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

Revision as of 16:40, 7 August 2019

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 1

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 $60x+2y$

Just 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 take $N=30x+y$ $\mod{25}$ 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 careful inspection, this is true 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 $5* 5 = \boxed{\textbf{(E) }25}$

Solution 2 (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 $5\cdot 5=\boxed{\textbf{(E) }25}$ .

Solution 3

Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits.

Then let $a_i$ , $b_i$ denotes the digits of $N_5$ , $N_6$ , respectively such that $0\le a_i<5,0\le b_i<6$ Thus we have $N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4$ $625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4$

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.

Revision as of 16:40, 7 August 2019 (view source) Nafer (talk \| contribs) (→‎Solution 3) ← Older edit		Revision as of 16:40, 7 August 2019 (view source) Nafer (talk \| contribs) (→‎Solution 3) Newer edit →
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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}</math>.		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>.
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	== Solution 3 ==		== Solution 3 ==

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