Multiplication trick

Let me introduce you all to a different way of multiplication.
Suppose you want to multiply 401 with 25
Enter them in 2 columns -the smaller number 25 preferably in the first column
In the first column write down one by one values after dividing by 2(If there is a remainder ignore it ).
In the second column write down one by one values after multiplication by 2 ,corresponding to each entry in the first column.
Select values in first column which are odd numbers and write down the corresponding entries in the second column.Add these entries of the second column.The total  will be the required product.-See below the working-
25-------401
12...........802
6............1604
3............3208
1............6416
The odd numbers in the first column are 25,3and 1.The corresponding entries in the second column are 401,3208 and 6416.Add these you get 10025 which is the product of 25 and 401.
Another Example-27 to multiply with 48
27...........48
13...........96
6............192
3............384
1............768..The required product   48+96+384+768=1296.
You can also enter the bigger number in the first column but this will only result in more entries in each column..

Another general method-divisibility tests

I am now introducing details another general method for divisibility tests.
Let me show the way for conducting a test for divisibility by 19.
Number to be tested-4783915
Break this into 2 parts-478391 and 5
Multiply the second part by 2 and add to first part-478391+10=478401
Repeat the procedure-47840+2=47842
Repeat again-4784+4=4788
Repeat 478+16=494-
Test the result for divisibility.--494 is divisible by 19 and so is therefore the original start number 4783915
Another Example-Number to be tested-15857799
1585779+18=1585797....158579+14=158593...15859+6=15865..1586+10=1596
159+12=171...171 is divisible by 19 and so is original start number 15857799
Similar  method can be adopted for conducting test for divisibility by other numbers like 17,23,29 ,79,89 etc
However the number to be used for multiplying the second part in each case is different-they are shown below-
For 17 it is 12
For 23 it is 7
For 29 it is 3
For 79 it is 8
For 89 it is 9
You will notice that these are the same numbers which can be worked out based on my blog Multipliers-Rejoinder posted a couple of days back.-though with reference to cyclic numbers.
Try numbers 159327944.....,65482725,....45327029331.-for testing divisibility by 17 

Multiplers-Rejoinder

This has reference to my previous blog posted some hours ago today.
I was worried the process mentioned would be difficult to understand.
There are two ways of looking at the issue of movement of digits in the cyclic number 142857.
Let us consider the movement of the first digit to the right now(instead of the last digit to the front as considered earlier).Thus we now have 42857l,285714,857142,571428 and 714285.The multipliers required now are 3,2,6,4 and 5(These you can see are the same multipliers considered in the reverse order)
This new set of multipliers are more easier to find.
They are all the successive remainders when you take the reciprocal of 7
While taking the reciprocal of 7 we first add zero to 1 getting us 10 and divide the same by 7.The remainder is 3 which is the first multiplier now.Next we add zero to 3 getting us 30 and divide the same by 7.The remainder is 2 which is the second multiplier now.Next we add zero to 2 getting us 20 and divide the same by 7.The remainder  is 6 which is  the third multiplier now.The process can be continued to get the next multipliers viz 4 and 5.
You have to understand the necessity of the 2 processes viz.the one in the previous blog and the one now shown.
This will be clear when you consider a cyclic number of too many digits
For instance the cyclic number for the prime 97 has 96 digits.The first 10 of them are 0103092783 and the last 10 of them are 0206185567.If you want to know what multiplier will make the last digit 7 go to the front you have to perform the difficult process of finding the reciprocal right till the end. as per this blog.But if you adopt the process in the previous blog you will see that the value  of 'f' is 68 and this multiplier will make the last digit 7 to go to the front.
You may choose in every case whichever process you consider easier and best. 

Multipliers for cyclic numbers

In a previous blog I had mentioned that by using the multipliers 5,4,6,2,and 3 on the cyclic number 142857,one can see the last digit at right moving successively to the front viz producing the results 714285,571428,857142,285714 and 428571,the other digits remaining in the same order in each case.
The multipliers 5,4,6,2 and 3 can be found as follows-
I had mentioned that the cyclic number 142857 was found by taking the reciprocal of prime number 7.We can call 7 as the 'generator'(g) of the cyclic number.
The first multiplier (f) was shown by me as 5
The relation between 'f' and 'g' is that (10f-1) should be a multiple of 'g'
In the case of g=7,we have f=5 as 50-1=49 is a multiple of 7.
For the subsequent multipliers you have to repeatedly multiply 5 by itself and  find modulus 7 of the result.
(I now have to explain the term modulus.The modulus of any number say 45 with  reference to another number say 7 is the remainder when 45 is divided by 7.In this case the remainder is 3 and so modulus 7 of 45 is 3)
For our case we have 5 multiplied by 5 gives us 25 and modulus 7 of 25 is 4 which is the second multiplier.
Next 4 multiplied by 5 gives us 20 and modulus 7 of 20 is 6,which is the third multiplier.
Next 6 multiplied by 5 gives us 30 and modulus 7 of 30 is 2 which is the fourth multiplier.
Next 2 multiplied by 5 gives us 10 and modulus 7 of 10 is 3 which is the fifth multiplier.
The process stops here as 3 multiplied by 5 gives us 15 and the modulus 7of 15 is 1.
You  may find the processes a bit difficult to understand but read again you will understand the same .
Now let me give you  the problem -whether you will be able to work out how the multipliers 4,3,12,9 and 10 can be found for the cyclic number 076923  relating to g=13(The cyclic number 076923 can be found by taking the reciprocal of 13).