Difference between revisions of "2014 AMC 10B Problems/Problem 14"
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<math> \textbf {(A) } 26 \qquad \textbf {(B) } 27 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 37 \qquad \textbf {(E) } 41</math> | <math> \textbf {(A) } 26 \qquad \textbf {(B) } 27 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 37 \qquad \textbf {(E) } 41</math> | ||
− | ==Solution== | + | ==Solution 1== |
+ | Let <math>h</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 15</math>, then <math>\text{min}\{c\}\ge\frac{9+5(15)}{9}>9</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>14</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>a=1</math>, so <math>b=0</math>, and <math>c=6</math>. Thus <math>a^2+b^2+c^2=1+0+36=\fbox{37 \textbf{(D)}}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Danica drives <math>m</math> miles, such that <math>m>0</math> and <math>m</math> is a multiple of 55. Therefore, <math>m</math> must have an units digit of either <math>0</math> or <math>5.</math> If the units digit of <math>m</math> is <math>0,</math> then <math>a=c</math> which would imply that Danica did not drive at all. Thus, <math>c>a.</math> Therefore, <math>|a-c|=5,</math> and because <math>a+b+c\leq7, c>a,</math> we have <math>(a,c)=(1,6).</math> Finally, <math>b</math> then must be <math>0</math> due to <math>a+b+c\leq 7,</math> and <math>a^2+b^2+c^2=1^2+0^2+6^2=\fbox{\textbf{(D) }37}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We can set up an algebraic equation for this problem. | ||
+ | |||
+ | From what's given, we have that <math>100c+10b+a=55x+100a+10b+c</math> | ||
+ | |||
+ | This simplifies to be <math>0=55x+99a-99c\implies -55x=99a-99c</math> | ||
+ | |||
+ | Factoring, we get that <math>-55x=99(a-c)\implies x=-\frac{9(a-c)}{5}</math> | ||
+ | |||
+ | Hence, notice that we want <math>a-c=-5</math> so that <math>x=9</math> | ||
+ | |||
+ | The only pair that works for this problem that satisfies the original requirements is <math>(1,6)</math> | ||
+ | |||
+ | Hence, <math>a=1, b=0, c=6</math> | ||
+ | |||
+ | Checking, we have that <math>106+55(9)=601\implies 601=601</math> | ||
+ | |||
+ | Hence, the answer is <math>1^2+0^2+6^2=37\implies\boxed{D}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/C0erYBsw5KI | ||
+ | |||
+ | ~savannahsolver | ||
==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}} |
Latest revision as of 17:20, 30 January 2021
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 1
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
Solution 2
Danica drives miles, such that
and
is a multiple of 55. Therefore,
must have an units digit of either
or
If the units digit of
is
then
which would imply that Danica did not drive at all. Thus,
Therefore,
and because
we have
Finally,
then must be
due to
and
Solution 3
We can set up an algebraic equation for this problem.
From what's given, we have that
This simplifies to be
Factoring, we get that
Hence, notice that we want so that
The only pair that works for this problem that satisfies the original requirements is
Hence,
Checking, we have that
Hence, the answer is
Video Solution
~savannahsolver
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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