Difference between revisions of "2001 AMC 12 Problems/Problem 6"
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<math>\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8</math> | <math>\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8</math> | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | We start by noting that there are <math>10</math> letters, meaning there are <math>10</math> digits in total. Listing them all out, we have <math>0, 1, 2, 3, 4, 5, 6, 7, 8, 9</math>. Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that. | ||
+ | |||
+ | Case 1: <math>G</math>, <math>H</math>, <math>I</math>, and <math>J</math> are <math>7</math>, <math>5</math>, <math>3</math>, and <math>1</math> respectively. | ||
+ | |||
+ | A cursory glance allows us to deduce that the smallest possible sum of <math>A + B + C</math> is <math>11</math> when <math>D</math>, <math>E</math>, and <math>F</math> are <math>8</math>, <math>6</math>, and <math>4</math> respectively, so this is out of the question. | ||
+ | |||
+ | Case 2: <math>G</math>, <math>H</math>, <math>I</math>, and <math>J</math> are <math>3</math>, <math>5</math>, <math>7</math>, and <math>9</math> respectively. | ||
+ | |||
+ | A cursory glance allows us to deduce the answer. Clearly, when <math>D</math>, <math>E</math>, and <math>F</math> are <math>6</math>, <math>4</math>, and <math>2</math> respectively, <math>A + B + C</math> is <math>9</math> when <math>A</math>, <math>B</math>, and <math>C</math> are <math>8</math>, <math>1</math>, and <math>0</math> respectively, giving us a final answer of <math>\boxed{\textbf{(E)}\ 8}</math> | ||
+ | |||
+ | == Solution 2== | ||
+ | |||
+ | The ten letters must all correspond to ten distinct digits, so every digit is used in the telephone number. | ||
+ | |||
+ | We note that <math>1</math>, <math>3</math>, <math>5</math>, <math>7</math>, <math>9</math> are the odd numbers, and possible sequences for <math>GHIJ</math> are either <math>1</math>, <math>3</math>, <math>5</math>, <math>7</math> or <math>3</math>, <math>5</math>, <math>7</math>, <math>9</math>. | ||
+ | |||
+ | <math>3</math>, <math>5</math>, and <math>7</math> are included in both possible sequences so that we can rule out the possibilities of <math>5</math> and <math>7</math> for <math>A</math>, so <math>A</math> can only be <math>4</math>, <math>6</math>, or <math>8</math>. Since every possible sequence of DEF also contains <math>4</math>, we can rule out <math>4</math> as well. | ||
+ | |||
+ | Testing A=<math>6</math> means <math>D</math>, <math>E</math>, <math>F</math> is <math>0</math>, <math>2</math>, <math>4</math>, respectively. However, <math>6</math> and <math>8</math> must be part of <math>A</math>, <math>B</math>, or <math>C</math>, and they already sum to more than <math>10</math>, so this leaves <math>\boxed{\textbf{(E)}\ 8}</math> as our answer. | ||
+ | |||
+ | ~megaboy6679 | ||
+ | |||
+ | ==Video Solution by Daily Dose of Math== | ||
+ | |||
+ | https://youtu.be/z7o_BiWLDlk?si=9aZ0zIx2lkh_8CV3 | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
== See Also == | == See Also == |
Latest revision as of 13:02, 12 October 2024
- The following problem is from both the 2001 AMC 12 #6 and 2001 AMC 10 #13, so both problems redirect to this page.
Problem
A telephone number has the form , where each letter represents a different digit. The digits in each part of the number are in decreasing order; that is, , , and . Furthermore, , , and are consecutive even digits; , , , and are consecutive odd digits; and . Find .
Solution
We start by noting that there are letters, meaning there are digits in total. Listing them all out, we have . Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.
Case 1: , , , and are , , , and respectively.
A cursory glance allows us to deduce that the smallest possible sum of is when , , and are , , and respectively, so this is out of the question.
Case 2: , , , and are , , , and respectively.
A cursory glance allows us to deduce the answer. Clearly, when , , and are , , and respectively, is when , , and are , , and respectively, giving us a final answer of
Solution 2
The ten letters must all correspond to ten distinct digits, so every digit is used in the telephone number.
We note that , , , , are the odd numbers, and possible sequences for are either , , , or , , , .
, , and are included in both possible sequences so that we can rule out the possibilities of and for , so can only be , , or . Since every possible sequence of DEF also contains , we can rule out as well.
Testing A= means , , is , , , respectively. However, and must be part of , , or , and they already sum to more than , so this leaves as our answer.
~megaboy6679
Video Solution by Daily Dose of Math
https://youtu.be/z7o_BiWLDlk?si=9aZ0zIx2lkh_8CV3
~Thesmartgreekmathdude
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 |
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.