Number System Question 1 - CAT 2021

Step-by-Step Explanation

• Simplify the main inequality:

Start by subtracting 2 from all parts of the inequality to isolate the power of 2:

2.25 - 2 ≤ 2ⁿ⁺² ≤ 202 - 2

0.25 ≤ 2ⁿ⁺² ≤ 200

• Find the boundaries for n:

The decimal value is the same as the fraction 1/4, which can be written as 2⁻². Let's rewrite the inequality using a base of 2:

2⁻² ≤ 2ⁿ⁺² ≤ 200

Now look at the exponents to find the limits:

• Lower limit: The exponent n + 2 must be at least - 2.

n + 2 ≥ - 2 ⇒ n ≥ - 4

• Upper limit: The term 2ⁿ⁺² must stay under 200. Checking standard powers of 2 shows that 2⁷ = 128 and 2⁸ = 256. Since 256 goes past 200, the highest integer value the exponent can take is 7.

n + 2 ≤ 7 ⇒ n ≤ 5

Putting these together, n can be any integer from - 4 to 5:

n ∈ {- 4, - 3, - 2, - 1,0,1,2,3,4,5}

• Apply the final integer constraint:

The question asks for the number of integer values of 3 + 3ⁿ⁺¹. For this total expression to be a whole number, the part 3ⁿ⁺¹ must be a whole number.

If the exponent (n + 1) is negative, it turns into a fraction (such as 3⁻ ¹ = (1)/(3)), which is not an integer. Therefore, the exponent must be 0 or positive:

n + 1 ≥ 0 ⇒ n ≥ - 1

Filtering our initial list of n values to keep only numbers that are - 1 or larger gives:

n ∈ {- 1,0,1,2,3,4,5}

Counting these up, we get a total of 7 values.

Correct Answer: 7

💡 FundaMakers Tip:

Avoid using long logarithms for boundary checks. Simply compare the values with standard powers of 2. Knowing that 2⁷ = 128 is under 200 and 2⁸ = 256 is over 200 instantly tells you that the maximum value for the exponent is 7.