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[x] is the greatest integer less than or equal to the

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by Vincen » Tue Dec 05, 2017 9:42 am
[x] is the greatest integer less than or equal to the real number x. How many natural numbers n satisfy the equation $$\left[\sqrt{n}\right]=17?$$ A. 17
B. 34
C. 35
D. 36
E. 38

The OA is C.

How can I know how many numbers satisfy the equation? Should I make the list? Experts, I'd appreciate your help.
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by EconomistGMATTutor » Tue Dec 05, 2017 10:32 am
Hello Vincen.

Let's see.

According to the definition of [x], we are interested in find all the values n such as $$17\le\left[\sqrt{n}\right]<18.$$ This is equivalent to find the values n such as $$289\le\left[\sqrt{n}\right]<324\ or\ 289\le\left[\sqrt{n}\right]\le323.$$ Now, in this range there are 323-289+1=35 numbers.

So the correct answer is C.

I hope this may help you.

Feel free to ask me again if you have a doubt.

Regards.
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by Scott@TargetTestPrep » Thu Jan 11, 2018 4:40 pm
Vincen wrote:[x] is the greatest integer less than or equal to the real number x. How many natural numbers n satisfy the equation $$\left[\sqrt{n}\right]=17?$$ A. 17
B. 34
C. 35
D. 36
E. 38
We see that the smallest value n can be is 17^2 = 289 since √289 = 17 and [√289] = [17] = 17.

The largest value n can be is 1 less than 18^2, i.e., n = 18^2 - 1 = 323.

We see that since √323 is still between 17 and 18, we have [√323] = 17 (but √324 = 18 and [√324] = [18] = 18).

Thus the number of natural numbers n can be is all the natural numbers between 289 and 323, inclusive.

There are 323 - 289 + 1 = 35 natural numbers between 289 and 323, inclusive.

Answer: C

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