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

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== See also ==
== See also ==
{{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}}
{{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}}
{{AMC10 box|year=2013|ab=B|num-b=24|num-a=Last Question}}
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Revision as of 14:47, 23 December 2014

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


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 $2N$?

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


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 $b < 5$, 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 $0, 1, 2, 3, 4$, five choices; 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 $5* 5 = \boxed{\textbf{(E) }25}$

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 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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