Difference between revisions of "2004 AMC 10B Problems/Problem 2"
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We use complementary counting. The complement of having at least one <math>7</math> as a digit is having no <math>7</math>s as a digit. | We use complementary counting. The complement of having at least one <math>7</math> as a digit is having no <math>7</math>s as a digit. | ||
− | We have <math>9</math> digits to choose from for the first digit and <math>10</math> digits for the second. This gives a total of <math>9 \ | + | We have <math>9</math> digits to choose from for the first digit and <math>10</math> digits for the second. This gives a total of <math>9 \dot 10 = 90</math> two-digit numbers. |
But since we cannot have <math>7</math> as a digit, we have <math>8</math> first digits and <math>9</math> second digits to choose from. | But since we cannot have <math>7</math> as a digit, we have <math>8</math> first digits and <math>9</math> second digits to choose from. |
Revision as of 19:06, 24 January 2015
Contents
Problem
How many two-digit positive integers have at least one as a digit?
Solution
Ten numbers have as the tens digit. Nine numbers have it as the ones digit. Number is in both sets.
Thus the result is .
Solution 2
We use complementary counting. The complement of having at least one as a digit is having no s as a digit.
We have digits to choose from for the first digit and digits for the second. This gives a total of two-digit numbers.
But since we cannot have as a digit, we have first digits and second digits to choose from.
Thus there are two-digit numbers without a as a digit.
(The total number of two-digit numbers) (The number of two-digit numbers without a ) .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 10 Problems and Solutions |
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