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Hello AnirudhaS
Could you please explain why x should be 35 and y 48?
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[√x] = 5
hence, 5< or =√x <6
hence 25< or = x <36
since x is an integer, 25< or = x < or = 35
greatest value of x = 35

Similarly 36< or = y<49
since y is an integer, 36< or = y< or = 48
greatest value of y = 48

the greatest value of (x+y) =35+48 = 83
Answer D

Bunuel
If [x] is the greatest integer less than or equal to x, and [√x] = 5 and [√y] = 6, where x and y are positive integers, what is the greatest possible value of x + y ?

A. 61
B. 81
C. 82
D. 83
E. 85


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[√x] = 5
hence, 5< or =√x <6
hence 25< or = x <36
since x is an integer, 25< or = x < or = 35

similar for y.
Hope, it is clear.


Ioan554
Hello AnirudhaS
Could you please explain why x should be 35 and y 48?
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Hello AnirudhaS
Could you please explain why x should be 35 and y 48?
Of course buddy.

Assume x=36 (just for argument sake).

Then what is \([\sqrt{x}]\) according to the problem?
\([\sqrt{x}]=6\), but see the question says \([\sqrt{x}]=5\), which means x cannot be 36. Also x is a positive integer. Therefore the maximum value of x = 36-1 = 35
Similarly, extending the same logic, y=49-1 =48

i hope it is a little more clearer now.
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Bunuel
If [x] is the greatest integer less than or equal to x, and [√x] = 5 and [√y] = 6, where x and y are positive integers, what is the greatest possible value of x + y ?

A. 61
B. 81
C. 82
D. 83
E. 85


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[√x] = 5 i.e. 5 ≤ √x <6 i.e 25 ≤ x < 36

[√y] = 6 i.e. 6 ≤ √y <7 i.e. 36 ≤ y < 49

\(x_{max} = 35\) and \(y_{max} = 48\)

i.e. \((x+y)_{max.} = 35+48 = 83\)

Answer: Option D
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max value of [√x] = 35 and [√y] = 48
sum will be 83
IMO D


Bunuel
If [x] is the greatest integer less than or equal to x, and [√x] = 5 and [√y] = 6, where x and y are positive integers, what is the greatest possible value of x + y ?

A. 61
B. 81
C. 82
D. 83
E. 85


Are You Up For the Challenge: 700 Level Questions
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