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

Revision as of 20:39, 8 July 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.

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

(Add 1 to include 72)

~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

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

-paixiao

Video Solution

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

~Math-X

Video Solution

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

@@ Line 1: / Line 1: @@
+==Problem==
 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 formed by reversing the digits of <math>N</math> is divisble by <math>5</math>.
+*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==
-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. <math>\boxed{(B)}</math>.
+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 ~andliu766
+<math>85 - 72 + 1 = 14</math>. <math>\boxed{\textbf{(B) } 14}</math>.
+(Add 1 to include 72)
+~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*}
+&\equiv N \:(\text{mod }7)\\
+&\equiv500+10a+b\:(\text{mod }7)\\
+&\equiv3+3a+b\:(\text{mod }7)\\
+a+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==
+https://www.youtube.com/watch?v=4uKo5NR2o9Y
+-paixiao
+==Video Solution ==
+https://youtu.be/N2lyYRMuZuk?si=6B-mTB070UP2yuDF&t=435
+~Math-X
+==Video Solution ==
+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}}

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

Revision as of 20:39, 8 July 2024

Contents

Problem

Solution 1

Solution 2 (solution 1 but more thorough)

Solution 3 (modular arithmetic)

Solution 4

Video Solution

Video Solution

Video Solution

Video Solution

Note

See Also