2004 AMC 10B Problems/Problem 2

Revision as of 19:05, 24 January 2015 by Warrenwangtennis (talk | contribs) (Solution 2)

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

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 $9 * 10 = 90$ 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 $8 * 9 = 72$ 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
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All AMC 10 Problems and Solutions

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