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Difference between revisions of "2013 AIME II Problems/Problem 1"

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==Solution==
 
==Solution==
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There are <math>24*60=1440</math> normal minutes in a day , and <math>10*100=1000</math> metric minutes in a day. The ratio of normal to metric minutes in a day is <math>\frac{1440}{1000}</math>, which simplifies to <math>\frac{36}{25}</math>. This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to <math>\text{6:36}</math> AM, <math>6*60+36=396</math> normal minutes pass. This can be viewed as <math>\frac{396}{36}=11</math> cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding <math>25*11=275</math> to <math>\text{0:00}</math> gives <math>\text{2:75}</math>, so the answer is <math>\boxed{275}</math>
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There are <math>24*60=1440</math> normal minutes in a day , and <math>10*100=1000</math> metric minutes in a day. The ratio of normal to metric minutes in a day is <math>\frac{1440}{1000}</math>, which simplifies to <math>\frac{36}{25}</math>. This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to <math>\text{6:36}</math> AM, <math>6*60+36=396</math> normal minutes pass. This can be viewed as <math>\frac{396}{36}=11</math> cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding <math>25*11=275</math> to <math>\text{0:00}</math> gives <math>\text{2:75}</math>, so the answer is <math>\boxed{275}</math>.
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== See also ==
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{{AIME box|year=2013|n=II|before=First Problem|num-a=2}}

Revision as of 16:34, 6 April 2013

Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$, where $\text{A}$, $\text{B}$, and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$.

Solution

There are $24*60=1440$ normal minutes in a day , and $10*100=1000$ metric minutes in a day. The ratio of normal to metric minutes in a day is $\frac{1440}{1000}$, which simplifies to $\frac{36}{25}$. This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to $\text{6:36}$ AM, $6*60+36=396$ normal minutes pass. This can be viewed as $\frac{396}{36}=11$ cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding $25*11=275$ to $\text{0:00}$ gives $\text{2:75}$, so the answer is $\boxed{275}$.

See also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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