Sums of 2 square numbers

There are not many cases of 3 consecutive numbers where all three of them can be expressed as a sum of 2 squares
There are only 7 cases with numbers below 1000,which satisfy this requirement.These are shown below with the squares involved in each case shown within brackets-
72(36+36)..73(9+64)...74(25+49)
288(144+144)..289(64+225)..290(289+1)..This  of course can be rejected as one of the numbers 289 is itself a square.
520(196+324)...521(121+400)...522(81+441)
584(100+484)...585(144+441)..586(225+361)
800(400+400)...801(225+576)...802(361+441)
232(36+196)...233(64+169)...234(9+225)
808(324+484)...809(25+784)...810(81+729).
Taking into account cases similar to 288,289,290 which have to be rejected,there are 8 cases of numbers between 1000 and 2000 and  5 cases relating to numbers between 2000 and 3000.Can you find them?

Telephone number puzzle

A person described the telephone number (of his friend) consisting of  7 digits as follows-
The initial 3 digit part is 447 more than the last 3 digits of the four digit second part,but less than the first 3 digits of the second part by 447.Find the telephone number

Harshad numbers-Rejoinder.-

I had mentioned about a Multiple Harshad Number in my previous blog .Same can be called MHN-12 as the division by the sums of the digits can be repeated 12 times.
The same is 20165 02858 57988 44661 76-The sum of the digits being 108,you get the result of division
18671 32276 46285 59872-sum now is 99 which can be used for the next division.
Proceeding repeatedly you will get the final two numbers 392 and 28.
In general  numbers of the form-10080,100800,1008000,10080000.. etc are all such multiple numberss

Diagonals-rectangular cube

Imagine any rectangular cube say with side lengths x,y and z
There will be 3 diagonals -we can call them as a,b and c
It will be very difficult to find values of x,y and z so that the diagonal lengths
a,b and c have integral values.
There is a complicated formula to find such values-
Here are some examples of such values-
Example 1) side lengths 44,117 and 240 .The diagonals have values  125,244 and 267
Example 2) side lengths 352,  936 and 1920.  The diagonals have values 1000,1952 and 2136.
Example 3) side lengths 828,2035 and 3120.The diagonals have values 2197,3228 and 3725.  

Perfect triangles

Triangles for which the perimeter and the area have the same integral values are called
perfect triangles-Some examples below-
1)Sides 6,8,10,-perimeter and area=24
2)Sides 5,12,13-perimeter and area=30
3)Sides 9,10,17-perimeter and area=36
4)Sides 7,15,20-perimeter and area=42
5)Sides 6,25,29-perimeter and area=60
You can see all the integral values are multiples of 6.
For your information the areas can be calculated using the following formula-
Find the perimeter 'p' and divide it by 2 to get 's' the semi-perimeter.
Assuming the sides are designated by a, b and c find the product of s,s-a,s-b and s-c
The square root of the product is the area
Example (4) above......a=7,b=15,c=20.....p=42....s=21
s-a=14,s-b=6,s-c=1.Hence the product is 21*14*6*1=1764...Sq root=42=area

Connection between numbers 39 and 1200

You can split up the number 39 into 3 parts in 3different ways,and in each case the products of the 3 parts will be the same viz 1200 as shown below-
25+8+6=39...25*8*6=1200
24+10+5=39....24*10*5=1200
20+15+4=39....20*15*4=1200

Using odd numbers for squares/cubes

Here are ways for obtaining squares/cubes using odd numbers-
Consider the following set of odd numbers-
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37etc
For getting squares you simply add them consecutively-
1=1*1
1+3=4=2*2
1+3+5=9=3*3
1+3+5+7=16=4*4etc
1+3+5+.......37=361=19*19 and further squares.
For getting cubes you add as follows-
1=cube of 1
3+5=8=cube of 2
7+9+11=27=cube of 3
13+15+17+19=64=cube of 4
21+23+25+27+29=125=cube of 5 etc
Considering  the above groups as the 1st,2nd,3rd,4th &5th, the 'n'th group will be as follows-
First term n^2-n+1
Last term n^2+n-1
No of terms-n
Sum of all the terms=n*average of the first &last term=n*n^2=n^3=cube of n
For example the 30th group will have 871 as the first term,929 as the last term and the sum of
all the terms will thus be 30*(871+929)/2=30*900=27000=cube of 30