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...) |
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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{ | + | 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?
Solution
Ten numbers () have as the tens digit. Nine numbers () have it as the ones digit. Number is in both sets.
Thus the result is .
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 |