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

Revision as of 20:06, 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 ~boppitybop

@@ Line 64: / Line 64: @@
 ==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!
+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.
+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.
-x 70 = 490 so the first multiple is 7 x 72
+x 70 = 490, so the first multiple is 7 x 72.
-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)
+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
+Now, we just have to count 7x72, 7x73, 7x74, ..., 7x85.
 We have a set that numbers 85-71 = <math>\boxed{\textbf{(B) 14}}</math>
-~Dilip
+~Dilip ~boppitybop
 ==See Also==
 {{AMC10 box|year=2023|ab=A|num-b=11|num-a=13}}
 {{MAA Notice}}

2023 AMC 10A (Problems • Answer Key • Resources)
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