The question is about Multiples of 6. Another question on possible values . We need to find out the possible value of 'n' for which the given term is a multiple of 6. Dealing with digits and base system of a number is a vital component in CAT Number Systems. A range of CAT questions can be asked based on this simple concept in the CAT Exam.

Question 5: n^{2} + 5n + 6 is a multiple of 6. n is natural number less than 100. How many values can n take?

- 33
- 65
- 66
- 67

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n^{2} + 5n + 6

(n + 2) (n + 3)

(n + 2) and (n + 3) are two consecutive numbers.

One of (n + 2) or (n + 3) has to be even.

We need n such that (n + 2) (n + 3) is a multiple of 3.

(n + 2) or (n + 3) should be a multiple of 3.

n leaves a remainder of 0 or 1 when divided by 3.

n could be 1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18 ....

So, from among the first 99 natural numbers, N cannot take 2, 5, 8, 11, 14…….98.

There are 33 numbers in this list – we are effectively listing all numbers {3 × 0 + 2, 3 × 1 + 2, 3 × 3 + 2, ….3 × 32 + 2}.

So, N can take 99 – 33 = 66 values.

The question is **"How many values can n take?"**

Choice C is the correct answer.

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