Fun with Mathematics

Intra numeral mixing with Pythagorean triples




A most effective and quick way of getting R^2-relation
                                                              2 2

is intra numeral mixing with Pythagorean triple. This technique is useful
only for numeral relations, where equal numbers of squares with equal
number of digits are equated. If there is any difference in the number of
squares, then that much 0^2 must be added in the shortage side.

Similarly, if all the root numbers do not have equal number of digits, then
they can be made equal without affecting its

value by affixing sufficient number of zeros in front of them.



For example 3^2 + 4^2 = 5^2 is written as

3^2 + 4^2 = 5^2 + 0^2

Intra numeral mixing means the number in one side, is simply

affixed in the front or prefixed at the end of the number in the other side.
In intra numeral mixing pairing can be interchanged, but pairing of
numbers must be same for both sides

For the given example,

Suffixing : 35^2 + 40^2 = 53^2 + 4^2 = 2825

Prefixing : 30^2 + 45^2 = 54^2 + 3^2 = 2925

By repeating the procedure one can generate more and more relations.

35^2 +40^2 = 53^2 + 04^2 gives

5335^2+440^2 = 3553^2 + 4004^2

435^2+ 5340^2 = 4053^2 + 3504^2

30^2+45^2= 54^2+ 03^2 gives,

5430^2+ 345^2= 3054^2 +4503^2

330^2+5445^2= 3054^2+ 4503^2

If we consider the Pythagorean triple (5,12,13), the number of digits
in the root numbers is not equal. Hence before making intra numeral mixing,
it is arranged as

05^2 + 12^2 = 13^2 + 00^2

It gives the following numeral relations,

1305^2+12^2= 513^2+1200^2 = 1703169

1312^2+5^2 = 500^2+1213^2 = 1721369

The mathematics behind it gives more insight on such relations.

The intra numeral mixing made on a^2+b^2=c^2+d^2
(for Pythagorean triple d=0) gives,

(10c+a)^2+(10d+b)^2= (10a+c)^2+ (10b+d)^2
                                         = 101(a^2+b^2)+20(ac+bd)
or

(10c+b)^2+(10d+a)^2= (10a+d)^2+(10b+c)^2
                                          = 101(a^2+b^2)+20(ad+bc).

It is noted that in the intra numeral mixing, the numbers in the same
side cannot be paired up, as they do not preserve the balanced condition
of the relation.



In this intra numeral mixing, we simply add with each number 10 times
the paired root number. It is found that the balance

of the new relations is not affected when the mixing is done with different
proportions of the two paired numbers.

For a^2+b^2=c^2+d^2, we have

(ma+nc)^2+(mb+nd)^2= (mc+na)^2+(md+nb)^2
                                             =(m^2+n^2)(a^2+b^2)+2mn(ac+db)

or

(ma+nd)^2+(mb+nc)^2= (mc+nb)^2+ (md+na)^2
                                           = (m^2+n^2)(a^2+b^2)+ 2mn(ad+bc)

creative thoughts-17

Creative thought -17

Equal product of two numbers with sum in one pair equals to the difference
 in the other pair.

It is very simple to everyone to show that a sum of two numbers is equal to a
sum of two other numbers and a product of two numbers is equal to a product
of two other numbers separately. It can be proved that same two pairs of numbers
will not satisfy both the conditions simultaneously. Mathematically it can be
 stated if a + b = c +d ,then ab =/= cd and if ab = cd then a+b =/= c+d.
If we assume supernaturally that they coexist, then the equality would become
true for all powers of the numbers in the pairs and it leads to an impossible relation

  n    n       n       n
a + b   = c +   d

where n = 1,2,3,,4…….. In fact such relation will be true under trivial condition
where the numbers in the pairs are equal.

However, a sum and product of two numbers may be made equal
with a difference and product of two other numbers respectively. e.g.,

2+3 = 5 = 6 – 1
2x3 = 6 = 6x1
3+10 = 13 = 15-2
3x10 = 30 = 15x2

For a same sum and difference, one can have two or more pairs of numbers
to give different products. e.g.,

4+21 = 25 = 28-3 ; 4 x 21 = 84 = 28x3
10+15 = 25 = 30-5 ; 10x15 = 150 = 30x5

As the pair of such numbers has substantial influence on the equal sum of
like powers , it is worth to investigate a method of identifying such pairs. For
a given sum (or difference) S and product P , the pairs (a,b) and (c,d) are
related by the following relations

a+b = S = c-d
and axb = P = cxd

Substitute for d (= P/c) in S = c-d ,we get,

(cxc) –Sc – P = 0

The roots of the quadratic equation give the value of c and d

c = [S+ √ (SxS) +4P]/2 and d = [S- √ (SxS)+4P]/2

We derived the solutions for a and b ( See-Creative thought-15)

a = [S + √ (SxS)-4P]/2 and b = [ S- √ (SxS)- 4P]/2

To have integral solutions for c and d , (SxS)+4P must be a square number
and for a and b (SxS)-4P must be a square number .

