Harshad numbers

A number which is divisible by the sum of its digits has been given the name as  a Harshad number.Examples are given below-
153 divisible by 1+5+3=9 giving the result 17
133 divisible by 1+3+3=7 giving the result 19.
There are 23 numbers below 100 and 26 numbers between 100 and 199
Can  you locate them?
There are also multiple Harshad numbers.An example is 6804.It is called a multiple Harshad no for the reason given below-
6804 divided by the sum of its digits 18 gives the result 378.
378 divided by the sum of its digits 18 gives the result 21.
21 divided by the sum of its digits 3 gives the result 7.
As process is repeated 3 times for 6804 it is called Grade 3 Multiple Harshad number.
There is a Grade 12 Multiple Harshad number which I will reveal if anyone is interested.

One more cyclic number

In the cyclic number 142857  the digits at right gradually shift to the left on multiplication by 5,4,6,2 and 3.viz producing 714285,571428,857142,285714 and 428571.This cyclic number is actually found by taking the reciprocal of the prime number 7.viz in the reciprocal the group of digits 1,4,2,8,5,7 repeat continuously..
You can find more cyclic numbers by finding similarly the reciprocals of more prime numbers or multiples of prime numbers.  
If you take the reciprocals of numbers having 2 digits or more you will face with one or more zeros at the beginning,like 076923 for 13.......,012345679 for 81.....00729927 for 137 etc and you should allow the zeros to remain while making the multiplications.
Now I am presenting a cyclic number with six zeros in the beginning.
The same is 00000048269001213 and this can be found by taking the reciprocal of the prime 2071723.
If you multiply this with the number 621517,you will find the last digit at right viz 3 moving to the front viz producing 30000004826900121.You can see that except for 3 all the other digits appear in the product in the same order.
I shall inform you later how this number 621517 used for multiplication has been found.I am in the process of finding more numbers which when used for multiplication allow the other digits at right to gradually move to the left.There is a method for finding them but because the numbers are large,the process is not easy and will take time.

Countdown

This is based on a TV programme shown by BBC and French TV.
You are given a set of numbers and you are required to arrive at a particular result using those numbers and the the usual mathematical signs relating to addition,subtraction,multiplication and division.Although the TV programmes had provided a small time period for working out the solution,I am not suggesting any period.Two examples below for you to try.
1) Numbers 6,1,25,8,5,8---result required 640
2) Numbers 4,7,9,25,6,4....result required 758
Except where there is already a repetition like 8 in example 1 and 4 in example 2,no repetition of numbers are allowed.  

Arrange digits

Here is a small puzzle to solve-
Arrange the digits 1,2,3,4,5,6,7,8,9 to get a fraction with equivalent value 1/3