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counting and probability intermediate

Problem

Let equal the number of four digit odd numbers. Let equal the number of four digit multiples of 5. Find .
Solution
For an odd number, there are 5 choices for the units digit, coming from the set . There will be 10 choices for the tens digit, 10 choices for the hundreds digit, and 9 choices for the thousands digit, which can not be zero. This is a total of: Multiples of 5 must end in 0 or 5. So, there are two possibilities for the units digit, and the same number of possibilities for remaining digits. This gives: Therefore, .
Final answer
6300