Help needed to solve math problems

[lukasz324]

1) Last-but-one: If all the integers from 1 up to n are multiplied together, this is called the factorial of n, and is written as n!. For example, 2!= 2x1=2, and 4!=4x3x2x1=24. This means 4! has last-but-one digit 2. What is the last-but-one digit of 99!?

2) Adjacent factors: Peter wishes to write down a list of different positive integers less than or equal to 10 in such a way that for each pair of adjacent numbers one of the numbers is divisible by the other. What is the length of the longest list that Peter could write down?

3) Prime order: How many of the three digit numbers that can be made from all of the digits 1, 3 and 5 (used only once each) are prime?

4) Divisible by its digit sum: The year 2010 is one in which the sum of the digits is a factor of the year itself. What is the next year that has the same property?

Thank you for help

[shyamjayakannan]

1) All factorials greater than 10! will be a multiple of 100 and so their last 2 digits will always be 00. So, the answer is 0.

2) The longest list is of length 9. Ex: 9 3 6 1 5 10 2 8 4

3) 0 because every 3 digit number made from 1,3 and 5 without repetition is divisible by 3 (1+3+5 is divisible by 3)

4) 2013