Collecting numbers for particular sums-2

 This is  a rejoinder to my earlier blog with same title.
There has been no response although it is a simple problem of logic
For exnmple instead of collecting 6 numbers from group A,you can consider removing 2 numbers from that group. Same logic with group B
Following are the solutions
Group A-total 220- by removing 55 and 11....group B-total 110- by removing 38 and 49
Group A-total 228-by removing 17 and 41....group B-total 114-by removing  45 and 38
Group A-total 234-by removing 11 and 41....group B-total 117-by removing  31 and 49
Is it not simple?

Finding square numbers-2

This is a rejoinder to my blog same title
I suggested you find a square number starting with five 2s
A number of that type is 2222219285521 which is a square of 1490711
You will find the square of 1490712 is 2222222266944
Also the square of 1490713 is 2222225248369 

Prime pyramid-adjacent primes-2

This is with reference to my earlier blog-same title
I had shown row 8 as follows-1,6,7,4,3,2,5,8 where each pair of adjacent numbers add upto a prime viz 1+6=7,6+7=13,7+4=11,4+3=7,3+2=5,2+5=7,5+8=13
The further rows can be formed as follows-
a)As 8+9=17 (a prime) ,we can simply add 9 to get 1,6,7,4,3,2,5,8,9
b)As 9+10=19(a prime),we can simply add 10 to get 1,6,7,4,3,2,5,8 ,9,10
c)As 10+11=21 which is not a prime we have to modify numbers in row (b)
For this we follow the same position upto 3,add the nos from 2 to 10 in the reverse way and then add 11
We will get 1,6,7,4,3,(10,9,8,5,2),11
d)12 can be added as 11+12=23,giving us 1,6,7,4,3,10,9,8,5,2,11,12
e)13 cannot be added as 12+13=25 is not a prime.The change is made as follows-
1,6,7,(12,11,2,5,8,9,10,3,4),13
Further rows are filled up in the same fashion-
f) 1,6,7,12,11,2,5,8,9,10,(13,4,3),14
h)1,6,7,12,11,2,5,8,9,10,13,4,3,14,15
j)1,6,7,12,11,2,5,8,9,10,13,4,3,14,15,16
k)1,6,7,12,11,2,5,8,9,10,13,4,3,(16,15,14),17
l)1,6,7,12,11,2,(17,14,15,16,3,4,13,10,9,8,5),18
m)1,6,7,12,11,2,17,14,15,16,3,4,13,10,9,8,5,18,19
n) 1,6,7,12,11,2,17,14,15,16,3,4,13,10,(19,18,5,8,9),20
We can follow the same procedure for further rows.  

Division one number by another-probability-rejoinder

 This is with reference to my previous blog where we consider any number A between 10 and 1000-remove the last digit to get number B-We wanted the probability of B evenly dividing A.
Total number of choices possible for A----1000-9=991
Following give correct divisibility by B-
1)All numbers ending with zero (like 450)---------109
2)All multiples of 11 below 100 (like 22,33 etc)-----8
3) Six other cases-24,26,28,36,39,48---------------6
Total----------------------------------------------123
The probability is hence 123/991

Finding square numbers

Everyone knows how to extract the square root of any number
My proposal here is to find a number which has certain specialities but is also a perfect square number
For instance you are asked to find a number starting with six 2's and which is a perfect square.
With a little work of trial and error you can find it.
My research gave the number 222222674025 which is the square of 471405
But suppose I ask you to find a square number starting with five 2's.Try to do a little math and find the number.
Will it be smaller than 222222674025 shown by me earlier?

