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Difference between revisions of "2023 AMC 10A Problems/Problem 12"

(→‎Solution 2 (solution 1 but more thorough + alternate way))
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Multiples of <math>5</math> always end in <math>0</math> or <math>5</math> and since it is a three-digit number (otherwise it would be a two-digit number), it cannot start with 0. All possibilities have to be in the range from <math>7 \cdot 72</math> to <math>7 \cdot 85</math> inclusive.  
 
Multiples of <math>5</math> always end in <math>0</math> or <math>5</math> and since it is a three-digit number (otherwise it would be a two-digit number), it cannot start with 0. All possibilities have to be in the range from <math>7 \cdot 72</math> to <math>7 \cdot 85</math> inclusive.  
  
−
<math>85 - 72 + 1 = 14</math>. <math>\boxed{(B)}</math>.
+
<math>85 - 72 + 1 = 14</math>. <math>\boxed{\textbf{(B) } 14}</math>.
  
 
~walmartbrian ~Shontai ~andliu766 ~andyluo
 
~walmartbrian ~Shontai ~andliu766 ~andyluo
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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>.  
 
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{\text{(B) }14.}</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>  
  
 
Alternate solution:  
 
Alternate solution:  
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If <math>a=9</math>, then <math>b</math> must be <math>5</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{\text{(B) }14.}</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>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2023|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2023|ab=A|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:17, 9 November 2023

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 divisble 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 $5$ always end in $0$ or $5$ and since it is a three-digit number (otherwise it would be a two-digit number), it cannot start with 0. 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

Solution 2 (solution 1 but more thorough + alternate way)

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 $\boxed{\textbf{(B) } 14.}$

Alternate solution:

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 $1+2+1+2+1+1+2+1+2+1=\boxed{\textbf{(B) } 14.}$

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

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