Difference between revisions of "2020 AMC 8 Problems/Problem 19"
Sevenoptimus (talk | contribs) m (Removed unnecessary and irrelevant attribution) |
|||
(8 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem== | + | ==Problem 19== |
A number is called flippy if its digits alternate between two distinct digits. For example, <math>2020</math> and <math>37373</math> are flippy, but <math>3883</math> and <math>123123</math> are not. How many five-digit flippy numbers are divisible by <math>15?</math> | A number is called flippy if its digits alternate between two distinct digits. For example, <math>2020</math> and <math>37373</math> are flippy, but <math>3883</math> and <math>123123</math> are not. How many five-digit flippy numbers are divisible by <math>15?</math> | ||
Line 9: | Line 9: | ||
==Solution 2 (variant of Solution 1)== | ==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>). | 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== | ==Video Solution== | ||
− | https://youtu.be/ | + | https://youtu.be/VnOecUiP-SA |
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=18|num-a=20}} | {{AMC8 box|year=2020|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:22, 26 February 2021
Contents
Problem 19
A number is called flippy if its digits alternate between two distinct digits. For example, and are flippy, but and are not. How many five-digit flippy numbers are divisible by
Solution 1
A number is divisible by precisely if it is divisible by and . The latter means the last digit must be either or , and the former means the sum of the digits must be divisible by . If the last digit is , the first digit would be (because the digits alternate), which is not possible. Hence the last digit must be , and the number is of the form . If the unknown digit is , we deduce . We know exists modulo because 2 is relatively prime to 3, so we conclude that (i.e. the second and fourth digit of the number) must be a multiple of . It can be , , , or , so there are options: , , , and .
Solution 2 (variant of Solution 1)
As in Solution 1, we find that such numbers must start with and alternate with (i.e. must be of the form ), where the two digits between the s need to be the same. Call that digit . For the number to be divisible by , the sum of the digits must be divisible by ; since the sum of the three s is , which is already a multiple of , it must also be the case that is a multiple of . Thus, the problem reduces to finding the number of digits from to for which is a multiple of . This leads to , , , or , so there are possible numbers (namely , , , and ).
Video Solution by WhyMath
~savannahsolver
Video Solution
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.