Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 4)
(→‎Video Solution)
(11 intermediate revisions by 8 users not shown)
Line 13: Line 13:
  
 
<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.
 
  
 
~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS
 
~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS
Line 65: Line 63:
  
 
==Solution 4==
 
==Solution 4==
−
 
−
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>.
  
−
So the problem is simply counting the number of multiples of <math>7 in the 500</math>s.
+
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 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>.
+
<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>.
 
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 85-71 = <math>\boxed{\textbf{(B) 14}}</math>
+
We have a set that numbers <math>85-71=\boxed{\textbf{(B) 14}}</math>
  
−
~Dilip ~boppitybop
+
~Dilip ~boppitybop ~ESAOPS (LaTeX)
  
 
==Video Solution==
 
==Video Solution==
Line 87: Line 83:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~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==
 
==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 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_12&oldid=213529"