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Difference between revisions of "2023 AMC 10A Problems/Problem 12"
(→‎Solution 2 (solution 1 but more thorough + alternate way))
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We then use modular arithmetic:  
 
We then use modular arithmetic:  
  
−
\begin{align*}
+
[[Image:2023_10a_12.png]]
−
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>:  
 
We know that <math>0\le a,b<10</math>. We then look at each possible value of <math>a</math>:  

Revision as of 19:38, 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:

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 $1+2+1+2+1+1+2+1+2+1=\boxed{\text{(B) }14.}$

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