Collecting numbers for particular sums

Given below are 2 groups of numbers-
Group A-17,32,55,41,26,67,11,37
Group B-27,45,31,38,49,7
Collect 6 numbers from Group A for a sum S1 and 4 numbers from Group B for a sum S2,where S1/S2=2
You can find 3 solutions at least.Proceed

Prime pyramid-Adjacent primes

You are to work out a number pyramid starting with 1 number in the first row,2 in the second,3 in the third,4 in the fourth row and so on.Each row should start with digit 1 and end with the digit represented by the row,for example the 7th row should end with 7.The digits/numbers to be filled up in between should be such that each pair of adjacent numbers should sum up to a prime.A few rows are shown filled up below-
1-                       Sums involved  -nil
1,2                                             3
1,2,3                                          3,5
1,2,3,4                                       3,5,7
1,4,3,2,5                                    5,7,5,7
1,4,3,2,5,6                                 5,7,5,7,11
1,4,3,2,5,6,7                              5,7,5,7,11,13
1,6,7,4,3,2,5,8                           7,13,11,7,5,7,13
Try filling up subsequent rows.There could sometimes be more ways in respect the digits/numbers involved in the middle of each row.
It may not be easy when you proceed upto row 17,19 etc.

Surprise in sudoku-rejoinder

Were you able to fill up the sudoku grid shown by me?
The completed grid will be as shown below-(row-wise)
418 239 657
695 781 342
273 546 819
961 853 274
354 627 981
782 914 563
546 192 738
839 475 126
127 368 495
I have deliberately shown the digits in groups of three
There is a  surprise in rows 1,3,5,7 and 9 where you will notice an addition sum as follows-
418+239=657
273+546=819
354+627=981
546+192=738
127+368=495 

Probability for division of one number by another

Choose any number (A)between 10 and 1000.Remove the last digit in the same to get another number (B).For example if A is 456 ,B will be 45.What is the probability that A will be divisible by B?

A sudoku with a surprise

Here is the grid for filling up by you-should not be difficult-The cells which were left blank have been marked b-
bbbbb96bb
6b5b8bbb2
bbbb4bbb9
b6185bb7b
b5bbbbb8b
b8bb1456b
5bbb9bbbb
8bbb7b1b6
bb73bbbbb
Fill it up and find out what the surprise is. I cannot reveal it now as you will be able to fill up the grid more easily.

Sums of 2 squares reading to totals 'c' and '2c'

If a number 'c' is a sum of 2 squares of say 'x' and 'y',can '2c' be also expressed as a sum of 2 different squares?
The answer is yes.
You have to work out (x+y) and (x-y) ,find their squares and add up.
We already have c=x^2+y^2
We now have (x+y)^2+(x-y)^2= (x^2+2xy+y^2)+(x^2-2xy+y^2)=2(x^2+y^2)=2c
Example-Let x=17 and y=9.So c=289+81=370
We also have (x+y)=26 and (x-y)=8.Their squares are 676 and 64 giving a total of 740 which is the same as '2c'.
As a corollary if we have three numbers x,y and z and their squares add upto 'c',we can express '4c' as the sum of 6 squares viz of x+y..x-y...y+z...y-z....z+x and z-x.

Kaprekar numbers- 7 digits

There is no single no of 7 digits but a series of 8 numbers as follows-
8429652...7619733..8439552...7509843..9529641..8719722..8649432...7519743...8429652(rpt)
Few of them explained below-
a)7619733. Highest number 9776331...Lowest number.1336779..Difference 8439552
b)8719722 .Highest number 9877221..Lowest number  1227789..Difference 8649432
c)7519743  Highest number 9775431..Lowest number  1345779..Difference 8429652

Kaprekar numbers-6digits

In my previous post of 26 oct 2012,I had shown 2 series of Kaprekar numbers of 5 digits as follows-
74943,62964,71973,83952,74943(repeat)
61974,82962,75933,63954,61974(repeat)
To enable you to recollect,the process is explained below-
74943.-This consists of digits 3,4,4,7,9
Find the largest and smallest numbers with these and find the difference.You will get the next number in the series viz 62964 and so on.
viz 97443-34479=62964
Now I will show two 6 digit numbers and one series of 6 digit numbers-
1)Number  631764-the process results in the same number..766431-134667=631764
2)Number 549945-behaves like 631764
3) Series-851742,750843,840852,860832,862632,642654,420876,851742(repeat)

Walk the talk-2(Rejoinder)

This is with reference to my blog of 25th April,where I mentioned an imaginary conversation between me and a 10 digit number which explained itself about it's composition-viz the number of times each digit from 0 to 9 occurs in it-viz the first digit shows how many times zeros occur,the second digit how many times 1 occurs,the third digit how many times 2 occurs etc etc,right till the end showing how many times 9 occurs etc
The number is actually 6210001000.You can verify the composition as expressed by itself.

