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

(New page: ==Problem== How many two-digit positive integers have at least one 7 as a digit? <math> \mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qq...)
 
m (Solution)
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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=\boxed{18}</math>.
+
Thus the result is <math>10+9-1=18 \Rightarrow</math> <math> \boxed{(B)}</math>.
  
 
== See also ==
 
== See also ==
  
 
{{AMC10 box|year=2004|ab=B|num-b=1|num-a=3}}
 
{{AMC10 box|year=2004|ab=B|num-b=1|num-a=3}}

Revision as of 19:41, 29 December 2011

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

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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