[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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[x] is the greatest integer less than or equal to the
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EconomistGMATTutor
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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.
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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Scott@TargetTestPrep
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We see that the smallest value n can be is 17^2 = 289 since √289 = 17 and [√289] = [17] = 17.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
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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