Difference between revisions of "2020 AMC 8 Problems/Problem 19"

Revision as of 17:22, 26 February 2021

Problem 19

A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$

$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$

Solution 1

A number is divisible by $15$ precisely if it is divisible by $3$ and $5$ . The latter means the last digit must be either $5$ or $0$ , and the former means the sum of the digits must be divisible by $3$ . If the last digit is $0$ , the first digit would be $0$ (because the digits alternate), which is not possible. Hence the last digit must be $5$ , and the number is of the form $5\square 5\square 5$ . If the unknown digit is $x$ , we deduce $5+x+5+x+5 \equiv 0 \pmod{3} \Rightarrow 2x \equiv 0 \pmod{3}$ . We know $2^{-1}$ exists modulo $3$ because 2 is relatively prime to 3, so we conclude that $x$ (i.e. the second and fourth digit of the number) must be a multiple of $3$ . It can be $0$ , $3$ , $6$ , or $9$ , so there are $\boxed{\textbf{(B) }4}$ options: $50505$ , $53535$ , $56565$ , and $59595$ .

Solution 2 (variant of Solution 1)

As in Solution 1, we find that such numbers must start with $5$ and alternate with $5$ (i.e. must be of the form $5\square 5\square 5$ ), where the two digits between the $5$ s need to be the same. Call that digit $x$ . For the number to be divisible by $3$ , the sum of the digits must be divisible by $3$ ; since the sum of the three $5$ s is $15$ , which is already a multiple of $3$ , it must also be the case that $x+x=2x$ is a multiple of $3$ . Thus, the problem reduces to finding the number of digits from $0$ to $9$ for which $2x$ is a multiple of $3$ . This leads to $x=0$ , $3$ , $6$ , or $9$ , so there are $\boxed{\textbf{(B) }4}$ possible numbers (namely $50505$ , $53535$ , $56565$ , and $59595$ ).

Video Solution by WhyMath

https://youtu.be/8nVSeTx5rro

~savannahsolver

Video Solution

https://youtu.be/VnOecUiP-SA

@@ Line 9: / Line 9: @@
 ==Solution 2 (variant of Solution 1)==
 As in Solution 1, we find that such numbers must start with <math>5</math> and alternate with <math>5</math> (i.e. must be of the form <math>5\square 5\square 5</math>), where the two digits between the <math>5</math>s need to be the same. Call that digit <math>x</math>. For the number to be divisible by <math>3</math>, the sum of the digits must be divisible by <math>3</math>; since the sum of the three <math>5</math>s is <math>15</math>, which is already a multiple of <math>3</math>, it must also be the case that <math>x+x=2x</math> is a multiple of <math>3</math>. Thus, the problem reduces to finding the number of digits from <math>0</math> to <math>9</math> for which <math>2x</math> is a multiple of <math>3</math>. This leads to <math>x=0</math>, <math>3</math>, <math>6</math>, or <math>9</math>, so there are <math>\boxed{\textbf{(B) }4}</math> possible numbers (namely <math>50505</math>, <math>53535</math>, <math>56565</math>, and <math>59595</math>).
+==Video Solution by WhyMath==
+https://youtu.be/8nVSeTx5rro
+~savannahsolver
 ==Video Solution==

2020 AMC 8 (Problems • Answer Key • Resources)
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