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

Revision as of 20:18, 9 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 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 $7 \cdot 72$ to $7 \cdot 85$ 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:

$\begin{align*} 0&\equiv N \:(\text{mod }7)\\ &\equiv500+10a+b\:(\text{mod }7)\\ &\equiv3+3a+b\:(\text{mod }7)\\ 3a+b&\equiv-3\:(\text{mod }7)\\ &\equiv4\:(\text{mod }7)\\ \end{align*}$

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

@@ Line 1: / Line 1: @@
+==Problem==
 How many three-digit positive integers <math>N</math> satisfy the following properties?
@@ Line 24: / Line 25: @@
 We then use modular arithmetic:
-[[Image:2023_10a_12.png]]
+<cmath>\begin{align*}
+&\equiv N \:(\text{mod }7)\\
+&\equiv500+10a+b\:(\text{mod }7)\\
+&\equiv3+3a+b\:(\text{mod }7)\\
+a+b&\equiv-3\:(\text{mod }7)\\
+&\equiv4\:(\text{mod }7)\\
+\end{align*}</cmath>
 We know that <math>0\le a,b<10</math>. We then look at each possible value of <math>a</math>:
@@ Line 49: / Line 56: @@
 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>
+==See Also==
+{{AMC10 box|year=2023|ab=A|num-b=11|num-a=13}}
+{{MAA Notice}}

2023 AMC 10A (Problems • Answer Key • Resources)
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
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Difference between revisions of "2023 AMC 10A Problems/Problem 12"

Revision as of 20:18, 9 November 2023

Contents

Problem

Solution 1

Solution 2 (solution 1 but more thorough + alternate way)

See Also