Difference between revisions of "2004 AMC 10B Problems/Problem 2"

Revision as of 20:01, 22 July 2014

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

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

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

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 10 Problems and Solutions

@@ Line 7: / Line 7: @@
 ==Solution==
-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{(B)}</math>.
+Thus the result is <math>10+9-1=18 \Rightarrow</math> <math>\boxed{\mathrm{(B)}\ 18}</math>.
 == See also ==