If (SXS)+4P = (ZxZ) and (SXS) -4P = (YxY),where Z and Y are integral
numbers ,then its sum and difference give

2(SxS) = (ZxZ)+ (YxY) and 8P = (ZxZ)- (YxY)

It implies that the pairs of numbers whose sum and difference are equal
and their products are equal are closely related with the

numeral relation representing the sum of two squares is equal to twice that
of an another square. e.g.,

(1x1) + (7x7) = 2 x (5x5)

It gives S = 5 and P = 6 and therefore we can derive the pairs as (2,3)and (1,6).

(7x7)+ (17x17) = 2 x(13x13)

It gives S=13 and P =30 and the pairs are (10,3) and (15,2).

The numbers expressed by the relation 2(nxn)+2n +1 ,where n is a number ,
generate such numeral relation where the sum of two squares is equal to twice
 the square of that number 2(nxn)+2n + 1

It can be further derived,

(cxc)+(dxd) = (SxS)+2P
and
(axa) + (bxb) = (SxS) – 2P

(axa)+(bxb)+(cxc)+(dxd) = 2 (SxS)= 2 [(a+b)(a+b)]=2[(c-d)(c-d)] Few solutions are

(2,3),(1,6); (2x2) + (3x3) + (1x1) + (6x6) = 2 (5x5)
(3,10),(2,15); (3x3) +(10x10) + (2x2) + (15x15) = 2 (13x13)
(5,12),(3,20); (5x5)+(12x12)+(3x3)+(20x20)= 2x(17x17)
(14,15)(6,35); (14x14)+(15x15)+(6x6)+(35x35)= 2(29x29)

From (ZxZ) +(YxY) = 2 (SxS), one can derive a,b,c and d,the members
of the pairs

a = √ [(ZxZ)+(YxY)]/8 + Y/2 and b = √ [(ZxZ)+(YxY)]/8 – y/2
c = Z/2 + √ [(ZxZ)+(YxY)]/8 and d = Z/2 - √ [(ZxZ)+(YxY)]/8

creative thoughts-16

creative thought-16




Difference of two numbers equals to its product

The difference between two numbers may be equal to its product.

Mathematically it is expressed as

a-b = d and ab = d

hence a – b = ab, where a > b .The dependency between these two variables becomes

a = b/(1-b) ; b = a/(a+l)

It shows that for ‘a’ to be positive, ’b’ must be less than unity but

greater than zero ,that is ‘b’ must be a fraction. ‘a’ may be a whole

number or a fraction ,but ‘b’ will always be a fraction. If ‘b’ is given integral value, ‘a’
becomes negative. Thus there is no solution with whole integral value for both ‘a’and ‘b’.

When ‘a’ takes a whole integral value n , ‘b’ becomes a fraction

n/(n+1) and when ‘a’ takes a fractional value x/y, ‘b’ becomes

x/(x+y).Few typical solutions are given in Table.1.



a is an integer        a is a frac tion

  a      b                     a       b

  1     ½                   1/3      ¼

  2    2/3                 2/5      2/7

  3    ¾                  3/8      3/11

…   …                  …         ….

 n   n/(n+1)          N/n     N/(N+n)



The pair of numbers whose difference d and product are equal can be related to d.
By eliminating one of the dependent variables (either ‘a’ or ‘b’) ,one can obtain
a quadratic equation

(axa) – d a – d = 0 . The positive root of the equation is

a = [d + √ (dxd) + 4d ]/2

which gives,

b = a – d = [ -d + √ (dxd) +4d]/2

The positive values of the pair of numbers under the given condition for a given d
are shown in Table.2.

d         a or b

6       √ 15 ± 3

8        √ 24 ± 4

10      √ 35 ± 5

….        …….

2n      √ n(n+2) ± n

Under the given condition ,the fractional pair of numbers may be with numerator or
the denominator identical. When the numerators are same, the pair of numbers is
assumed as x/b and x/a so that

(x/b) – (x/a) = (x/b).(x/a)

It gives an additional condition x = a-b. Thus the pair becomes [(a-b)/b, (a-b)/a].
The pair of numbers whose denominators are same can be derived directly from
the pair of numbers whose numerators are identical. By multiplying both the numerator
and the denominator of a fractional pair by the denominator of the other same . e.g.,

a   b     a-b/b      a-b/a       a(a-b)/ab     b(a-b)/ab

7   3      4/3        4/7          28/21           12/21

3   5     2/3        2/5          10/15            6/15

7  11    4/7       4/11         44/77           28/77

Creative thoughts-15

Creative thought-15



Sum of two numbers equals to its product


When the addition and multiplication tables are memorized, the very fact that strikes one’s mind
is that the addition and multiplication of 2 with 2 give same resultant.

