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1) The product of n counting numbers is 2010. The sum of these counting number is also 2010. What is the greatest possible value of n?

2) Alex from Ayden and Bob from Beaufort are driving towards each other at the same time. They meet at a first point 32km away from Ayden. After Alex arrives at Beaufort and Bob arrives ay Ayden, they turn around right away. They meet at a second point 64km away from Ayden. Find the distance from Ayden to Beaufort.

 Jun 10, 2019
 #1
avatar+105488 
+4

2) Alex from Ayden and Bob from Beaufort are driving towards each other at the same time. They meet at a first point 32km away from Ayden. After Alex arrives at Beaufort and Bob arrives ay Ayden, they turn around right away. They meet at a second point 64km away from Ayden. Find the distance from Ayden to Beaufort.

 

Let the distance be D

Let Alex's Rate = A

Let Bob's Rate = B

Distance / Rate = Time

 

When they meet the first time  Alex has traveld 32 km  and Bob has traveled (D - 32) km

 

So.....from the start until they meet the first time we have that the times are equal

 

32/ A  =  (D - 32) / B    ⇒   A/B  = 32 / (D - 32)

 

 

And after the first meeting.....Alex has traveled (D - 32 + D -64)  = (2D - 96)

And Bob has traveled (32 + 64) = 96

 

So......equating the time again

 

(2D -96) /A  = (96) /B ⇒   A/B  = (2D - 96) / 96

 

And A/B  =A/B   so

 

32 / (D -32)  = (2D -96) / 96     cross-multiply

 

96 * 32  = (2D - 96) ( D - 32)     simplify

 

96*32  = 2D^2 - 160D + 96*32

 

2D^2 - 160D = 0

 

2D ( D - 80)  = 0     so

 

2D = 0                                 D - 80  = 0  

D = 0   reject                       D = 80km

 

 

cool cool cool

 Jun 10, 2019
 #2
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Another way:
Cphill: Does this question remind you of a famous puzzle by Sam Lloyd about a river's width?
Let the total distance between A and B =D
Alex drove 32 km                Bob drove D - 32
When they met the 2nd time they had already covered 3 times the distance between the 2 towns. Let us take Bob's first distance, or D - 32.
3 x (D - 32) - 64(the distance covered when they met the 2nd time) =D - the distance between the 2 towns.Or:
3D -96 -64 =D
2D =160
D = 80 km - the distance between A and B.

 Jun 11, 2019
 #3
avatar+105488 
+3

THX, guest....

 

It's EXACTLY the same problem as Sam Lloyd's  !!!!!

 

[ I noticed that, too ..... ]

 

cool cool laugh

CPhill  Jun 11, 2019
edited by CPhill  Jun 11, 2019
 #4
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1) The product of n counting numbers is 2010. The sum of these counting number is also 2010. What is the greatest possible value of n?

 

Your question doesn't seem to make sense:

2010 = 2 * 3 * 5 * 67

The SUM of the above 4 prime factors CANNOT = 2010.

However, 2010 has 16 DIVISORS =1 | 2 | 3 | 5 | 6 | 10 | 15 | 30 | 67 | 134 | 201 | 335 | 402 | 670 | 1005 | 2010 .

You can add up 3 of them to make 2010 =1005 + 670 + 335 =2010.

 Jun 11, 2019
 #5
avatar+18 
+3

It does make sense since we could add a bunch of ones and the product would still be the same but the sum would go up.

hailereco  Jun 16, 2019

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