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

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also that <math>N \equiv b \pmod{5}</math>
 
also that <math>N \equiv b \pmod{5}</math>
  
−
Substituting these equations into the question and setting the units digits of <math>2N</math> and <math>S</math> equal to each other, it can be seen that <math>a=b</math>, and <math>b < 5</math>, (otherwise <math>a</math> and <math>b</math> always have different parities) so
+
Substituting these equations into the question and setting the units digits of <math>2N</math> and <math>S</math> equal to each other, it can be seen that <math>b < 5</math> (because otherwise <math>a</math> and <math>b</math> will have different parities), and thus <math>a=b</math>.
 
<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>,
Line 36: Line 36:
  
 
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, 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.
 +
<cmath>N \equiv 5x \pmod{6}</cmath>
 
<cmath>N \equiv 6x \equiv x \pmod{5}</cmath>  
 
<cmath>N \equiv 6x \equiv x \pmod{5}</cmath>  
−
<cmath>N \equiv 5x \pmod{6}</cmath>
 
 
<cmath>2N\equiv 6x \pmod{10}</cmath>
 
<cmath>2N\equiv 6x \pmod{10}</cmath>
  
Line 68: Line 68:
 
<math>5* 5 = \boxed{\textbf{(E) }25}</math>
 
<math>5* 5 = \boxed{\textbf{(E) }25}</math>
  
−
==Solution 2 (Shortcut)==
+
==Solution 2==
 
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>.
 
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>.
  
Line 85: Line 85:
 
Notice that <math>N_5</math> ranges from <math>3</math> to <math>5</math> digits and <math>N_6</math> ranges from <math>3</math> to <math>4</math> digits.
 
Notice that <math>N_5</math> ranges from <math>3</math> to <math>5</math> digits and <math>N_6</math> ranges from <math>3</math> to <math>4</math> digits.
  
−
Then let <math>a_l&, </math>b_l<math> denotes the digits of </math>N_5<math>, </math>N_6$, respectively such that <cmath>0\le a_i<5, 0\le b_i<6</cmath>
+
Then let <math>a_i</math>, <math>b_i</math> denotes the digits of <math>N_5</math>, <math>N_6</math>, respectively such that <cmath>0\le a_i<5,0\le b_i<6</cmath> Thus we have <cmath>N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4</cmath>
 +
<cmath>625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4</cmath>
 +
Now we are given <cmath>2N \equiv S \equiv N_5+N_6\pmod{100}</cmath>
 +
<cmath>2(625a_1+125a_2+25a_3+5a_4+a_5) \equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\pmod{100}</cmath>
 +
<cmath>1250a_1+250a_2+50a_3+10a_4+2a_5 \equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\pmod{100}</cmath>
 +
<cmath>50a_1+50a_2+50a_3+10a_4+2a_5 \equiv 10a_4+10b_3+a_5+b_4\pmod{100}</cmath>
 +
Canceling out <math>a_5</math> left with
 +
<cmath>50a_1+50a_2+50a_3+10a_4+a_5 \equiv 10a_4+10b_3+b_4\pmod{100}</cmath>
 +
 
 +
Since <math>a_5</math>, <math>b_4</math> determine the unit digits of the two sides of the congruence equation, we have <math>a_5=b_4=0,1,2,3,4</math>. Thus,
 +
 
 +
<cmath>50a_1+50a_2+50a_3+10a_4 \equiv 10a_4+10b_3\pmod{100}</cmath> canceling out <math>10a_4</math>, we have
 +
<cmath>50a_1+50a_2+50a_3 \equiv 10b_3\pmod{100}</cmath>
 +
<cmath>5a_1+5a_2+5a_3 \equiv b_3\pmod{10}</cmath>
 +
<cmath>5(a_1+a_2+a_3) \equiv b_3\pmod{10}</cmath>
 +
 
 +
Thus <math>b_3</math> is a multiple of <math>5</math>.
 +
 
 +
Now going back to our original equation
 +
<cmath>625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4</cmath>
 +
Since <math>a_5=b_4</math>,
 +
<cmath>625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3</cmath>
 +
<cmath>5(125a_1+25a_2+5a_3+a_4)=6(36b_1+6b_2+b_3)</cmath>
 +
<cmath>5(125a_1+25a_2+5a_3+a_4)=6[6(6b_1+b_2)+b_3]</cmath>
 +
 
 +
Since the left side is a multiple of <math>5</math>, then so does the right side. Thus <math>5\mid6(6b_1+b_2)+b_3</math>.
 +
 
