8 2 6 0 4 }\textcolor{red}{\text{8 2}}\text{ 6 0 4 8 2 6 0 4 8 2 }\textcolor{red}{\text{6}}$

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 $5\cdot 5=\boxed{\textbf{(E) }25}$ .

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

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

@@ Line 69: / Line 69: @@
 ==Shortcut==
-Notice that there are exactly <math>1000-100=900=5^2\cdot 6^2</math> possible values of <math>n</math>. This means, from <math>100\le n\le 999</math>, every possible combination of <math>2</math> digits will happen exactly once. We know that <math>n=900,901,902,903,904</math> works because <math>900\equiv\dots00_5\equiv\dots00_6</math>.
+Notice that there are exactly <math>1000-100=900=5^2\cdot 6^2</math> possible values of <math>n</math>. This means, in <math>100\le n\le 999</math>, every possible combination of <math>2</math> digits will happen exactly once. We know that <math>n=900,901,902,903,904</math> works because <math>900\equiv\dots00_5\equiv\dots00_6</math>.
 We know for sure that the units digit will add perfectly every <math>30</math> added or subtracted, because <math>\text{lcm }5,6=30</math>. So we only have to care about cases of <math>n</math> every <math>30</math> subtracted. In each case, <math>2n</math> subtracts <math>6</math>/adds <math>4</math>, <math>n_5</math> subtracts <math>1</math> and <math>n_6</math> adds <math>1</math> for the <math>10</math>'s digit.

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

Revision as of 08:42, 5 February 2017

Problem

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