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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.
  
−
Usually, 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> <math>2</math>-digit numbers.
+
Usually, 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.
  
 
But since we cannot have a <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 a <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> <math>2</math>-digit numbers without a <math>7</math> as a digit.
+
Thus there are <math>72</math> two-digit numbers without a <math>7</math> as a digit.
  
−
<math>90</math> (The total number of <math>2</math>-digit numbers) - <math>72</math> (The number of <math>2</math>-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>.
  
 
== See also ==
 
== See also ==

Revision as of 18:01, 24 January 2015

Problem

How many two-digit positive integers have at least one $7$ as a digit?

$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 30$

Solution

Ten numbers $(70,71,\dots,79)$ have $7$ as the tens digit. Nine numbers $(17,27,\dots,97)$ have it as the ones digit. Number $77$ is in both sets.

Thus the result is $10+9-1=18 \Rightarrow$ $\boxed{\mathrm{(B)}\ 18}$.

Solution 2

We use complementary counting. The complement of having at least one $7$ as a digit is having no $7$s as a digit.

Usually, we have $9$ digits to choose from for the first digit, and $10$ digits for the second. This gives a total of $90$ two-digit numbers.

But since we cannot have a $7$ as a digit, we have $8$ first digits and $9$ second digits to choose from.

Thus there are $72$ two-digit numbers without a $7$ as a digit.

$90$ (The total number of two-digit numbers) - $72$ (The number of two-digit numbers without a $7$) $= 18 \Rightarrow$ $\boxed{\mathrm{(B)}\18}$ (Error compiling LaTeX. Unknown error_msg).

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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