IMAT 2011 Q78 [Integers]

How many different integers, n, are there such that the difference between
2 \sqrt{n} and 7 is less than 1?

A. 0

B. 2

C. 4

D. 6

E. 8

We construct the inequality:

|2 \sqrt {n} -7| <1

And we know that if we apply the knowledge of

if |X|< D , then -D < X < D

We can reassume the inequality to become:

-1 < 2 \sqrt{n} - 7 < 1

And so we begin simplifying-

Let’s add 7 to each respective side, it becomes:

6 < 2 \sqrt {n} < 8

Next we can divide by 2:

3 < \sqrt{n} < 4

We can raise to the square of 2:

3^2 < n < 4^2 which we can quickly re-write as 9 < n < 16

With this simple inequality, we can list out our possible values, and count how many they are:

n = { 10, 11, 12, 13, 14, 15 } which make for 6 values

Our answer is, therefore, D.