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"
Line 11: Line 11:
  
 
~walmartbrian ~Shontai ~andliu766
 
~walmartbrian ~Shontai ~andliu766
 +
 +
==Solution 2 (solution 1 but more thorough + alternate way)==
 +
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>
 +
 +
Alternate solution:
 +
 +
We first proceed as in the above solution, up to <math>N=500+10a+b</math>.
 +
We then use modular arithmetic:
 +
 +
\begin{align*}
 +
0&\equiv N~(\mod7) \\
 +
&\equiv500+10a+b~(\mod7) \\
 +
&\equiv3+3a+b~(\mod7) \\
 +
3a+b&\equiv-3~(\mod7) \\
 +
&\equiv4~(\mod7). \\
 +
\end{align*}
 +
 +
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{\text{(B) }14.}</math>

Revision as of 19:21, 9 November 2023

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 $5$.


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. $\boxed{(B)}$.

~walmartbrian ~Shontai ~andliu766

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

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~(\mod7) \\ &\equiv500+10a+b~(\mod7) \\ &\equiv3+3a+b~(\mod7) \\ 3a+b&\equiv-3~(\mod7) \\ &\equiv4~(\mod7). \\ \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 $1+2+1+2+1+1+2+1+2+1=\boxed{\text{(B) }14.}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_12&oldid=201160"