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==
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==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.
  
 
Thus the result is <math>10+9-1=18 \Rightarrow</math> <math>\boxed{\mathrm{(B)}\ 18}</math>.
 
Thus the result is <math>10+9-1=18 \Rightarrow</math> <math>\boxed{\mathrm{(B)}\ 18}</math>.
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==Solution 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.
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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.
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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.
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Thus there are <math>8 \times 9 = 72</math> two-digit numbers without a <math>7</math> as a digit.
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<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 ==

Latest revision as of 19:10, 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 1

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.

We have $9$ digits to choose from for the first digit and $10$ digits for the second. This gives a total of $9 \times 10 = 90$ two-digit numbers.

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

Thus there are $8 \times 9 = 72$ 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 $7$) $= 18 \Rightarrow$ $\boxed{\mathrm{(B)}\ 18}$.

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
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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