Arithmetical Progressions

I wish to familiarise the topic Arithmetical Progressions
Any sequence of numbers starting with a particular number and progressively increasing with equal additions can be stated to be forming such progressions.
The start number can be called the first term-generally denoted by 'a'
The additions in equal values can be called the 'common difference-generally denoted by 'd'
'a' can be anything like 5,    100      ,3/7    ,4.05,    6^2   etc
'd' can also be similarly anything-can also have negative values.
A few examples shown below-
100,103,106,109,112......
(4.5)   .(5.2)    (5.9)    (6.6)....
7/45......9/45    11/45...13/45....15/45    17/45
1000,992,984,976,968...  etc
Given the first term  like 32 and common difference like 5,we have methods to find out the 16th or 45th etc terms.Also the sums of the various terms upto the 16th term or 45th term.
I will explain the methods in my next blog.

Remainders

I am posing a question for you to solve.
I want you to find the smallest number which gives a remainder 2 when divided by 3,a remainder 3 when divided by 5 and a remainder 5 when divided by 7., 
How can you find more such numbers?

Clock face-Times..Rejoinder.

No one has yet given me a reply.The solution is based on the following facts-
After 12 noon,the hands first merge a little after 01-05
Next a little after 02-10...Next a little after 03-15 etcetc.
In fact,they will be together again several times and for the 11th time they will be at 12 midnight.
Thus in 12 hours,11 intervals occur
Each interval will be {1 +(1/11)}hours or 1 hour and {5+(5/11)} minutes.
The times of merger are thus {5+(5/11)} minutes after 1 PM...{10+(10/11)}minutes after 2PM,{16+(4/11}minutes after 3PM..,{21+(9/11)}minutes after 4PM  etc etc .

Forming fractions

Suppose we have 2 fractions like 2/7 and 4/9.
We may like to find more fractions with values between these two.There is a method for this.In general if a/b and c/d are 2 fractions,then the fraction (a+c)/(b+d) will have values between those two.You can thus find as many fractions as you want between 2/7 and 4/9.
I have shown some of them below.
Let A=2/7 and B=4/9.The first fraction in between C=(2+4)/(7+9)=6/16=3/8
Next D between 2/7 and 3/8=5/15=1/3.Next E between 3/8 and 4/9=7/17and so on.The following is a more detailed list-
2/7,5/17,3/10,4/13,1/3,5/14,4/11,7/19,3/8,5/13,2/5,9/22,7/17,18/43,11/26,3/7,4/9.You can expand the list to whatever length you wish.

Clock face-times when the hour hand and minute hand merge

This is a problem for you.
Both the hands on a clock face merge at 12noon or midnight.
After 12noon,they again merge a little later than  01-05 and again a little later than 02-10.
Can you find the actual times in each case?
Also subsequent stages.

Number 153

I am now introducing a new black hole number 153.
Take any number divisible by 3 like 423.
Find the cubes of digits in 423 and add.You get 99.
Repeat the same process with 99.You get 1458.
Repeat the processes again.
1458 gives you 702...702 gives you 351...351 gives you 153..and finally 153 gives you again 153.
The same result will be arrived at with any number divisible by 3
Second example   63.The repeat processes give you 243,99 and again as in previous case 153.
Third example 168.The results are 729,378,882,1032,36 and again 153 like for 63
You can consider even larger numbers divisible by 3 also.More repetitions may be required.

Recurring decimals

You must have seen decimals with some digits occurring in them like those shown below-
0.23232323.....
0.236236236236....
0.238383838...
0.2538383838...
The first two examples show all digits recurring while the next 2 show only some of latter digits recurring.
I now wish to show how you can convert these into ordinary fractions.
The first will be 23/99.The second will be 236/999
The third will be 2/10+38/990 which will get reduced to 118/495
The fourth will be 25/100+38/9900 which will get reduced to 2513/9900.
Try understanding the process.
Try working out 0.437878787878...etc