__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

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