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

Revision as of 16:14, 27 March 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 $a=b$ , 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 $5\cdot 5 = 25 \implies \boxed{(E)}$

@@ Line 15: / Line 15: @@
 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
+After some inspection, it can be seen that <math>a=b</math>, and <math>b < 5</math>, so
-<math>N \equiv a \pmod{6}</math>
+<math>N \equiv a \pmod{6}</math>,
-<math>N \equiv  a \pmod{5}</math>
+<math>N \equiv  a \pmod{5}</math>,
-<math>\implies N=a \pmod{30}</math>
+<math>\implies N=a \pmod{30}</math>,
 <math>0 \le a \le 4 </math>
-//===============================================================================
+Therefore, <math>N</math> can be written as  <math>30x+y</math>
+and <math>2N</math> can be written as <math>60x+2y</math>
-Therefore, N can be written as  30x+y
+Keep in mind that <math>y</math> can be <math>0, 1, 2, 3, 4</math>, five choices;
-and 2N can be written as 60x+2y
+Also, we have already found which digits of <math>y</math> will add up into the units digits of <math>2N</math>.
-Keep in mind that y can be 0, 1, 2, 3, 4, five choices;
+Now, examine the tens digit, <math>x</math> by using <math>\mod{25}</math> and <math>\mod{36}</math> to find the tens digit (units digits can be disregarded because <math>y=0,1,2,3,4</math> will always work)
-Also, we have already found which digits of y will add up into the units digits of 2N.
+Then we see that <math>N=30x+y</math> and take it <math>\mod{25}</math> and <math>\mod{36}</math> to find the last two digits in the base <math>5</math> and <math>6</math> representation.
-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>
@@ Line 38: / Line 36: @@
 (<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, since <math>y=0,1,2,3,4</math> will always work if <math>x</math> works, then we can treat <math>x</math> as a units digit instead of a tens digit in the respective bases and decrease the mods so that <math>x</math> is now the units digit.
-now the units digit :)
 <cmath>N \equiv 6x \equiv x \pmod{5}</cmath>
 <cmath>N \equiv 5x \pmod{6}</cmath>
@@ Line 60: / Line 56: @@
 From inspection, when
-n=0, m=6
+<math>n=0, m=6</math>
-n=1, m=6
+<math>n=1, m=6</math>
-n=2, m=2
+<math>n=2, m=2</math>
-n=3, m=2
+<math>n=3, m=2</math>
-n=4, m=4
+<math>n=4, m=4</math>
-This gives you 5 choices for x, and 5 choices for y, so the answer is
+This gives you <math>5</math> choices for <math>x</math>, and <math>5</math> choices for <math>y</math>, so the answer is
 <math>5\cdot 5 = 25 \implies \boxed{(E)}</math>
+== See also ==
+{{AMC10 box|year=2013|ab=B|num-b=24|after=Last question}}

2013 AMC 10B (Problems • Answer Key • Resources)
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Difference between revisions of "2013 AMC 10B Problems/Problem 25"

Revision as of 16:14, 27 March 2013

Problem

Solution

See also