2+2 = 2x2 = 4

This is the only answer with whole integral numbers. Are there any pairs of numbers whose
sum and product yield same result? Infact, there are many solutions to this puzzle, if we
allow fractional numbers in the pairs.

If the sum and product of two numbers x and y are same,

x+y = xy or x = y/(y-1)

By giving a value to y arbitrarily ,the corresponding value of x can be predicted.
Few pairs are given in Table.1.

Table.1.

y         x

2         2

3        3/2

4        4/3

5        5/4
. .
. .
n       n/(n-1)

The pair of numbers under the given condition can be related separately with its
sum S (or its product S)

x+y =S and xy = S

By solving these two relations,

(x.x) –Sx + S = 0

or x = [S ± √ (SxS)-4S]/2

When x and y are interchanged, it does not alter the given condition and the sum S.
Hence the two roots correspond to x and y.

x = [S + √ (SxS)-4S]/2 and y = [S- √ (SxS)-4S]/2

With the help of these two expressions, one can find out the proper pair of values (x,y)
for any given S. The simplest solutions with S in the form 2n ,where n is any number are
tabulated in Table.2

Table.2.

S        x or y

6        3 ± √ 3

8       4 ± √ 8

10     5 ± √ 15

12     6 ± √ 24
…        ….

2n     n ± √n(n-2)

The pair of numbers whose sum and product are same has a particular importance
in the field of electrical network. If they represent the electrical resistance of two resistors,
its parallel combination will always yield an effective resistance of one unit.

The pair of numbers under the given condition may have fractional value. If they
are (x/a,y/b), then

x/a + y/b = (x/a)(y/b)

or x= (ya)/(y-b) and y = (xb)/(x-a)

The fractional pair can be shown as [ y/(y-b), y/b ],where y and b are two independent
variables. It is noted that either the numerators or the denominators of the fractional pairs
are identical. e.g.,

y     b     y/b       y/y-b

7     3    7/3        7/4

7     5     7/5       7/2

11   3     11/3     11/8

11   7     11/7     11/4

…   …    …        …

N     n     N/n     N/N-n

Another way of getting the fractional pairs instantaneously is the utilization of
Pythagorean triples (x,y,z) where x ≤ y ≤ z . If N is taken as the square of the
greatest of the triple (z.z) and n is the square of another number of the triple
(x.x or y.y) ,then the fractional pairs becomes [(z.z)/(x.x) , (z.z)/(y.y)]

Table.4
Pythagorean fractional triples          pairs

          (3,4,5)                            25/9, 25/16

        (5,12,13)                      169/25, 169/144

        (8,15,17)                     289/64, 289/225

In such fractional pairs both the numerators and the denominators are square
 numbers and the sum of the numbers in the denominators of the fractional pairs
 is equal to its identical numerator. Hence they can simply be written as (a+b)/a
and (a+b)/b, where a and b are any two independent variables.

Instead of numerator, the denominators of the fractional pairs may be made same.
Let them be (x/a, y/a).According to the given condition (sum = product),

(1/a)(x+y) = xy/(axa) or a = xy/(x+y)

The fractional pairs with identical denominators then becomes [ x(x+y)/xy , y(x+y)/xy] ,
where x and y do not depend upon each other.

Table.5

x      y        (x+y)/x        (x+y)/y       y(x+y)/xy       x(x+y)/xy

2      3         5/2              5/3             15/6                10/6

3      5         8/3              8/5             40/15              24/15

5      7        12/5            12/7            84/35              60/35

In general the product of two numbers will be greater than its sum and hence one
can modulate the given condition as the product of two numbers equals to twice
or thrice or in general n times that of its sum.

xy = n(x+y)

or x = ny/(y-n)

To avoid negative numbers y≥ (n+1). Few solutions for n= 2 and n=3 are given in Table.6

Table.6

________________________

y                x

________________
            n=2 n=3

________________________

3           6     …

4          4      12

5        10/3   15/2

6          3        6

7       14/4     21/4

…       …       …

n       2n/n-2   3n/n-3

creative thoughts-13

Creative thoughts-13

Natural series of even numbers


It is interesting to find the sum of even or odd numbers only in the given natural
series. The sum of the even numbers from 2 to 2n containing n even numbers
is given by

