Difference between revisions of "2008 iTest Problems/Problem 81"

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Of the 10 digits, four of them are composite (4, 6, 8, 9), so six of the digits are not composite. There are five ways to pick a non-composite digit for the first digit (since the number can not start with a 0) and six ways to pick a non-composite digit for the last digit. Thus, there are <math>5 \cdot 6 \cdot 10^5 = 3000000</math>.
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Of the 10 digits, four of them are composite (4, 6, 8, 9), so six of the digits are not composite. There are five ways to pick a non-composite digit for the first digit (since the number can not start with a 0) and six ways to pick a non-composite digit for the last digit. Thus, there are <math>5 \cdot 6 \cdot 10^5 = 3000000</math> numbers with seven digits that do not start or end with a composite number.
  
 
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Latest revision as of 19:28, 21 November 2018

Problem

Compute the number of $7$-digit positive integers that start $\textit{or}$ end (or both) with a digit that is a (nonzero) composite number.

Solution

Rather than counting the seven-digit positive integers that start or end with a nonzero composite digit, we can use complementary counting by counting the total number of seven-digit integers and the number of seven-digit integers that do not start and end with nonzero composite digits.


Of the 10 digits, four of them are composite (4, 6, 8, 9), so six of the digits are not composite. There are five ways to pick a non-composite digit for the first digit (since the number can not start with a 0) and six ways to pick a non-composite digit for the last digit. Thus, there are $5 \cdot 6 \cdot 10^5 = 3000000$ numbers with seven digits that do not start or end with a composite number.


The total number of 7-digit numbers is $9 \cdot 10^6 = 9000000$, so the total number of 7-digit numbers that start or end with a composite digit is $9000000 - 3000000 = \boxed{6000000}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 80
Followed by:
Problem 82
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