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Difference between revisions of "1985 AJHSME Problems/Problem 15"

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<math>\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148</math>
 
<math>\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148</math>
  
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==Solution==
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==Solution 1==
  
 
This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.
 
This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.

Revision as of 15:28, 6 September 2018

Problem

How many whole numbers between $100$ and $400$ contain the digit $2$?

$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148$

Solution 1

This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.

If you ever learned about complementary counting, this would be the best time to utilize it. Instead of counting how many DO have 2's, why don't we count how many that DON'T?

So let's find the number of numbers. Obviously, we'd start by subtracting 100 from 400, getting us 300, but we're not done. Since just subtracting includes the number 400, we must subtract one (because 400 isn't allowed - it says between), getting us 299.

So how many numbers are there that DON'T have a 2? Well, we have 2 possibilities for the hundreds digit (1, 3, note that 2 is not allowed), 9 possibilities for the tens digit (1, 3, 4, 5, ... , 9, 0), and 9 possibilities for the ones digit. $2 \times 9 \times 9 = 162$. However, one of the numbers we counted is $100$, which isn't allowed, so there are $162-1=161$ numbers without a 2.

Since there are 299 numbers in total and 161 that DON'T have any 2's, $299 - 161 = 138$ numbers WILL have at least one two.

$\boxed{\text{C}}$

Solution using PIE

As in the previous solution, get rid of the $400$ to make things easier. This gives us the numbers from $100$ to $399$.

Let $A$ be the event that the first digit is $2$, $B$ be the event that the second digit is $2$, and $C$ be the event that the third digit is $2$. Then, PIE says that our answer will be $|A|+|B|+|C|-|A\cup B|-|A\cup C|-|B\cup C|+|A\cup B\cup C|$.

We have that $|A|$ is just $1\times10\times10=100$, $|B|=3\times1\times10=30$, and $|C|=3\times10\times1=30$.

Next, $|A\cup B|$ is just having something in the form $22\_$, so there are $10$ ways. Similarly, $|A\cup C|$ means that we have $2\_2$, so there are again $10$ ways. Finally, $|B\cup C|$ means that our number is like $\_22$, so there are $3$ ways.

Finally, $|A\cup B\cup C|$ counts the number of three digit numbers with all three digits $2$, which there is only $1$ of: $222$.

Putting this together, our answer will be $100+30+30-10-10-3+1=\boxed{\text{(C)}\ 138}$.

See Also

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 AJHSME/AMC 8 Problems and Solutions


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