S(even) = 2+4+6+8+ …….. 2n = 2(1+2+3+4+……n)

The terms inside the bracket are in natural series up to n. Hence its sum becomes

S(even) = 2 x n(n+1)/2 = n(n+1),where n is the number of even numbers added up
which is equal to half of the last even number in the given even series. Thus the sum,
in terms of the last even number N of the series is

S(even) = (2n) x(2n +2)/4 = N(N+2)/4

Natural series of odd numbers

The sum of the odd numbers from 1 to N (=2n-1) in the natural series is given by

S(odd) = 1+3+5+7+9+….. (2n-1),where n = (N+1)/2

But S(odd) = S(natural ) – S (even),the difference between the total sum of the natural
series from 1 to (2n-1) and the series of all even numbers from 2 to (2n-2).
The sum of the natural series from 1 to (2n-1) is n(2n-1) and in terms of the last
number of the series ,it is N(N+1)/2.Similarly ,the sum of all even numbers from 2 to
2(n-1) is n(n-1) and in terms of last number N, it is [(NxN)-1]/4. Hence the sum of all
odd numbers from 1 to (2n-1) becomes,

S(odd) = n(2n-1)- n(n-1) = nxn

Thus the sum of natural series of n odd numbers from 1 is always a square number nxn.
In terms of N ,the last odd number of the series

S(odd) = [N(N+1)/2] – [(NxN)-1]/4 = (1/4)[(N+1)(N+1)]

If the natural series has a set of numbers from N(L),the smallest number to N(H),
the greatest number, its sum is given by


S = N(H)[N(H)+1]/2 – N(L)[N(l)+1]/2

   = [N(H)-N(L)][N(H)+N(L)+1]/2

In the case of both natural even series and natural odd series , it is

S= [N(H)-N(L)][N(H)+N(L)+2]/4

Series of square numbers


It is noted that the sum of squares of all odd numbers from 1 to (2n-1) in natural
series is 1/6 times the product of three successive numbers (2n-1),2n and (2n+1).e.g.,

1x1 = 1x2x3/6 = 1
1x1 + 3x3 = 3x4x5/6 = 10
1x1+3x3+5x5 = 5x6x7/6 = 35
1x1+3x3+5x5+7x7 = 7x8x9/6 = 84

for the product of any three successive numbers in natural series ,6 will
invariably be a factor. Hence with the help of Algebra, it can be shown as

S(odd square) = 1x1 +3x3+5x5 +……. (2n-1)x(2n-1)

                      = [(2n-1)2n(2n+1)]/6 = n (4nxn – 1)/3

In terms of the last number of the given natural series of square numbers, it becomes,

S(odd square) = N(N+1)(N+2)/6

Like this the sum of squares of all even numbers from 2 to 2n(=N) in the
natural series is 1/6 times the product of three successive numbers 2n,2n+1,2n+2, e.g.,

2x2 = 2x3x4/6 = 4
2x2 +4x4 = 4xx6/6 = 20
2x2 +4x4+6x6 = 6x7x8/6 = 56
and in general

S(even squares) = 2x2 + 4x4 + 6x6 + …… 2nx2n

                         = 2n(2n+1)(2n+2)/6 = (2/3) n(n+1)(2n+1)

The sum of squares of all numbers from 1 to N in the natural series
can also be expressed in a similar fashion.

S(natural squares) = S(odd squares) + S(even squares)

                            = N(N+1)(N+2)/6 + (N-1)N(N+1)/6

                            = N(N+1)(2N+1)/6

The sum is (1/6) times the product of two successive numbers N and N+1
which is multiplied with its sum (2N+1)

1x1+2x2+3x3+4x4 +…… NxN = N(N+1)(2N+1)/6

Creative thoughts-12

Creativethoughts-12

We know the sum of n numbers from 1 in the natural series is n(n+1)/2

S = 1+2+3+4 +5+6+7+8+………. n = n(n+1)/2

If n is an even number, then the first and last numbers, second and the last but one
numbers of the series ……… added together each sum will be equal to (n+1).
As there are n/2 such pairs, the total sum will be (n/2)(n+1). If n is odd, we get the
same result.

Here all the number can be paired up except the central number i.e., (n+1)/2,
the number of pairs will be (n-1)/2 and the sum of each pair is (n+1) . Hence its sum will be

S = [(n-1)/2][n+1] + (n+1)/2 = n(n+1)/2

It predicts that for twice the sum of n numbers from 1 in natural series , n and (n+1)
are factors. With numerical examples,

1+2 = 3 1 x3
1+2+3 = 6 = 2 x3
1+2+3+4 = 10 = 2x5
1+2+3+4+5 = 15 = 3 x 5
1+2+3+4+5+6 = 21 = 3 x 7
1+2+3+4+5+6+7 = 28 = 4x 7

I f the number of numbers added together is even ,then its sum can be represented
as a product of two numbers (n/2)and (n+1) ,they are n and [(n+1)/2],if it is odd.

