Difference between revisions of "2018 AMC 12B Problems/Problem 15"
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== Problem == | == Problem == | ||
− | + | For how many digits <math>C</math> is the positive three-digit number <math>1C3</math> a multiple of 3? | |
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== Solution 1 (For Dummies) == | == Solution 1 (For Dummies) == |
Revision as of 19:33, 14 December 2019
Problem
For how many digits is the positive three-digit number a multiple of 3?
Solution 1 (For Dummies)
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
Note that this isn't a great solution, but a more practical one to achieve the answer.
Notice that there are numbers that have digits and are divisible by (from to ). Now one by one apply the restrictions.
The restriction for only odd numbers would mean that half the numbers are taken out .
Next, apply the restriction of no s. For the units digit, that would mean multiplying by (remember that now you only have odd numbers to choose from).
For the tens that would mean multiplying by , and for the hundreds that would mean multiplying by (because you cant have 0 here).
Thus, we get , which is .
Sol by IronicNinja~
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 |
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