2018 AMC 12B Problems/Problem 15

Problem

How many odd positive -digit integers are divisible by but do not contain the digit ?

Solution 1

Let be one such odd positive -digit integer with hundreds digit tens digit and ones digit Since we need

As and there are possibilities for and possibilities for Note that each ordered pair determines the value of modulo so can be any element in one of the sets or We conclude that there are always possibilities for

By the Multiplication Principle, the answer is

~Plasma_Vortex ~MRENTHUSIASM

Solution 2

Let be one such odd positive -digit integer with hundreds digit tens digit and ones digit Since we need

Solution 3

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 - and -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

Solution 4

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, we have 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 5

We need to take care of all restrictions. Ranging from to there are odd -digit numbers. Exactly of these numbers are divisible by which is Of these numbers, have in their ones (units) digit, have in their tens digit, and have in their hundreds digit. Thus, the total number of -digit integers is

~mathpro12345

Video Solution

https://youtu.be/mgEZOXgIZXs?t=448

~ pi_is_3.14

See Also

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