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

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==Solution==
 
==Solution==
 +
First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.
 +
 +
Say that <math>N \equiv a \pmod{6}</math>
 +
 +
also that <math>N \equiv b \pmod{5}</math>
 +
 +
After some inspection, it can be seen that b=d, and <math>b < 5</math>, so
 +
<math>N \equiv a \pmod{6}</math>
 +
<math>N \equiv  a \pmod{5}</math>
 +
<math>\implies N=a \pmod{30}</math>
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<math>0 \le a \le 4 </math>
 +
 +
//===============================================================================
 +
 +
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 36 and 25 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 36 to find the last two digits in the base 5 and 6 representation.
 +
<cmath>N \equiv 30x \pmod{36}</cmath>
 +
<cmath>N \equiv 30x \equiv 5x \pmod{25}</cmath>
 +
Both of those must add up to
 +
<cmath>2N\equiv60x \pmod{100}</cmath>
 +
 +
(<math>33 \ge x \ge 4</math>)
 +
 +
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 :)
 +
<cmath>N \equiv 6x \equiv x \pmod{5}</cmath>
 +
<cmath>N \equiv 5x \pmod{6}</cmath>
 +
<cmath>2N\equiv 6x \pmod{10}</cmath>
 +
 +
Say that <math>x=5m+n</math> (m is between 0-6, n is 0-4 because of constraints on x)
 +
Then
 +
 +
<cmath>N \equiv 5m+n \pmod{5}</cmath>
 +
<cmath>N \equiv 25m+5n \pmod{6}</cmath>
 +
<cmath>2N\equiv30m + 5n \pmod{10}</cmath>
 +
 +
and this simplifies to
 +
 +
<cmath>N \equiv n \pmod{5}</cmath>
 +
<cmath>N \equiv m+5n \pmod{6}</cmath>
 +
<cmath>2N\equiv 5n \pmod{10}</cmath>
 +
 +
From inspection, when
 +
 +
n=0, m=6
 +
 +
n=1, m=5
 +
 +
n=2, m=4
 +
 +
n=3, m=5
 +
 +
n=4, m=4
 +
 +
This gives you 5 choices for x, and 5 choices for y, so the answer is
 +
<math>5\cdot 5 = 25 \implies \boxed{(E)}</math>

Revision as of 02:13, 24 February 2013

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 $N \equiv b \pmod{5}$

After some inspection, it can be seen that b=d, 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 36 and 25 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 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

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

and this simplifies to

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

From inspection, when

n=0, m=6

n=1, m=5

n=2, m=4

n=3, m=5

n=4, m=4

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

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