Difference between revisions of "2018 AMC 12B Problems/Problem 15"
Mathpro12345 (talk | contribs) (→Solution 4 (easy)) |
Pi is 3.14 (talk | contribs) (→Solution 4 (easy)) |
||
Line 25: | Line 25: | ||
~mathpro12345 | ~mathpro12345 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/mgEZOXgIZXs?t=448 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Revision as of 22:21, 17 January 2021
Contents
Problem
How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?
Solution 1
Analyze that the three-digit integers divisible by start from . In the 's, it starts from . In the 's, it starts from . We see that the units digits is and
Write out the 1- and 2-digit multiples of starting from and Count up the ones that meet the conditions. Then, add up and multiply by , since there are three sets of three from to Then, subtract the amount that started from , since the 's ll contain the digit .
We get:
This gives us:
Solution 2
There are choices for the last digit (), and choices for the first digit (exclude ). We know what the second digit mod is, so there are choices for it (pick from one of the sets ). The answer is (Plasma_Vortex)
Solution 3
Consider the number of -digit numbers that do not contain the digit , which is . For any of these -digit numbers, we can append or to reach a desirable -digit number. However, , and thus we need to count any -digit number twice. There are total such numbers that have remainder , but of them contain , so the number we want is . Therefore, the final answer is .
Solution 4 (easy)
We need to take care of all restrictions. Ranging from to , there are odd 3-digit numbers. Exactly of these numbers are divisible by 3, which is . Of these 150 numbers, have 3 in their ones (units) digit, have 3 in their tens digit, and have 3 in their hundreds digit. Thus, the total number of 3 digit integers are , or
~mathpro12345
Video Solution
https://youtu.be/mgEZOXgIZXs?t=448
~ pi_is_3.14
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.