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.

hailereco Jun 10, 2019

#1**+3 **

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

CPhill Jun 10, 2019

#2**+3 **

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.**

Guest Jun 11, 2019

#4**+1 **

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.

Guest Jun 11, 2019