Numbers Questions

Divisor = 

           = 

Let the required fraction be x. Then, (1 / x )- x = 9/20
1 - x^(2) / x = 9 / 20  =>  20 - 20 * x^(2) = 9 * x.
20 * x^(2) + 9 *x - 20 = 0.
=> (4 * x + 5) (5 * x - 4) = 0.
=> x = 4 / 5.

The sum of three consecutive integers can be written as n + (n + 1) + (n + 2) = 3n + 3

If the sum is 24, we need to solve the equation 3n + 3 = 24;

=> 3n = 21;

=> n = 7

The greatest of the three numbers is therefore 7 + 2 = 9

Here a = 3 and r = 6/3 = 2. Let the number of terms be n.
Then, t = 384 => a * r^(n-1) = 384
=> 3 * 2^(n-1) = 384 => 2^(n-1) = 128 = 2^(7)

=> n-1 = 7 => n = 8.

132 = 4 x 3 x 11, So if the number is divisible by all three numbers 4,3 and 11,then the number is divisible by 132 also.
264   => 4,3,11(/)
396   => 4,3,11(/)
462   => 11,3
792   => 4,3,11(/)
968   => 11,4
2178 => 11,3
5184 => 3,4
6336 => 4,3,11(/)
Required number of numbers=4.

The square of a natural number never ends in 7.
42437 is not the square of a natural number

Let the number be x.Then

60% of 35of x=36.

=> 60100*35*x=36

=> 9x25=36 => x=25*369=100

Required number is 100.

LCM of 2, 3, 7 is 42.

=> (700 – 300)/42 = 9  22/42 => 9 Numbers.