20 Minutes Test: Number Systems 2 - GMAT MCQ

The resulting number is always divisible by 9. e.g.,

Consider 47.

Sum of its digits is 4+7=11

So,47−11=36 which is exactly divisible by 9

a = 1/ (3-2√2) 

b =  1/ ( 3+2√2) 

∴ ab =  1/(3-2√2) * 1/ (3+2√2) 

=>  1/ (9-8) 

=> 1 

We know; 

(a+b)² = a² + b² + 2ab

And, a²+ b² = (a+b)² - 2ab

=> [ 1/(3-2√2) + 1/(3+2√2) ]² -2*1

=> [3+2√2+3-2√2 / 1]^2 -2 

=> (6)² - 2 

=>  36 - 2
=>  34

Step 1: Let x Represent the Repeating Decimal

Let:
x = 5.2222... (where 2 repeats forever)

Step 2: Multiply by 10 to Shift the Decimal Point

Multiply both sides by 10 to move the decimal point one place to the right:
10x = 52.2222...

Step 3: Subtract the Original Equation from the New One

Now subtract the original equation (x = 5.2222...) from this new equation:
10x - x = 52.2222... - 5.2222...
9x = 47

Step 4: Solve for x

x = 47 / 9

Final Answer

The exact value of 5.2222... (or ) is 47/9

63 is divisible by 7 for any power, so required remainder will depend on the power of 4
Required remainder 

find remainder when 99/4 ( by 4 because to check periodicity)

There are 10 prime numbers between 0 and 30 which are: 2,3,5,7,11,13,17,19,23,29. Number one is not a prime number.
Now, for 1, the number of positive divisors or factors is only one i.e. 1 itself.