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Difference between revisions of "2023 AMC 10A Problems/Problem 12"
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==Problem==
 
How many three-digit positive integers <math>N</math> satisfy the following properties?
 
How many three-digit positive integers <math>N</math> satisfy the following properties?
  
 
*The number <math>N</math> is divisible by <math>7</math>.
 
*The number <math>N</math> is divisible by <math>7</math>.
  
−
*The number formed by reversing the digits of <math>N</math> is divisble by <math>5</math>.
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*The number formed by reversing the digits of <math>N</math> is divisible by <math>5</math>.
  
 +
<math>\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17</math>
  
−
Solution 1
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==Solution 1==
  
−
Multiples of 5 always end in 0 or 5 and since it is a three digit number, it cannot start with 0. All possibilities have to be in the range from 7 x 72 to 7 x 85 inclusive. 85 - 72 + 1 = 14 (B).
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Multiples of <math>5</math> will always end in <math>0</math> or <math>5</math>, and since the numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with <math>5</math>. Since the numbers must be divisible by 7, all possibilities have to be in the range from <math>7 \cdot 72</math> to <math>7 \cdot 85</math> inclusive.  
  
−
~walmartbrian ~Shontai
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<math>85 - 72 + 1 = 14</math>. <math>\boxed{\textbf{(B) } 14}</math>.
 +
 
 +
~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS
 +
 
 +
==Solution 2 (solution 1 but more thorough)==
 +
Let <math>N=\overline{cab}=100c+10a+b.</math> We know that <math>\overline{bac}</math> is divisible by <math>5</math>, so <math>c</math> is either <math>0</math> or <math>5</math>. However, since <math>c</math> is the first digit of the three-digit number <math>N</math>, it can not be <math>0</math>, so therefore, <math>c=5</math>. Thus, <math>N=\overline{5ab}=500+10a+b.</math> There are no further restrictions on digits <math>a</math> and <math>b</math> aside from <math>N</math> being divisible by <math>7</math>.
 +
 
 +
The smallest possible <math>N</math> is <math>504</math>. The next smallest <math>N</math> is <math>511</math>, then <math>518</math>, and so on, all the way up to <math>595</math>. Thus, our set of possible <math>N</math> is <math>\{504,511,518,\dots,595\}</math>. Dividing by <math>7</math> for each of the terms will not affect the cardinality of this set, so we do so and get <math>\{72,73,74,\dots,85\}</math>. We subtract <math>71</math> from each of the terms, again leaving the cardinality unchanged. We end up with <math>\{1,2,3,\cdots,14\}</math>, which has a cardinality of <math>14</math>. Therefore, our answer is <math>\boxed{\textbf{(B) } 14.}</math>
 +
 
 +
~ Technodoggo
 +
 
 +
==Solution 3 (modular arithmetic)==
 +
 
 +
We first proceed as in the above solution, up to <math>N=500+10a+b</math>.
 +
We then use modular arithmetic:
 +
 
 +
<cmath>\begin{align*}
 +
0&\equiv N \:(\text{mod }7)\\
 +
&\equiv500+10a+b\:(\text{mod }7)\\
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&\equiv3+3a+b\:(\text{mod }7)\\
 +
3a+b&\equiv-3\:(\text{mod }7)\\
 +
&\equiv4\:(\text{mod }7)\\
 +
\end{align*}</cmath>
 +
 
 +
We know that <math>0\le a,b<10</math>. We then look at each possible value of <math>a</math>:
 +
 
 +
If <math>a=0</math>, then <math>b</math> must be <math>4</math>.
 +
 
 +
If <math>a=1</math>, then <math>b</math> must be <math>1</math> or <math>8</math>.
 +
 
 +
If <math>a=2</math>, then <math>b</math> must be <math>5</math>.
 +
 
 +
If <math>a=3</math>, then <math>b</math> must be <math>2</math> or <math>9</math>.
 +
 
 +
If <math>a=4</math>, then <math>b</math> must be <math>6</math>.
 +
 
 +
If <math>a=5</math>, then <math>b</math> must be <math>3</math>.
 +
 
 +
If <math>a=6</math>, then <math>b</math> must be <math>0</math> or <math>7</math>.
 +
 
 +
If <math>a=7</math>, then <math>b</math> must be <math>4</math>.
 +
 
 +
If <math>a=8</math>, then <math>b</math> must be <math>1</math> or <math>8</math>.
 +
 
 +
If <math>a=9</math>, then <math>b</math> must be <math>5</math>.
 +
 
 +
Each of these cases are unique, so there are a total of <math>1+2+1+2+1+1+2+1+2+1=\boxed{\textbf{(B) } 14.}</math>
 +
 
 +
~ Technodoggo
 +
 
 +
==Solution 4==
 +
 
 +
The key point is that when reversed, the number must start with a <math>0</math> or a <math>5</math> based on the second restriction. But numbers can't start with a <math>0</math>.
 +
 
 +
So the problem is simply counting the number of multiples of <math>7</math> in the <math>500</math>s.
 +
 
 +
<math>7 \times 72 = 504</math>, so the first multiple is <math>7 \times 72</math>.
 +
 
 +
<math>7 \times 85 = 595</math>, so the last multiple is <math>7 \times 85</math>.
 +
 
 +
Now, we just have to count <math>7\times 72, 7\times 73, 7\times 74,\cdots, 7\times 85</math>.
 +
 
