Difference between revisions of "2004 AMC 10B Problems/Problem 2"
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<math> \mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 30 </math> | <math> \mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 30 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
Ten numbers <math>(70,71,\dots,79)</math> have <math>7</math> as the tens digit. Nine numbers <math>(17,27,\dots,97)</math> have it as the ones digit. Number <math>77</math> is in both sets. | Ten numbers <math>(70,71,\dots,79)</math> have <math>7</math> as the tens digit. Nine numbers <math>(17,27,\dots,97)</math> have it as the ones digit. Number <math>77</math> is in both sets. | ||
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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>90</math> two-digit numbers. | + | 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 \times 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. | ||
| − | Thus there are <math>72</math> two-digit numbers without a <math>7</math> as a digit. | + | Thus there are <math>8 \times 9 = 72</math> two-digit numbers without a <math>7</math> as a digit. |
<math>90</math> (The total number of two-digit numbers) <math>- 72</math> (The number of two-digit numbers without a <math>7</math>) <math>= 18 \Rightarrow</math> <math>\boxed{\mathrm{(B)}\ 18}</math>. | <math>90</math> (The total number of two-digit numbers) <math>- 72</math> (The number of two-digit numbers without a <math>7</math>) <math>= 18 \Rightarrow</math> <math>\boxed{\mathrm{(B)}\ 18}</math>. | ||
Latest revision as of 18:10, 24 January 2015
Contents
Problem
How many two-digit positive integers have at least one 
as a digit?
Solution 1
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
