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Two trains travel directly toward each other. One of the trains travels at a rate of 12 km/h while the other travels at a rate of 20 km/h. When the trains are 72 km apart a conductor at the front of one of the trains releases the insane pigeon Hyde. Hyde flies first from the slower of the two trains to the faster train at which point Hyde doubles back toward the slower train. Hyde continues to fly back and forth between the trains as they approach, always at a constant speed of 48 km/h. Assuming the trains never change speed until they meet and magically stop, how many kilometers has Hyde flown when the trains meet?
The terms of a particular sequence are determined according to the following rules:
* If the value of a given term is an odd positive integer s, then the value of the following term is 3s - 9
* If the value of a given term is an even positive integer t, then the value of the following term is 2t - 7.
Suppose that the terms of the sequence alternate between two positive integers (a,b,a,b,...). What is the sum of the two positive integers?
Given positive integers
and such that and , what is the smallest possible positive value for ?Help would be very appreciated. Thanks!
Last edited by mathster (2015-10-18 08:04:47)
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Hi;
Given positive integers
and such that and , what is the smallest possible positive value for ?
I am getting 49 as the smallest.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi mathster,
Q1. The trains are approaching at a relative speed of (12 + 20) Kph and the distance travelled is 72Km. So calculate the time until they meet and apply this to the pigeon's travel.
Q2. Suppose that a is odd. Then
b = 3a-9 which will be even and so
a = 2b-7
You can solve for a and b from this.
I'll leave you to consider the case where a is even.
Q3. I did this:
A sketch http://www.mathsisfun.com/data/function … 12/(x-12)) shows that after x = 12 the graph has a minimum somewhere between x = 20 and 30. You can use calculus to get the exact point.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks so much guys! Question 3 was quite confusing, although I don't know calculus, apparently you can solve this with SFFT (Simon's Favorite Factoring Trick). Quite clever.
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hi mathster,
"Simon's Favorite Factoring Trick". I've not met this before. Looked up on AoPS but still none the wizer.
Please would you post how you used it.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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hi mathster,
"Simon's Favorite Factoring Trick". I've not met this before. Looked up on AoPS but still none the wizer.
Please would you post how you used it.
Bob
Multiplying
by gives .Using the trick, it's factorable to
.We would probably want to set
and to , but we remember that .The next closest factors are
and .This gives
and . Therefore and . Finally we come to .Last edited by mathster (2015-10-19 04:14:11)
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This is an Egyptian fraction so even Simon is not needed here.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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How do you finish question number 1?
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hi Wangotango,
Two trains travel directly toward each other. One of the trains travels at a rate of 12 km/h while the other travels at a rate of 20 km/h. When the trains are 72 km apart a conductor at the front of one of the trains releases the insane pigeon Hyde. Hyde flies first from the slower of the two trains to the faster train at which point Hyde doubles back toward the slower train. Hyde continues to fly back and forth between the trains as they approach, always at a constant speed of 48 km/h. Assuming the trains never change speed until they meet and magically stop, how many kilometers has Hyde flown when the trains meet?
From the viewpoint of one of the trains, the other is 72Km away and approaching at (12+22) Km/h. So using time = distance/speed you can calculate the time until the trains meet. That is also the time that Hyde flies, so use distance = speed x time to calculate the distance Hyde has travelled.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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(2) a+b=11
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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