Which of the following can be used to illustrate that not all prime numbers are odd? 

 

   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.

Let the two consecutive odd integers be (2x + 1) and (2x + 3)

Then,

(2x + 3)² - (2x + 1)²

= (2x + 3 + 2x + 1) (2x + 3 - 2x - 1)

= (4x + 4)(2)
= 8 (x + 1), which is always divisible by 8

Two digit numbers: The two digits can be 2 and 9: Two possibilities 29 and 92.

Three-digit numbers: The three digits can be 1, 2 and 9 => 3! Or 6 possibilities.

We cannot have three digits as (3, 3, 2) as the digits have to be distinct.

We cannot have numbers with 4 digits or more without repeating the digits.

So, there are totally 8 numbers.

We know that am×an=am+n 

∴3131-27 = 314 = 961 x 961.

Smallest number of 6 digits is 100000

on dividing 100000 by 111 we get 100 as remainder 

 Number to be added = (111 -100) = 11

 Required Number  = 100011

2n+2n22x2n - 2n = 2n1 + 122n(2-1) = 32

Let the numbers be x and (x-11)

   x = 28

 Another Number is (28 - 11) = 17

Number = ( 779 x a) + 47, where "a" is the quotient

           = (19 x 41 x a) + (19 x 2) + 9

           = 19 x (41a + 2) + 9

           = 19 x (New quotient) + 9

  Required remainder = 9