Divisibility tests for all prime numbers upto 100-prime 31

I had dealt with prime 23 in my previous blog.
I now deal with another prime 31,where there is a small difference in the procedure.
Here I choose the number P as 992 which has 31 as a factor.This is of the form (1000m-n) instead of the form (1000m+n) as in the case of prime 23 and hence the need for the change in procedure
No to be tested is the same viz 26105069.(This number was specifically chosen as it is a multiple of 31).
Step 1-We have  the same groups of 3 nos-A 069,B 105 and C 26.
We also now have m=1 and n=8
Step 2 Division by 'm' and multiplication by 'n' of C-
26/1=26.....26*8=208
Step 3-The change in procedure occurs here.We now have to add this result to B instead of deducting as in the case of prime 23.
105+208=313.
Step 4 Process repeated   viz 313/1=313....313*8=2504
Step 5 Result is added to A  viz 069+2504=2573
Step 6-The result in step 5 is the final result to be tested for divisibility by 31.
As 2573 is divisible by 31....(2573=31*83) we conclude the number under test 26105069 is divisible by 31.Actually the quotient is 842099.
In all cases where P is of the form (1000m-n) the change in procedure in step 3 viz addition as done above is to be followed.

Divisibility tests for all primes upto 100

I have devised another general method for testing divisibility of any number by all the primes below 100.
I will cover cases of each prime one after another.The method is based on a procedure similar to that for 17,19 and 59 shown in my earlier blogs.
In each case a suitable number -call it P- for which the prime considered is a factor should first be selected of the form (1000m+n) or(1000m-n)
Let be show the procedure for the next prime after 19 viz 23
The number P chosen for 23 is 2001 which has 23 as a factor.In this case m=2 and n=1
Let us consider the number 26105069 for test.
Step 1-This is broken up into groups of 3 digits starting from right calling them A,B and C.
We thus have A=069,B=105, and C=26.
Step2-We divide C by m and then multiply by n
We get 26/2=13 and 13*1=13
Step3-The result is subtracted from B .We get 105-13=92
Step4-Process in Step 2 is now repeated with the result in previous step.
We get 92/2=46 and 46*1=46
Step5-The result is subtracted from A.We get 069-46=23
Step6-The final result is tested for divisibility by 23.If it is divisible, the number under test can be considered as divisible
In our case the final result is itself 23-divisible obviously by 23 and hence the number under test viz 26105069 is so divisible
The values of P ,m and n vary for each prime otherwise the process is more or less similar.

Palindrome formation

There is a very easy method to form palindromes,described below-
Take for instance 2 digit numbers like  56,57,59 or a 3 digit number like125-
In each case you have to reverse the digits to form a new number and add it to the original.You may have to repeat the same step if a palindrome is not formed-
56+65=121  a palindrome.
57+75=132....132+231=363 a palindrome
59+95=154.....154+451=605....605+506=1111 a palindrome
125+521=646 a palindrome
You now have some surprises-
Try with 89..You will need 24 iterations(repeats of steps) and then only you get a palindrome shown below-
8813200023188.
Similarly for  the start number 10911,you will need 55 repetititions to get the palindrome of 28 digits viz  46687.31596.6842.2486.69513.78664
Biggest surprise, the number 1,186,060,307,891,929,990 will need 261 repetititions to get a palindrome of 119 digits.
Another surprise-the number 196 was tested.-No palindrome could be found even after more than 300 million steps.There are some more such numbers-897....1997....7059 etc