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Difference between revisions of "2002 AMC 12P Problems/Problem 7"

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{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #7]] and [[2002 AMC 10P Problems|2002 AMC 10P #20]]}}
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== Problem ==
 
== Problem ==
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How many three-digit numbers have at least one <math>2</math> and at least one <math>3?</math>
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How many three-digit numbers have at least one <math>2</math> and at least one <math>3</math>?
  
 
<math>
 
<math>
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== Solution ==
 
== Solution ==
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
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We can do this problem with some simple case work.
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Case 1: The hundreds place is not <math>2</math> or <math>3.</math>
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This means that the tens place and ones place must be <math>2</math> and <math>3</math> respectively or <math>3</math> and <math>2</math> respectively. This case covers <math>1, 4, 5, 6, 7, 8,</math> and <math>9,</math> so it gives us <math>2 \cdot 7 = 14</math> cases.
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Case 2: The hundreds place is <math>2.</math>
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This means that <math>3</math> must be in the tens place or ones place. Starting with cases in which the tens place is not <math>3</math>, we get <math>203, 213, 223, 243, 253, 263, 273, 283,</math> and <math>293.</math>  With cases in which the tens place is <math>3</math>, we have <math>230-239</math>, or <math>10</math> more cases. This gives us <math>9 + 10=19</math> cases.
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Case 3: The hundreds place is <math>3.</math>
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This case is almost identical to the second case, just swap the <math>2</math>s with <math>3</math>s and <math>3</math>s with <math>2</math>s in the reasoning and its the same, giving us an additional <math>19</math> cases.
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Adding up all of these cases gives <math>14+19+19=52</math> cases, or <math>\boxed{\textbf{(A) }52}.</math>
  
 
== See also ==
 
== See also ==
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{{AMC10 box|year=2002|ab=P|num-b=19|num-a=21}}
 
{{AMC12 box|year=2002|ab=P|num-b=6|num-a=8}}
 
{{AMC12 box|year=2002|ab=P|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:20, 14 July 2024

The following problem is from both the 2002 AMC 12P #7 and 2002 AMC 10P #20, so both problems redirect to this page.

Problem

How many three-digit numbers have at least one $2$ and at least one $3$?

$\text{(A) }52 \qquad \text{(B) }54 \qquad \text{(C) }56 \qquad \text{(D) }58 \qquad \text{(E) }60$

Solution

We can do this problem with some simple case work.

Case 1: The hundreds place is not $2$ or $3.$ This means that the tens place and ones place must be $2$ and $3$ respectively or $3$ and $2$ respectively. This case covers $1, 4, 5, 6, 7, 8,$ and $9,$ so it gives us $2 \cdot 7 = 14$ cases.

Case 2: The hundreds place is $2.$ This means that $3$ must be in the tens place or ones place. Starting with cases in which the tens place is not $3$, we get $203, 213, 223, 243, 253, 263, 273, 283,$ and $293.$ With cases in which the tens place is $3$, we have $230-239$, or $10$ more cases. This gives us $9 + 10=19$ cases.

Case 3: The hundreds place is $3.$ This case is almost identical to the second case, just swap the $2$s with $3$s and $3$s with $2$s in the reasoning and its the same, giving us an additional $19$ cases.

Adding up all of these cases gives $14+19+19=52$ cases, or $\boxed{\textbf{(A) }52}.$

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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
2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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