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...) |
(→Solution) |
||
(14 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | How many two-digit positive integers have at least one 7 as a digit? | + | How many two-digit positive integers have at least one <math>7</math> as a digit? |
<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== | + | ==Solution 1== |
− | Ten numbers | + | 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{\mathrm{(B)}\ 18}</math>. |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Thus there are <math>8 \times 9 = 72</math> two-digit numbers without a <math>7</math> as a digit. | ||
+ | |||
+ | <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 == | ||
{{AMC10 box|year=2004|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2004|ab=B|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:10, 24 January 2015
Contents
Problem
How many two-digit positive integers have at least one as a digit?
Solution 1
Ten numbers have
as the tens digit. Nine numbers
have it as the ones digit. Number
is in both sets.
Thus the result is
.
Solution 2
We use complementary counting. The complement of having at least one as a digit is having no
s as a digit.
We have digits to choose from for the first digit and
digits for the second. This gives a total of
two-digit numbers.
But since we cannot have as a digit, we have
first digits and
second digits to choose from.
Thus there are two-digit numbers without a
as a digit.
(The total number of two-digit numbers)
(The number of two-digit numbers without a
)
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.