The sum of numbers in natural series from smaller number n(s) to higher number n(h)

S = n(s) + [n(s)+1] + [(n(s)+2] + ………… n(h)

n(h) = n(s) + N-1,where N is the number of numbers in the series

The sum of numbers in natural series from 1 to n(h) is [(n(h)][(n(h)+1)]/2 and the
sum of number in natural series from 1 to n(s) – 1 is [n(s)-1][n(s)]/2

The difference between these two sums gives the required sum and is given by,

S = S(h) – S(L-1) = [n(h)][(n(h)+1)/2] – [n(s)-1][n(s)]/2

=[ n(h)xn(h) + n(h) – n(s)xn(s) + n(s)]/2

{[n(h)-n(s)][n(h)+n(s)] + [n(h) + n(s)]}/2

= [n(h)+n(s)][n(h)-n(s)+1]/2

where n(h) + n(s) is the sum of the initial and final terms of the series and
 n(h)-n(s) + 1 denotes the number of terms added up in the given series.

It opens an avenue for quite a large number of mathematical puzzles. The sum
of a set of n numbers in natural series is S ,find the possible natural series with
same sum but with different number of terms. We know,

Twice of the sum = (sum of the first and last terms) x (number of terms in the series)

= [n(s) + n(h) ] N

If the twice of the sum is divisible by N-x , then there will be one or more another
solutions. For example,

S = 11+12+13+14+15+16 = 81

2S = 162 which is divisible by 6, the number of terms in the given series.
162/6 = 27 = n(s)+n(h) .It is not divisible by 5,4 but by 3.

162/3 = 54 = n(s) + n(h)

so there must be a series with three terms whose first and last terms give 54 on addition.

S = 26 + 27 + 28= 81

It is divisible by2 as well, so

S = 40+41 = 81

Creative thoughts-11

Creative thoughts-11

Fifth power of a number

The fifth power of a number is equal to the difference of two squares,the sum of its roots
is equal to cube of that number and the difference of its roots is equal to square of the
number.

2x2x2x2x2 = 32 = 6x6 – 2x2 ; 6 + 2 = 8 = 2x2x2 ; and 6 – 2 = 4 = 2x2
3x3x3x3x3 = 243 = 18x18 – 9x9 ; 18 + 9 = 27 = 3x3x3 and 18-9 = 9 = 3x3
4x4x4x4x4= 1024 = 40x40 – 24x24 ; 40 + 24 = 64 = 4x4x4 and 40 – 24 = 16 = 4x4
5x5x5x5x5 = 3125 = 75x75-50x50 ; 75+50 = 125 = 5x5x5 and 75- 50 = 25 = 5x5

In general, it can be shown as ,

NxNxNxNxN = (NxN)(NxNxN) = [(NxN)(N+1)/2][(NxN)(N+1)/2]-[(NxN )
(N-1)/2][(NxN)(N-1)/2]

The greater square root number in this relation is N times the sum of N numbers from 1
In the natural series and the smaller square root number is obtained by subtracting
NxN from it.

Sixth power of a number

The sixth power of a number N is equal to the difference of two squares ,the sum of its
roots is equal to fourth power of that number and the difference of its roots is equal to
the square of that number.

2x2x2x2x2x2 = 64 = 10x10 – 6x 6 ; 10 + 6 = 16 = 2x2x2x2 and 10 – 6 = 4 = 2x2
3x3x3x3x3x3 = 729 = 45x45 – 36x36 ; 45+36 = 8l = 3x3x3x3 and 45-36 = 9 = 3x3
4x4x4x4x4x4 = 4096 = 136x136 – 120x120 ;136+120=256=4x4x4x4 and
                                                                                                   136-120 =16 = 4x4 .
5x5x5x5x5x5 = 15625 = 325x325 – 300x300 ; 325+300=625 =5x5x5x5 and
                                                                                                    325-300 = 25=5x5 .
The general form of this type of relation is

NxNxNxNxNxN =(NxN)(NxNxNxN) = [(NxN +1)(NxN)/2][(NxN +1)(NxN)/2]
                                                                       - [(NxN-1)(NxN)/2]{(NxN-1)(NxN)/2]
If NxNxNxNxNxN is split into N and NxNxNxNxN ,then

