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

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==Solution 1==
 
==Solution 1==
  
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.  
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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>.

Revision as of 12:02, 10 November 2023

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

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

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!

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 500s.

7 x 70 = 490 so the first multiple is 7 x 72

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)

Now we just have to count 7x72, 7x73, 7x74, ..., 7x85

We have a set that numbers 85-71 = $\boxed{\textbf{(B) 14}}$

~Dilip

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
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

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