Numbers Questions

The best approach is to pick a number that satisfies the rules that x and y are greater than 0. So for example you could choose 25 for y.

The value of x before doubling is then 25 - 50/25 = 23

The value after doubling y will be 50 - 50/50 = 49, which is more than double.

Step 1 :2x + 5y

Step 2 :3(2x + 5y) = 6x + 15y

Step 3 :(6x + 15y) - (x +y) = 6x + 15y - x - y (watch that last minus sign!)

This gives 5x + 14y

   1 1
   8 A 9
+ 4 C 6
-  6 B 2
   7 2 3
We may represent the given sum, as shown below:

1 + A + C - B = 12
A + C - B = 11

Now,giving the maximum values to A and C, i.e.

A = 9 and C = 9, we get B = 7.

Given  65×359×113

⇒2×35×5×79×113 

⇒25×35×59×79×113

Thus, there are (5 + 5 + 9 + 9 + 3)= 31 prime numbers.

Let assume the value of a,b &c that will satisfy the equation.
so a=1,b=1,c= -1 or a=1,b= -1 &c=1 or a= -1,b=1,c=1
here all three cases satisfy the equation
assume any one a=1,b=1,c= -1

Hence (a+b)(b+c)(c+a)=(1+1)(1-1)(-1+1) = 0

Let
x = peacocks
y = deers

2x + 4y = 200
x + y = 72
y = 72 - x
2x + 288 - 4x = 200
x = 44 peacocks
y = 28 deers