NxNxNxNxNxN = [(N+1)(NxN) √N/2][(N+1)(NxN)√ N/2]
                                              – [(N-1)(NxN)√ N/2][(N-1)(NxN)√ N/2]

If N is an even square number ,then NxNxNxNxNxN can be expressed as a
difference of two squares in another way also e.g., when N = 4,

4x4x4x4x4x4 = 80x80 – 48x48

This idea can be extended to any power of a number .e.g.,

NxNxNxNxNxNxN = [(N+1)NxNxNx/2][(N+1)NxNxN/2]
                                                    - [(N-1)NxNxN/2]{(N-1)NxNxN/2]

NxNxNxNxNxNxNxN= [(NxN+1)NxNxN/2][(NxN+1)NxNxN/2]
                                                                -[(NxN-1)NxNxN/2][(NxN-1)NxNxN/2]

NxNxNxNxNxNxNxNxN= [(N+1)NxNxNxN/2][(N+1)NxNxNxN/2]
                                                          -[(N-1)NxNxNxNxN/2][(N-1)NxNxNxN/2]

Creative thoughts-10

Creative thoughts-10


Power of a number as the difference of two squares

Any power of a number can be expressed as the difference between two squares ,
where the sum and difference of its roots give higher and lower powers of that number,
so that the product of them is exactly equal to the initial power of the number. This can be
applied effectively to any power of any number.

Cubes

1x1x1 = 1 = 1x1 – 0x0 ; 1 + 0 = 1 = 1x1 ; 1-0 = 1

2x2x2 =8 = 3x3 – 1x1 ; 3 + 1 = 4 = 2x2 ; 3-1 = 2

3x3x3 = 27 = 6x6 – 3x3 ; 6 + 3 = 9 =3x3 ; 6 – 3 = 3

4x4x4 = 64 = 10x10 – 6x6 ; 10 + 6 = 16 = 4x4 ; 10-6 = 4

5x5x5 = 125 = 15x15 – 10x10 ; 15 +10 = 25 = 5x5 ; 15-10= 5

It is found that all the squares involved in these relations are related to triangular numbers
( as they can be represented by triangles). The first few triangular numbers are 1,3,6,10,
15,21,28,36,45,……… The n th triangular number is the sum of n natural numbers from
1 and it is equal to n(n+1)/2. The cube of a number N is found to be the difference
between two successive triangular numbers ,whose sum gives square of the cube
root and the difference ,the cube root itself.

NxNxN = [N(N+1)/2] x [N(N+1)/2] - [N(N-1)/2]x[N(N-1)/2]

If T(n) denotes the n th triangular number ,then

T(n) + T (n-1) = n x n
T(n) – T(n-1) = n

4 th power of a number

The fourth power of a number N is equal to the difference of two squares, the sum
of its roots is equal to cube of that number and the difference of its roots is equal to
the number Itself.

2x2x2x2 = 16 = 5x5 – 3x3 ; 5+3 = 8 = 2x2x2 ; 5-3 = 2

3x3x3x3 = 81 = 15x15 – 12x12 ; 15 + 12 = 27 = 3x3x3 ; 15-12 = 3

4x4x4x4 = 256 = 34x34 – 30x30 ; 34 +30 = 64 = 4x4x4 ; 34-30 = 4

5x5x5x5 = 625 = 65x65-60x60 ; 65+60 = 125 =5x5x5 ;; 65-60 = 5

If N is any number

NxNxNxN = (N/2)[(NxN+1)(NxN+1)] – (N/2)[(NxN-1)(NxN-1)]

The smaller square root number in thes relations is (N-1) times the sum of N numbers
 from 1 in the natural series and the greater square root number is obtained by adding N
with it.

Creative thoughts-9

Creative thought-9


Health is a kind of wealth. In fact it is more valuable than the conventional wealth.

Health is required not only to our physical body but also to our mental brain. Mental health is different than that of the physical health. As it is not exhausted on continuous use, there is no need for recharging. In fact it is strengthened up every time it is used up.

The real wealth is the acquisition of profound knowledge which cannot be stolen or snatched by others.

