Solution:

Charley Smallish and his best girl could trolley three times as fast as they could walk. Therefore, three-quarters of their outing time was spent in walking, and but one-quarter in riding, since they had to walk back.

They rode for two hours, going 18 miles, and walked back in six hours, thus consuming their eight hours.

Solution:

Starting with 1 marble in a bag, each bag contains double the number of marbles as the previous bag.
ie 1,2,4,8,16,32,64,128,256,512.

That way whatever number of marbles I would ask for you would just add up the boxes to equal the number asked.

Solution:

1) 8/(3-(8/3))
= 8/(1/3)
= 24

2)((8 x 3!)/3)+8
= ((8 × 3 × 2 × 1)/3)+8
= (48/3)+8
= (16)+8
= 24

3)(3!/)*8

4)(8-3)!/(8-3)
( × )!
( + )!
?(8×8×3×3)
8+(8×(3!/3))
((?(8+8) × (3/3))!
?(8+8) × (3+3)
(log base(3!/3) of 8) × 8
((log base(3!/3) of (8+8))!

Solution:

The complete addition looks as follows:

  167161 +
  664711 +
  162166 +
  164611
-------------
1158649

Solution:

9376

Looking at the last digit, the last digit must be either 0, 1, 5 or 6.
Then looking at the last two digits, the last two digits must be either 00, 01, 25 or 76.
Then looking at the last three digits, the last three digits must be either 000, 001, 625 or 376.
Then looking at the last four digits, the last four digits must be either 0000, 0001, 0625 or 9376.
Out of those, only 9376 is a 4 digit number

Solution:

33 seconds

The train takes 30 seconds to travel 1 mile, plus 3 seconds for the complete train to pass any point, making a total of 33 seconds.

Solution:

This is simply the numbers 2^0,2^1,2^2 ... that is 1,2,4,8,16... So for making 1000 kg we need up to 1, 2, 4, 8, 16, 32, 64, 128, and 512

Solution:

143547

Solution:

18 x 12 ÷ 2 + 3 = 111

Solution:

We can calculate out how many days there will be for the 13th on each month if we count from the beginning of the year (January 1). Then we divide total days by 7 to get the remainders. We also need to consider the leap year. Through the whole year we had all kinds of remainders, from 0 to 6. The minimum of occurence for all the unique remainders was 1. It means that we have at least one Friday on 13th. In a regular year, the best chance you can get 3 Fridays on 13th, which are in February, March and December because the remainders of these 3 months are 2. In a leap year, the best chance you also can get 3 Fridays on 13th, which are in January, April and July because the remainders of these 3 months are 6.