 +
Since we already know that <math>5\mid b_3</math>, then <math>5\mid6(6b_1+b_2)</math>, from where we also know that <math>5\mid6b_1+b_2</math>.
 +
 
 +
For <math>b_1,b_2<6</math>, there is a total of 7 ordered pairs that satisfy the condition. Namely,
 +
 
 +
<cmath>(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)</cmath>
 +
 
 +
Since <math>N_6</math> has at least <math>3</math> digits, <math>(b_1,b_2)=(0,0)</math> doesn't work. Furthermore, when <math>b_1=5</math>, <math>216b_1</math> exceeds <math>1000</math> which is not possible as <math>N</math> is a three digit number, thus <math>(b_1,b_2)=(5,0)</math> won't work as well.
 +
 
 +
Since we know that <math>a_i<5</math>, for each of the ordered pairs <math>(b_1,b_2)</math>, there is respectively one and only one solution <math>(a_1,a_2,a_3,a_4)</math> that satisfies the equation
 +
 
 +
<cmath>625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3</cmath>
 +
 
 +
Thus there are five solutions to the equation. Also since we have 5 possibilities for <math>a_5=b_4</math>, we have a total of <math>5\cdot5=25</math> values for <math>N</math>. <math>\boxed{\textbf{(E) }25}</math>
 +
 
 +
~ Nafer
 +
 
 +
==Solution 4==
 +
 
 +
Observe that the maximum possible value of the sum of the last two digits of the base <math>5</math> number and the base <math>6</math> number is <math>44+55=99</math>.
 +
Let <math>N \equiv a \pmod {25}</math> and <math>N \equiv b \pmod {36}</math>.
 +
 
 +
If <math>a < \frac{25}{2}</math>, <math>2N \equiv 2a \pmod {25}</math> and if <math>a > \frac{25}{2}</math>, <math>2N \equiv 2a - 25 \pmod {25}</math>.
 +
 
 +
Using the same logic for <math>b</math>, if <math>b < 18</math>, <math>2N \equiv 2b \pmod {36}</math>, and in the other case <math>2N \equiv 2b - 36 \pmod {36}</math>.
 +
 
 +
We can do four cases:
 +
 
 +
Case 1: <math>a + b = 2a - 25 + 2b - 36 \implies a + b = 61</math>.
 +
 
 +
For this case, there is trivially only one possible solution, <math>(a, b) = (25, 36)</math>, which is equivalent to <math>(a, b) = (0, 0)</math>.
 +
 
 +
Case 2: <math>a + b = 2a - 25 + 2b \implies a + b = 25</math>.
 +
 
 +
Note that in this case, <math>a \geq 13</math> must hold, and <math>b < 18</math> must hold.
 +
We find the possible ordered pairs to be: <math>(13, 12), (14, 11), (15, 10), ..., (24, 1)</math> for a total of <math>12</math> ordered pairs.
 +
 
 +
Case 3: <math>a + b = 2a + 2b - 36 \implies a + b = 36</math>.
 +
 
 +
Note that in this case, <math>b \geq 18</math> must hold, and <math>a < 13</math> must hold.
 +
We find the possible ordered pairs to be: <math>(24, 12), (25, 11), (26, 10), ..., (35, 1)</math> for a total of <math>12</math> ordered pairs.
 +
 
 +
Case 4: <math>a + b = 2a + 2b</math>.
 +
 
 +
Trivially no solutions except <math>(a, b) = (0, 0)</math>, which matches the solution in Case 1, which makes this an overcount.
 +
 
 +
By CRT, each solution <math>(a, b)</math> corresponds exactly one positive integer in a set of exactly <math>\text{lcm} (25, 36) = 900</math> consecutive positive integers, and since there are <math>900</math> positive integers between <math>100</math> and <math>999</math>, our induction is complete, and our answer is <math>1 + 12 + 12 = \boxed{\textbf{(E) }25}</math>.
 +
 
 +
~ fidgetboss_4000
 +
 
  
 +
==Video Solution==
  
 +
https://www.youtube.com/watch?v=tgCK-H5jsOE&t=497s
  
 +
==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}}

Latest revision as of 14:44, 31 August 2022

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 $b < 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$. $N \equiv a \pmod{6}$, $N \equiv a \pmod{5}$, $\implies N=a \pmod{30}$, $0 \le a \le 4$

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. \[N \equiv 5x \pmod{6}\] \[N \equiv 6x \equiv x \pmod{5}\] \[2N\equiv 6x \pmod{10}\]