 +
We have a set that numbers <math>85-71=\boxed{\textbf{(B) 14}}</math>
 +
 
 +
~Dilip ~boppitybop ~ESAOPS (LaTeX)
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/UYHCNlRDZBo
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 
 +
https://www.youtube.com/watch?v=7qWMFzq7Tfs
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=4uKo5NR2o9Y
 +
 
 +
-paixiao
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/N2lyYRMuZuk?si=6B-mTB070UP2yuDF&t=435
 +
 
 +
~Math-X
 +
 
 +
==Video Solution by Power Solve (easy to digest!)==
 +
https://www.youtube.com/watch?v=Mg6JUanYNJY
 +
 
 +
==Note==
 +
According to the official answer key, choice (B) is correct. However, some have argued that it is ambiguous whether the number <math>560</math> should be included in the count, since its reversal, <math>065</math>, has a leading zero. It is assumed that <math>065</math> denotes the two-digit number <math>65</math>, which is divisible by <math>5</math>, but MAA should have clarified what happens when a number with trailing zeros is reversed.
 +
 
 +
~A_MatheMagician ~ESAOPS ~sdpandit
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2023|ab=A|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Revision as of 06:59, 28 January 2024

Problem

How many three-digit positive integers $N$ satisfy the following properties?

  • The number $N$ is divisible by $7$.
  • The number formed by reversing the digits of $N$ is divisible by $5$.

$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$

Solution 1

Multiples of $5$ will always end in $0$ or $5$, and since the numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with $5$. Since the numbers must be divisible by 7, all possibilities have to be in the range from $7 \cdot 72$ to $7 \cdot 85$ inclusive.

$85 - 72 + 1 = 14$. $\boxed{\textbf{(B) } 14}$.

~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS

Solution 2 (solution 1 but more thorough)

Let $N=\overline{cab}=100c+10a+b.$ We know that $\overline{bac}$ is divisible by $5$, so $c$ is either $0$ or $5$. However, since $c$ is the first digit of the three-digit number $N$, it can not be $0$, so therefore, $c=5$. Thus, $N=\overline{5ab}=500+10a+b.$ There are no further restrictions on digits $a$ and $b$ aside from $N$ being divisible by $7$.

The smallest possible $N$ is $504$. The next smallest $N$ is $511$, then $518$, and so on, all the way up to $595$. Thus, our set of possible $N$ is $\{504,511,518,\dots,595\}$. Dividing by $7$ for each of the terms will not affect the cardinality of this set, so we do so and get $\{72,73,74,\dots,85\}$. We subtract $71$ from each of the terms, again leaving the cardinality unchanged. We end up with $\{1,2,3,\cdots,14\}$, which has a cardinality of $14$. Therefore, our answer is $\boxed{\textbf{(B) } 14.}$

~ Technodoggo

Solution 3 (modular arithmetic)

We first proceed as in the above solution, up to $N=500+10a+b$. We then use modular arithmetic:

\begin{align*} 0&\equiv N \:(\text{mod }7)\\ &\equiv500+10a+b\:(\text{mod }7)\\ &\equiv3+3a+b\:(\text{mod }7)\\ 3a+b&\equiv-3\:(\text{mod }7)\\ &\equiv4\:(\text{mod }7)\\ \end{align*}

We know that $0\le a,b<10$. We then look at each possible value of $a$:

If $a=0$, then $b$ must be $4$.

If $a=1$, then $b$ must be $1$ or $8$.

If $a=2$, then $b$ must be $5$.

If $a=3$, then $b$ must be $2$ or $9$.

If $a=4$, then $b$ must be $6$.

If $a=5$, then $b$ must be $3$.

If $a=6$, then $b$ must be $0$ or $7$.

If $a=7$, then $b$ must be $4$.

If $a=8$, then $b$ must be $1$ or $8$.

If $a=9$, then $b$ must be $5$.

Each of these cases are unique, so there are a total of $1+2+1+2+1+1+2+1+2+1=\boxed{\textbf{(B) } 14.}$

~ Technodoggo

Solution 4

The key point is that when reversed, the number must start with a $0$ or a $5$ based on the second restriction. But numbers can't start with a $0$.

So the problem is simply counting the number of multiples of $7$ in the $500$s.

$7 \times 72 = 504$, so the first multiple is $7 \times 72$.

$7 \times 85 = 595$, so the last multiple is $7 \times 85$.

Now, we just have to count $7\times 72, 7\times 73, 7\times 74,\cdots, 7\times 85$.

We have a set that numbers $85-71=\boxed{\textbf{(B) 14}}$

~Dilip ~boppitybop ~ESAOPS (LaTeX)

Video Solution

https://youtu.be/UYHCNlRDZBo

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=7qWMFzq7Tfs

Video Solution

https://www.youtube.com/watch?v=4uKo5NR2o9Y

-paixiao

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/N2lyYRMuZuk?si=6B-mTB070UP2yuDF&t=435

~Math-X

Video Solution by Power Solve (easy to digest!)

https://www.youtube.com/watch?v=Mg6JUanYNJY

Note

According to the official answer key, choice (B) is correct. However, some have argued that it is ambiguous whether the number $560$ should be included in the count, since its reversal, $065$, has a leading zero. It is assumed that $065$ denotes the two-digit number $65$, which is divisible by $5$, but MAA should have clarified what happens when a number with trailing zeros is reversed.

~A_MatheMagician ~ESAOPS ~sdpandit

See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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. AMC logo.png

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