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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 number has to be a three-digit number (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>. 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 (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.  
  
 
<math>85 - 72 + 1 = 14</math>. <math>\boxed{\textbf{(B) } 14}</math>.
 
<math>85 - 72 + 1 = 14</math>. <math>\boxed{\textbf{(B) } 14}</math>.
  
−
~walmartbrian ~Shontai ~andliu766 ~andyluo
+
(Add 1 to include 72)
 +
 
 +
~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS
  
 
==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 142 such numbers divisible by 7 since 1000 divided by 7 gives 142 with a remainder. But it's not relevant!
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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 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
  
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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.
+
<i> ~Education, the Study of Everything </i>
  
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So the problem is simply counting the number of multiples of 7 in the 500s.
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==Video Solution==
  
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7 x 70 = 490 so the first multiple is 7 x 72
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https://youtu.be/UYHCNlRDZBo
  
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7 x 80 = 560 so the first multiple more than 599 is 7 x 86 (since 7 x 6 = 42 and 560 + 42 is in the 600s)
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
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Now we just have to count 7x72, 7x73, 7x74, ..., 7x85
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==Video Solution ==
 +
https://www.youtube.com/watch?v=Mg6JUanYNJY
  
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We have a set that numbers 85-71 = <math>\boxed{\textbf{(B) 14}}</math>
+
==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.
  
−
~Dilip
+
~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}}

Revision as of 11:28, 1 October 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 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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