Difference between revisions of "2014 AMC 10B Problems/Problem 14"
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==Solution== | ==Solution== | ||
+ | Let <math>h\in\mathbb{N}</math> be the number of hours Danica drove. Note that <math>abc</math> can be expressed as <math>100\cdot a+10\cdot b+c</math>. From the given information, we have <math>100a+10b+c+55h=100c+10b+a</math>. This can be simplified into <math>99a+55h=99c</math> by subtraction, which can further be simplified into <math>9a+5h=9c</math> by dividing both sides by <math>11</math>. Thus we must have <math>h\equiv0\pmod9</math>. However, if <math>h\ge 16</math>, then <math>\text{min}\{c\}\ge\frac{9+5(16)}{9}\ge9</math>, which is impossible since <math>c</math> must be a digit. The only value of <math>h</math> that is divisible by <math>9</math> and less than or equal to <math>15</math> is <math>h=9</math>. | ||
+ | |||
+ | From this information, <math>9a+5(9)=9c\Rightarrow a+5=c</math>. Combining this with the inequalities <math>a+b+c\le7</math> and <math>a\ge1</math>, we have <math>a+b+a+5\le7\Rightarrow 2a+b\le2</math>, which implies <math>1\le a\le1</math>, so <math>a=1</math>, <math>b=0</math>, and <math>c=6</math>. Thus $a^2+b^2+c^2=1+0+36=\boxed{37 \textbf {(D) }} | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2014|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:44, 20 February 2014
Problem
Danica drove her new car on a trip for a whole number of hours, averaging miles per hour. At the beginning of the trip,
miles was displayed on the odometer, where
is a
-digit number with
and
. At the end of the trip, the odometer showed
miles. What is
?
Solution
Let be the number of hours Danica drove. Note that
can be expressed as
. From the given information, we have
. This can be simplified into
by subtraction, which can further be simplified into
by dividing both sides by
. Thus we must have
. However, if
, then
, which is impossible since
must be a digit. The only value of
that is divisible by
and less than or equal to
is
.
From this information, . Combining this with the inequalities
and
, we have
, which implies
, so
,
, and
. Thus $a^2+b^2+c^2=1+0+36=\boxed{37 \textbf {(D) }}
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
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