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Haven't quite understood the Question;Can someone help what is it exactly asking?
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Bunuel
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Haven't quite understood the Question;Can someone help what is it exactly asking?

"[x] is the greatest integer less than or equal to the real number x" means that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

Now, since \([\sqrt{n}] = 17\), then \(17\leq{\sqrt{n}}<18\) --> ANy number from this range when rounded down to the nearest integer is 17.

Hope it helps.

We are only told that "[x] is the greatest integer less than or equal to the real number .."

And the answer is given as 1, so this makes 17<= sqrt(n) < 18

Where does the 18 come from? why cant it be 17<= sqrt (n) < 20??

Thanks
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Bunuel
pavan2185
Haven't quite understood the Question;Can someone help what is it exactly asking?

"[x] is the greatest integer less than or equal to the real number x" means that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

Now, since \([\sqrt{n}] = 17\), then \(17\leq{\sqrt{n}}<18\) --> ANy number from this range when rounded down to the nearest integer is 17.

Hope it helps.

We are only told that "[x] is the greatest integer less than or equal to the real number .."

And the answer is given as 1, so this makes 17<= sqrt(n) < 18

Where does the 18 come from? why cant it be 17<= sqrt (n) < 20??

Thanks

\([x] = 17\) means that \(17\leq{x}<18\), because ANY number from this range rounded down to the nearest integer is 17. It cannot be \(17\leq{x}<20\), because if x is for example 19,5, then [19.5]=19 and not 17.

Hope it's clear.
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Bunuel
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[x] is the greatest integer less than or equal to the real number x. How many natural numbers n satisfy the equation \([\sqrt{n}] = 17?\)

A. 17
B. 34
C. 35
D. 36
E. 38

From above we have that \(17\leq{\sqrt{n}}<18\) --> \(289\leq{n}<324\). Thus n can take 35 integer values from 289 to 323, inclusive.

Answer: C.
Bunuel,

Shouldn't it be upto 17.5 then because you could round it off to 18 if it was 17.6?
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Bunuel
atalpanditgmat
[x] is the greatest integer less than or equal to the real number x. How many natural numbers n satisfy the equation \([\sqrt{n}] = 17?\)

A. 17
B. 34
C. 35
D. 36
E. 38

From above we have that \(17\leq{\sqrt{n}}<18\) --> \(289\leq{n}<324\). Thus n can take 35 integer values from 289 to 323, inclusive.

Answer: C.
Bunuel,

Shouldn't it be upto 17.5 then because you could round it off to 18 if it was 17.6?

No. "[x] is the greatest integer less than or equal to the real number x" means that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

17.6 rounded down to the nearest integer is also 17.

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Hope it helps.
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I had a slightly different approach. the Q says [x] is GREATEST integer is less than equal to x --> so it is approaching x value and we are to consider the lower values of any decimal/ non integer value given as the value of x.
now { sq rt n} = 17
we know sq root 16 = 4; so root of 17 is about 4.5 or little more than 4 but not 5 (approaching 5)
so we get 1 -2 (10 values in decimals); 2 - 3 (10 more values); 3 - 4 (10 more) and 4 - 5 (10 more) -- I did this because the it is approaching x and we are to consider values which are lesser than x.
Total 40 values --> 40 - 5 (5 for repeat/ overlap) = 35 values
Would this reasoning be correct?
I feel Bunuel's method was much more succinct though!
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Bunuel
atalpanditgmat
[x] is the greatest integer less than or equal to the real number x. How many natural numbers n satisfy the equation \([\sqrt{n}] = 17?\)

A. 17
B. 34
C. 35
D. 36
E. 38

From above we have that \(17\leq{\sqrt{n}}<18\) --> \(289\leq{n}<324\). Thus n can take 35 integer values from 289 to 323, inclusive.

Answer: C.

Hi Bunuel,

I was able to identify 289<n<=324, but I came with answer 36 because (324-289) + 1 = 36.

I think my mistake was I included 289 also in my calculation by adding 1. Whereas 289<n, so n should start from 290. Hence the OA is 35. Am I correct?
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Bunuel
atalpanditgmat
[x] is the greatest integer less than or equal to the real number x. How many natural numbers n satisfy the equation \([\sqrt{n}] = 17?\)

A. 17
B. 34
C. 35
D. 36
E. 38

From above we have that \(17\leq{\sqrt{n}}<18\) --> \(289\leq{n}<324\). Thus n can take 35 integer values from 289 to 323, inclusive.

Answer: C.

Hi Bunuel,

I was able to identify 289<n<=324, but I came with answer 36 because (324-289) + 1 = 36.

I think my mistake was I included 289 also in my calculation by adding 1. Whereas 289<n, so n should start from 290. Hence the OA is 35. Am I correct?

\(289\leq{n}<324\). So, n can take all integer value from 289 to 323 inclusive: 323-289+1 = 35.
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