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 2 (solution 1 but more thorough + alternate way))
m (→‎Solution 1)
Line 8: Line 8:
 
==Solution 1==
 
==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> always end in <math>0</math> or <math>5</math> and since it is a three-digit number (otherwise it would be a two-digit number), it cannot start with 0. All possibilities have to be in the range from <math>7 x 72</math> to <math>7 x 85</math> inclusive.  
  
−
~walmartbrian ~Shontai ~andliu766
+
<math>85 - 72 + 1 = 14</math>. <math>\boxed{(B)}</math>.
 +
 
 +
~walmartbrian ~Shontai ~andliu766 ~andyluo
  
 
==Solution 2 (solution 1 but more thorough + alternate way)==
 
==Solution 2 (solution 1 but more thorough + alternate way)==

Revision as of 19:55, 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 (otherwise it would be a two-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 ~andyluo

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:

2023 10a 12.png

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=201222"