Let the two no's be a and b;
Given product of the no's is p = ab;
If the each nos is increased by 2 then the new product will be
(a+2)(b+2) = ab + 2a + 2b + 4
= ab + 2(a+b) + 4
= p + 2(a+b) + 4
Hence the new product is (p+4) times greater than twice the sum of the two original numbers.

If a number to be divisile by 88, it should be divisible by both "8" and "11"

Check for '8' :
For a number to be divisible by "8", the last 3-digit should be divisible by "8"
Here 72x23y --> last 3-digit is '23y'
So y=2 [ (i.e) 232 is absolutely divisible by "8"]

Chech for '11' :
For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"
(7 + x + 3) - (2 + 2 + y)
(7 + x + 3) - (2 + 2 + 2)
(10 + x) - 6 should be divisible by "11"
for x = 7
=> 17 - 6 = 11 [ which is absolutely divisible by "11"]

So x = 7 , y= 2.

Lowest 4-digit number is 1000.
LCM of 3, 4 and 5 is 60.
Dividing 1000 by 60, we get the remainder 40. Thus, the lowest 4-digit number that exactly divisible by 3, 4 and 5 is 1000 + (60 - 40) = 1020.
Now, add the remainder 2 that's required. Thus, the answer is 1022.

Let the two-digit number be 10a + b
a + b = 12 --- (1)
If a>b, a - b = 6
If b>a, b - a = 6
If a - b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 - a = 3
Number would be 93.
if b - a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 - b = 3.
Number would be 39.
There fore, Number would be 39 or 93.

Given 

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 x 0 + 1 + 1

Using BODMAS Rule,

As multiplication precedes addition, 1 x 0 = 0, 

Now, 10 + 0 + 1 + 1 = 10 + 2 = 12.

Hence, 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 x 0 + 1 + 1 = 12.

Prime Numbers :: Numbers which are divisible by only 1 and itself are Prime Numbers.

It's answer will be 91.

Because 91 can be divisible by 7,13,91,1.

It is quite clear that prime number should be divisible only by itself and by 1.

First line will cut all other 14, similarly second will cut 13, and so on
Total = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105.

LCM of 4, 6, 8 and 10 = 120
120) 1000 (8
960
------
40

The least number of four digits which is divisible by 4, 6, 8 and 10 => 1000 + 120 - 40 = 1080.

Let the smallest number be x.
Then larger number = (x + 1365)
x + 1365 = 6x + 15
= 5x = 1350
x = 270
Smaller number = 270.