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

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 \[N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4\] \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Now we are given \[2N \equiv S \equiv N_5+N_6\pmod{100}\] \[2(625a_1+125a_2+25a_3+5a_4+a_5) \equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\pmod{100}\] \[1250a_1+250a_2+50a_3+10a_4+2a_5 \equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\pmod{100}\] \[50a_1+50a_2+50a_3+10a_4+2a_5 \equiv 10a_4+10b_3+a_5+b_4\pmod{100}\] Canceling out $a_5$ left with \[50a_1+50a_2+50a_3+10a_4+a_5 \equiv 10a_4+10b_3+b_4\pmod{100}\]

Since $a_5$, $b_4$ determine the unit digits of the two sides of the congruence equation, we have $a_5=b_4=0,1,2,3,4$. Thus,

\[50a_1+50a_2+50a_3+10a_4 \equiv 10a_4+10b_3\pmod{100}\] canceling out $10a_4$, we have \[50a_1+50a_2+50a_3 \equiv 10b_3\pmod{100}\] \[5a_1+5a_2+5a_3 \equiv b_3\pmod{10}\] \[5(a_1+a_2+a_3) \equiv b_3\pmod{10}\]

Thus $b_3$ is a multiple of $5$.

Now going back to our original equation \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Since $a_5=b_4$, \[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\] \[5(125a_1+25a_2+5a_3+a_4)=6(36b_1+6b_2+b_3)\] \[5(125a_1+25a_2+5a_3+a_4)=6[6(6b_1+b_2)+b_3]\]

Since the left side is a multiple of $5$, then so does the right side. Thus $5\mid6(6b_1+b_2)+b_3$.

Since we already know that $5\mid b_3$, then $5\mid6(6b_1+b_2)$, from where we also know that $5\mid6b_1+b_2$.

For $b_1,b_2<6$, there is a total of 7 ordered pairs that satisfy the condition. Namely,

\[(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)\]

Since $N_6$ has at least $3$ digits, $(b_1,b_2)=(0,0)$ doesn't work. Furthermore, when $b_1=5$, $216b_1$ exceeds $1000$ which is not possible as $N$ is a three digit number, thus $(b_1,b_2)=(5,0)$ won't work as well.

Since we know that $a_i<5$, for each of the ordered pairs $(b_1,b_2)$, there is respectively one and only one solution $(a_1,a_2,a_3,a_4)$ that satisfies the equation

\[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\]

Thus there are five solutions to the equation. Also since we have 5 possibilities for $a_5=b_4$, we have a total of $5\cdot5=25$ values for $N$. $\boxed{\textbf{(E) }25}$

~ Nafer

Solution 4

Observe that the maximum possible value of the sum of the last two digits of the base $5$ number and the base $6$ number is $44+55=99$. Let $N \equiv a \pmod {25}$ and $N \equiv b \pmod {36}$.

If $a < \frac{25}{2}$, $2N \equiv 2a \pmod {25}$ and if $a > \frac{25}{2}$, $2N \equiv 2a - 25 \pmod {25}$.

Using the same logic for $b$, if $b < 18$, $2N \equiv 2b \pmod {36}$, and in the other case $2N \equiv 2b - 36 \pmod {36}$.

We can do four cases:

Case 1: $a + b = 2a - 25 + 2b - 36 \implies a + b = 61$.

For this case, there is trivially only one possible solution, $(a, b) = (25, 36)$, which is equivalent to $(a, b) = (0, 0)$.

Case 2: $a + b = 2a - 25 + 2b \implies a + b = 25$.

Note that in this case, $a \geq 13$ must hold, and $b < 18$ must hold. We find the possible ordered pairs to be: $(13, 12), (14, 11), (15, 10), ..., (24, 1)$ for a total of $12$ ordered pairs.

Case 3: $a + b = 2a + 2b - 36 \implies a + b = 36$.

Note that in this case, $b \geq 18$ must hold, and $a < 13$ must hold. We find the possible ordered pairs to be: $(24, 12), (25, 11), (26, 10), ..., (35, 1)$ for a total of $12$ ordered pairs.

Case 4: $a + b = 2a + 2b$.

Trivially no solutions except $(a, b) = (0, 0)$, which matches the solution in Case 1, which makes this an overcount.

By CRT, each solution $(a, b)$ corresponds exactly one positive integer in a set of exactly $\text{lcm} (25, 36) = 900$ consecutive positive integers, and since there are $900$ positive integers between $100$ and $999$, our induction is complete, and our answer is $1 + 12 + 12 = \boxed{\textbf{(E) }25}$.

~ fidgetboss_4000


Video Solution

https://www.youtube.com/watch?v=tgCK-H5jsOE&t=497s

See Also

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

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