Difference between revisions of "2001 AMC 12 Problems/Problem 6"
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== Solution == | == 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. | + | 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. | + | 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. | ||
− | A cursory glance allows us to deduce | + | 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> | ||
− | + | ==Video Solution by Daily Dose of Math== | |
+ | https://youtu.be/z7o_BiWLDlk?si=9aZ0zIx2lkh_8CV3 | ||
− | + | ~Thesmartgreekmathdude | |
== See Also == | == See Also == |
Latest revision as of 21:43, 15 July 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
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.