Recreational Mathematics

(Higher power of a number)

The fifth power of a number can be expressed as

NxNxNxNxN = (NxNxN-1)(NxNxN+1) + 1 /N

and NxNxNxNxN – N = N(NxNxNxN-1) = (N-1)N(N+1)(NxN+1)

Since 5 is a factor to (NxNxNxN-1) and 6 is a factor to (N-1)N(N+1),any three successive numbers in natural series ,NxNxNxNxN –N will have 30 as factor for all values of N. Hence the fifth power of a number can be expressed as

2x2x2x2x2 = 1x30 +2 = 15 x 2 + 2 = 5x(3x2) +2 = (2x2+1)1x2x3 + 2

3x3x3x3x3= 8x30 + 3 = 80 x3 + 3 = 20 x(4x3) +3 = (3x3x3+1)2x3x4 + 3

4x4x4x4x4= 34 x 30 + 4 = 255 x 4 + 4 = 51 x(5x4) + 4 = (4x4 +1)3x4x5 + 4

5x5x5x5x5= 104 x30 + 5 = 624 x5 + 5 = 104 x (6x5)+5 = (5x5 +1)4x5x6 + 5

The sixth power of a number N can be expressed as

NxNxNxNxNxN = NxN (NxNxNxN-1) + NxN

The term NxN (NxNxNxN-1) has a factor 60 for all values of N, hence NxNxNxNxNxN

can be shown as the sum of NxN and a multiple of 60.


2x2x2x2x2x2 = 64 = 10 x 6 +4 = 1x 60 + 4

3x3x3x3x3x3 = 729 = 120 x 6 + 9 = 12 x 60 + 9

4x4x4x4x4x4 = 4096 = 680 x 6 +16 = 68 x60 + 16

5x5x5x5x5x5 = 15625 = 2600 x6 +25 = 260 x 60 +25

It shows that NxN (NxNxNxN -1) is always divisible by 60.

Seventh power of a number

In the case of 7 th power of a number

2x2x2x2x2x2x2 = 128 = 120 + 8

3x3x3x3x3x3x3= 2187 = 18 x 120 + 27

4x4x4x4x4x4x4 = 16384 = 136 x 120 + 64

5x5x5x5x5x5x5 = 78125 = 650 x 120 + 125


Thus NxNxN (NxNxNxN-1) is always divisible by 120. In general the term NxNxNx……(k times) (NxNxNxN-1) is divisible by 2x2x2……(k-1) times x 30 for k greater than equal to l .

Creative thoughts-8

Creative thoughts-8


If our goal and desire is pale and anemic, it will invariably be reflected in all our
activities. Yes, all the deeds done outside are simply the follow-up of the initial
and inherent ambitions inside. If the later is good, the former will also be good.
If we go after our subject with persistence we can reach the target like a
forward wave from a light source that reaches the obstacle at infinite distance
with non-stop propagation

Recreational Mathematics.

Fourth power of a number

* The fourth power of a number N can be expressed as the sum of the product of
(NxN -1) and (NxN + 1) and 1 i.e.,

NxN xN xN = (NxN-1)(NxN+1) + 1

* The fourth power of even numbers is equal to a multiple of 16 and one excess
over a multiple of 16 for all odd numbers.

1x1x1x1 =      1 =  0 x 16 +  1  ; 2x2x2x2 =     16 = 16 x 1 + 0
3x3x3x3 =    81 =  5 x 16 +  1  ; 4x4x4x4 =   256 = 16 x16 +0
5x5x5x5 =  625 = 39 x16 +  1  ; 6x6x6x6 = 1296 = 16 x 81 +0
7x7x7x7 = 2401 =150 x16 +1  ; 8x8x8x8 = 4096 = 16 x 256 + 0

It shows that NxNxNxN – 1 is completely divisible by 16 ,when N is add..Again,

NxNxNxN – 1 = (NxN+1)(N-1)(N+1)

* When N is odd, all these factors will be even and hence NxNxNxN – 1 will be
divisible by 8.

* The fourth power of a number which is not a multiple of 5, is equal to one excess
over a multiple of 5.

2x2x2x2 =   16 = 1 x 5 + 1  ; 6x6x6x6 =   1296 =   259 x5 +1
3x3x3x3 =   81 = 16x5 + 1  ; 7x7x7x7 =   2401 =   480 x5 + 1
4x4x4x4 = 256 = 51x5 + 1  ;  8x8x8x8 =  4096 =   819 x5 + 1
                                            ; 9x9x9x9 =   6561 = 1312 x 5 + 1

* The fourth power of a number N can be expressed as the sum of NxN odd numbers from 1.

1x1x1x1 = 1
2x2x2x2 = 1+3+5+7 = 16
3x3x3x3 = 1+3+5+7+9+11+13+15+17 = 81
4x4x4x4 = 1+3 ……….29 +31 = 256







NxN

NxNxNxN = ∑ (2n-1)

1

Creative thoughts-7

A thought to think
Why people forget the information often which they wanted to convey ?

There are two solid reasons- one is they don't know anything about it. To hide the real state of affair,they are supposed to pretend as if they forgot. Another reason is lack of concentration which causes a lack of confidence.

