Difference between revisions of "2001 AMC 12 Problems/Problem 6"
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The last four digits <math>\text{GHIJ}</math> are either <math>9753</math> or <math>7531</math>, and the other | The last four digits <math>\text{GHIJ}</math> are either <math>9753</math> or <math>7531</math>, and the other | ||
odd digit (<math>1</math> or <math>9</math>) must be <math>A</math>, <math>B</math>, or <math>C</math>. Since <math>A + B + C = 9</math>, that digit must be <math>1</math>. | odd digit (<math>1</math> or <math>9</math>) must be <math>A</math>, <math>B</math>, or <math>C</math>. Since <math>A + B + C = 9</math>, that digit must be <math>1</math>. | ||
− | Thus the sum of the two even digits in <math>\text{ABC}</math> is <math>8</math>. <math>\text{DEF}</math> must be <math>864</math>, <math>642</math>, or <math>420</math>, which respectively leave the pairs <math>2</math> and <math>0</math>, <math>8</math> and <math>0</math>, or <math>8</math> and <math>6</math>, as the two even digits in <math>\text{ABC}</math>. Only <math>8</math> and <math>0</math> has sum <math>8</math>, so <math>\text{ABC}</math> is <math>810</math>, and the required first digit | + | Thus the sum of the two even digits in <math>\text{ABC}</math> is <math>8</math>. <math>\text{DEF}</math> must be <math>864</math>, <math>642</math>, or <math>420</math>, which respectively leave the pairs <math>2</math> and <math>0</math>, <math>8</math> and <math>0</math>, or <math>8</math> and <math>6</math>, as the two even digits in <math>\text{ABC}</math>. Only <math>8</math> and <math>0</math> has sum <math>8</math>, so <math>\text{ABC}</math> is <math>810</math>, and the required first digit is <math>\boxed{(\text{E})8}</math>. |
== See Also == | == See Also == |
Revision as of 20:55, 16 November 2013
- 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
The last four digits are either
or
, and the other
odd digit (
or
) must be
,
, or
. Since
, that digit must be
.
Thus the sum of the two even digits in
is
.
must be
,
, or
, which respectively leave the pairs
and
,
and
, or
and
, as the two even digits in
. Only
and
has sum
, so
is
, and the required first digit is
.
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.