Difference between revisions of "2013 AIME II Problems/Problem 1"
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Suppose that the measurement of time during the day is converted to the metric system so that each day has <math>10</math> metric hours, and each metric hour has <math>100</math> metric minutes. Digital clocks would then be produced that would read <math>\text{9:99}</math> just before midnight, <math>\text{0:00}</math> at midnight, <math>\text{1:25}</math> at the former <math>\text{3:00}</math> AM, and <math>\text{7:50}</math> at the former <math>\text{6:00}</math> PM. After the conversion, a person who wanted to wake up at the equivalent of the former <math>\text{6:36}</math> AM would set his new digital alarm clock for <math>\text{A:BC}</math>, where <math>\text{A}</math>, <math>\text{B}</math>, and <math>\text{C}</math> are digits. Find <math>100\text{A}+10\text{B}+\text{C}</math>. | Suppose that the measurement of time during the day is converted to the metric system so that each day has <math>10</math> metric hours, and each metric hour has <math>100</math> metric minutes. Digital clocks would then be produced that would read <math>\text{9:99}</math> just before midnight, <math>\text{0:00}</math> at midnight, <math>\text{1:25}</math> at the former <math>\text{3:00}</math> AM, and <math>\text{7:50}</math> at the former <math>\text{6:00}</math> PM. After the conversion, a person who wanted to wake up at the equivalent of the former <math>\text{6:36}</math> AM would set his new digital alarm clock for <math>\text{A:BC}</math>, where <math>\text{A}</math>, <math>\text{B}</math>, and <math>\text{C}</math> are digits. Find <math>100\text{A}+10\text{B}+\text{C}</math>. | ||
==Solution== | ==Solution== | ||
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> | 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> | ||
Revision as of 17:00, 4 April 2013
Suppose that the measurement of time during the day is converted to the metric system so that each day has 
metric hours, and each metric hour has 
metric minutes. Digital clocks would then be produced that would read 
just before midnight, 
at midnight, 
at the former 
AM, and 
at the former 
PM. After the conversion, a person who wanted to wake up at the equivalent of the former 
AM would set his new digital alarm clock for 
, where 
, 
, and 
are digits. Find 
.
Solution
There are 
normal minutes in a day , and 
metric minutes in a day. The ratio of normal to metric minutes in a day is 
, which simplifies to 
. This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to AM, 
normal minutes pass. This can be viewed as 
cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding 
to gives 
, so the answer is 
