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>
  
==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.
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<math>\boxed{\text{C}}</math>
 
<math>\boxed{\text{C}}</math>
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==Video Solution==
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https://youtu.be/9APPn2mC8t8
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AJHSME box|year=1985|num-b=14|num-a=16}}
 
{{AJHSME box|year=1985|num-b=14|num-a=16}}
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 16:33, 6 June 2024

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.

$\boxed{\text{C}}$

Video Solution

https://youtu.be/9APPn2mC8t8

~savannahsolver

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
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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