If two or more pieces of information are stored simultaneously, they are registered randomly and usually not filed in order. Hence they often volatilize during the crucial time of usage

Recreational Mathematics
(Even squares)

4,16,36,64,100,144,196..... are few first even squares. They usually end with 00,4 or 6. .Even squares will always be divisible by 4.

The square of an even number N can be represented by (N/2)[(N-1)+(N+1)], where (N-1) and (N+1) are the two successive odd numbers ,the lower and higher neighbours to the given even number. Thus
2x2 = 1(1+3) = 4 ; 4 x 4 = 2(3+5) = 16; 6 x6 = 3 (5 + 7) = 36 and so on.

It is noted that both the odd and even squares are in the form of 5x or 5x + 1 or 5x - 1.
       odd squares                                             even squares
      2 x 5 - 1 =    9 = 3 x 3                           1 x 5 -  1 =  4  = 2 x 2
    10 x 5 - 1 =  49 = 7 x 7                           3 x 5 + 1 = 16 = 4 x 4
    16 x 5 + 1=  81 = 9 x 9                           7 x 5 + 1 = 36 = 6 x 6
    24 x 5 + 1=121 = 11 x 11                     13 x 5 -  1 = 64 = 8 x 8

Again it is observed that square of a numbe N (may be odd or even) is equal to sum of N successive odd numbers from 1 to (2N-1) in natural series.
                                              1 + 3 = 4 = 2 x 2
                                        1 + 3 + 5 = 9 = 3 x 3
                                1 + 3 + 5 + 7 = 16 = 4 x 4
                          1 + 3 + 5 + 7 + 9 = 25 = 5 x 5
It is seen that the sum of N odd numbers from 1 in the natural series is equal to the square of the mean of all the numbers so added up.
The sum of n even numbers in the natural series of even numbers also has an analogous property. The sum of n successive even numbers from 2 is equal to n times the mean of the even numbers added up.
                       2 = 2 = 1 x 2 = 1 x 1 + 1= 1(1+1)
                 2 + 4 = 6 = 2 x 3 = 2 x 2 + 2  = 2(2+1)
           2 + 4 + 6 = 12 = 3 x4 = 3 x 3 + 3 = 3(3+1)

It is found that the product of any two successive odd or even numbers in the odd or even natural series is equal to one less than the square of the mean of the two numbers

1 x 3 =   3 =   4 - 1 = 2 x 2 - 1   ;    0 x 2 = 0 =    1 - 1 = 1 x 1 - 1
3 x 5 = 15 = 16 - 1 = 4 x 4 - 1   ;    2 x 4 = 8 =    9 - 1 = 3 x 3 - 1
5 x 7 = 35 = 36 - 1 = 6 x 6 - 1   ;    4 x 6 = 24 = 25 - 1 = 5 x 5 - 1

Creative thoughts-6

1.A thought to think

Every challenging effort achieved seems to be impossible in the beginning.
Yes,there is a definite way to make a phase change from the state of impossible to a state of possible.
Impossible is of course, a zero energy state,where as possible is at higher energy state. The required phase change to take place, one must necessarily spend a quantum of energy.

2.Recreational mathematics
odd squares
A number multiplied by itself gives its square. If n is a number then its square is n x n. It is found that no square numbers ends with 2,3,7 and 8. The square numbers have digital roots of either 1,4,7 or 9 only.A signle digit number obtained after repeatedly adding the digits of a given number is called its digital root.Another important property of square numbers is that they have an odd number of divisors.

We know that 1,9,25,49,81,121,169,225,..... are the first few odd square numbers. They are all ending with 1 or 5 or 9. A kind of regularity is inherent among them. They are all one excess over 4 times the product of any two successive numbers and it is equal to the square of sum of such two successive numbers.
                                                 4(0x1) + 1 = 1 = (0+1)(0+1) = 1 x 1
                                                 4(1x2) + 1 = 9 = (1+2)(1+2) = 3 x 3
                                                 4(2x3) +1 = 25 = (2x3)(2x3) = 5 x 5
There is no strange at all when we seek the help of Algebra. The general form of an odd number is (2n-l), where n takes any value from 1 to n. The square of an odd number (2n-l) is given by 4(nxn) - 4n + 1=
 4n(n-1) + 1 and it explains the expression given above.`
The square of an odd number N can be expressed as,
                                                   NxN = (N/2)[(N-1) +(N+1)]
where (N-1) and (N+1) are the even numbers, the lower and higher neighbours to the given odd number.
Thus,
                               3x3 = (3/2) (2+4) = 3(1+2) = 9
                               5x5 = (5/2)(4+6) =  5(2+3) = 25
                               7x7 = (7/2)(6+8) = 7(3+4) = 49