Difference between revisions of "2014 AMC 12B Problems/Problem 10"
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Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, <math>abc</math> miles was displayed on the odometer, where <math>abc</math> is a 3-digit number with <math>a \geq{1}</math> and <math>a+b+c \leq{7}</math>. At the end of the trip, the odometer showed <math>cba</math> miles. What is <math>a^2+b^2+c^2?</math>. | Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, <math>abc</math> miles was displayed on the odometer, where <math>abc</math> is a 3-digit number with <math>a \geq{1}</math> and <math>a+b+c \leq{7}</math>. At the end of the trip, the odometer showed <math>cba</math> miles. What is <math>a^2+b^2+c^2?</math>. | ||
− | <math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D) | + | <math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41 </math> |
− | ==Solution== | + | ==Solution 1== |
We know that the number of miles she drove is divisible by <math>5</math>, so <math>a</math> and <math>c</math> must either be the equal or differ by <math>5</math>. We can quickly conclude that the former is impossible, so <math>a</math> and <math>c</math> must be <math>5</math> apart. Because we know that <math>c > a</math> and <math>a + c \le 7</math> and <math>a \ge 1</math>, we find that the only possible values for <math>a</math> and <math>c</math> are <math>1</math> and <math>6</math>, respectively. Because <math>a + b + c \le 7</math>, <math>b = 0</math>. Therefore, we have | We know that the number of miles she drove is divisible by <math>5</math>, so <math>a</math> and <math>c</math> must either be the equal or differ by <math>5</math>. We can quickly conclude that the former is impossible, so <math>a</math> and <math>c</math> must be <math>5</math> apart. Because we know that <math>c > a</math> and <math>a + c \le 7</math> and <math>a \ge 1</math>, we find that the only possible values for <math>a</math> and <math>c</math> are <math>1</math> and <math>6</math>, respectively. Because <math>a + b + c \le 7</math>, <math>b = 0</math>. Therefore, we have | ||
<cmath>a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}</cmath> | <cmath>a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let the number of hours Danica drove be <math>k</math>. Then we know that <math>100a + 10b + c + 55k</math> = <math>100c + 10b + a</math>. Simplifying, we have <math>99c - 99a = 55k</math>, or <math>9c - 9a = 5k</math>. Thus, k is divisible by <math>9</math>. Because <math>55 * 18 = 990</math>, <math>k</math> must be <math>9</math>, and therefore <math>c - a = 5</math>. Because <math>a + b + c \leq{7}</math> and <math>a \geq{1}</math>, <math>a = 1</math>, <math>c = 6</math> and <math>b = 0</math>, and our answer is <math>a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37</math>, or <math>\boxed{D}</math>. | ||
+ | |||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:46, 17 January 2021
Contents
Problem
Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, miles was displayed on the odometer, where
is a 3-digit number with
and
. At the end of the trip, the odometer showed
miles. What is
.
Solution 1
We know that the number of miles she drove is divisible by , so
and
must either be the equal or differ by
. We can quickly conclude that the former is impossible, so
and
must be
apart. Because we know that
and
and
, we find that the only possible values for
and
are
and
, respectively. Because
,
. Therefore, we have
Solution 2
Let the number of hours Danica drove be . Then we know that
=
. Simplifying, we have
, or
. Thus, k is divisible by
. Because
,
must be
, and therefore
. Because
and
,
,
and
, and our answer is
, or
.
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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.