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

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==Solution 1==
 
==Solution 1==
  
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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.  
+
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, it cannot start with 0 (otherwise it would be a two-digit number), 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(504 to 595).  
 +
 
 +
(Add 1 to include 72)
  
 
<math>85 - 72 + 1 = 14</math>. <math>\boxed{\textbf{(B) } 14}</math>.
 
<math>85 - 72 + 1 = 14</math>. <math>\boxed{\textbf{(B) } 14}</math>.
  
−
One thing to note is the number 560. When it is flipped, you get 065, and no one actually writes it like this. This problem doesn't specify whether or not 560 is allowed.
+
You can also take 497 away from each of the numbers(removing the hundreds digit and adding three to each of the numbers), resulting in the numbers {7, 14, 21..., 84, 91, 98}. Dividing each of them by 7, you get the numbers {1, 2, 3..., 12, 13, 14}. Therefore, the answer is <math>\boxed{\textbf{(B) 14}}</math>
  
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~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS
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~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS ~NXC
  
 
==Solution 2 (solution 1 but more thorough)==
 
==Solution 2 (solution 1 but more thorough)==
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==Solution 4==
 
==Solution 4==
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Initially, I thought of finding that there are <math>142</math> such numbers divisible by <math>7</math> since <math>1000</math> divided by <math>7</math> gives <math>142</math> with a remainder. But it's not relevant!
 
  
 
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>.
 
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>.
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Now, we just have to count <math>7\times 72, 7\times 73, 7\times 74,\cdots, 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>.
  
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We have a set that numbers 85-71 = <math>\boxed{\textbf{(B) 14}}</math>
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We have a set that numbers <math>85-71=\boxed{\textbf{(B) 14}}</math>
 +
 
 +
~Dilip ~boppitybop ~ESAOPS (<math>\LaTeX</math>)
 +
 
 +
== Solution 5 (Quick and Fast ⚡⚡⚡⚡) ==
 +
 
 +
We notice that the numbers have to be divisible by <math>5</math>, implying it ends with a <math>5</math> or a <math>0</math>. This means that when reversed it will start with a <math>0</math> and <math>5</math>. If a three-digit number starts with a <math>0</math>, it would be a two-digit number. This means that our number would have to start with a <math>5</math>. There are <math>99</math> total three-digit number that start with a <math>5</math> <math>(500 - 599)</math>. Since we want to find the numbers from <math>500 - 599</math> that are divisible by <math>7</math>, we do
 +
 
 +
<math>\lfloor{\dfrac{99}7} \rfloor = \boxed{\textbf{(C) } 14}</math>.
 +
 
 +
~yuvag
 +
 
 +
==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=xKgx8T-n-Y1ELLZF&t=2669
 +
~little-fermat
 +
 
 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=t4QMuoYyk2u5n64a&t=3140
 +
 
 +
~Math-X
  
−
~Dilip ~boppitybop ~ESAOPS (LaTeX)
+
==Video Solution ⚡️ 2 min solution ⚡️==
 +
https://youtu.be/YdaQIdxyBSg
 +
 
 +
<i> ~Education, the Study of Everything </i>
  
 
==Video Solution==
 
==Video Solution==
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==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==
 
==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}}

Latest revision as of 19:25, 10 November 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, it cannot start with 0 (otherwise it would be a two-digit number), 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(504 to 595).

(Add 1 to include 72)

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

You can also take 497 away from each of the numbers(removing the hundreds digit and adding three to each of the numbers), resulting in the numbers {7, 14, 21..., 84, 91, 98}. Dividing each of them by 7, you get the numbers {1, 2, 3..., 12, 13, 14}. Therefore, the answer is $\boxed{\textbf{(B) 14}}$

~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS ~NXC

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 ($\LaTeX$)

Solution 5 (Quick and Fast ⚡⚡⚡⚡)

We notice that the numbers have to be divisible by $5$, implying it ends with a $5$ or a $0$. This means that when reversed it will start with a $0$ and $5$. If a three-digit number starts with a $0$, it would be a two-digit number. This means that our number would have to start with a $5$. There are $99$ total three-digit number that start with a $5$ $(500 - 599)$. Since we want to find the numbers from $500 - 599$ that are divisible by $7$, we do

$\lfloor{\dfrac{99}7} \rfloor = \boxed{\textbf{(C) } 14}$.

~yuvag

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=xKgx8T-n-Y1ELLZF&t=2669 ~little-fermat

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

https://youtu.be/GP-DYudh5qU?si=t4QMuoYyk2u5n64a&t=3140

~Math-X

Video Solution ⚡️ 2 min solution ⚡️

https://youtu.be/YdaQIdxyBSg

~Education, the Study of Everything

Video Solution

https://youtu.be/UYHCNlRDZBo

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

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

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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