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# If [x] denotes the greatest integer less than or equal to x for any nu

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If [x] denotes the greatest integer less than or equal to x for any nu  [#permalink]

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31 Oct 2014, 02:43
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47% (01:05) correct 53% (01:48) wrong based on 86 sessions

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If [x] denotes the greatest integer less than or equal to x for any number x, is [a]+[b]=1?

(1) ab=2.

(2) 0<a<b<2.

Source: GCT M28-56

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Re: If [x] denotes the greatest integer less than or equal to x for any nu  [#permalink]

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31 Oct 2014, 03:45
Nice question!

[a]+[b]=1 can happen if one value is negative and one positive OR if one value is 0 and the other is 1.

A - ab = 2
if both a and b are negative - then [a]+[b]=1 is not possible

If a and b are positive - can it result in one value being 0 (i.e. 0<a<1) and the other 0 (1<b<2)
This is impossible as if 1<b<2 then it has to be multiplied by a number larger than 1 to give 2.
But if a is larger than one then [a]+[b]=2.

So if ab=2 then it cannot happen that [a]+[b]=1 and A is SUFFICIENT

B - 0<a<b<2
If 0<a<1 and 1<b<2 then [a]+[b]=1 is true
if 0<a<1 and a<b<1 then [a]+[b]=1 is not true

So we cannot conclude and B is NOT SUFFICIENT
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If [x] denotes the greatest integer less than or equal to x for any nu  [#permalink]

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31 Oct 2014, 03:54
If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that $$0\leq{a}<1$$ and $$1\leq{b}<2$$ (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

THIS QUESTION IS DISCUSSED HERE: new-set-number-properties-149775-60.html#p1205389
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If [x] denotes the greatest integer less than or equal to x for any nu  [#permalink]

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01 Nov 2014, 03:31
Bunuel wrote:
If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that $$0\leq{a}<1$$ and $$1\leq{b}<2$$ (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

THIS QUESTION IS DISCUSSED HERE: new-set-number-properties-149775-60.html#p1205389

I think Bunuel you missed something for statement 2 what if a=1 and b=2 in this case ab =2 and [a] + [b] =3

what if X IS ROOT of 2, root of 2 X root of 2 = 2

In that case [a] + [b] = 2 (a=b = 1.414)
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Re: If [x] denotes the greatest integer less than or equal to x for any nu  [#permalink]

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01 Nov 2014, 04:46
honchos wrote:
Bunuel wrote:
If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that $$0\leq{a}<1$$ and $$1\leq{b}<2$$ (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

THIS QUESTION IS DISCUSSED HERE: new-set-number-properties-149775-60.html#p1205389

I think Bunuel you missed something for statement 2 what if a=1 and b=2 in this case ab =2 and [a] + [b] =3

what if X IS ROOT of 2, root of 2 X root of 2 = 2

In that case [a] + [b] = 2 (a=b = 1.414)

Are you talking about statement 1 or 2?

Since you are using ab = 2 there, then I guess you are talking about statement 1. Still not clear what are you trying to say... I got that [a] + [b] cannot equal to 1, which means that we have a NO answer to the question (so statement 1 is sufficient). In your examples [a] + [b] also does not equal to 1. So, they do not contradict what I wrote...
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Re: If [x] denotes the greatest integer less than or equal to x for any nu  [#permalink]

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01 Nov 2014, 05:00
Bunuel wrote:
honchos wrote:
Bunuel wrote:
If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that $$0\leq{a}<1$$ and $$1\leq{b}<2$$ (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

THIS QUESTION IS DISCUSSED HERE: new-set-number-properties-149775-60.html#p1205389

I think Bunuel you missed something for statement 2 what if a=1 and b=2 in this case ab =2 and [a] + [b] =3

what if X IS ROOT of 2, root of 2 X root of 2 = 2

In that case [a] + [b] = 2 (a=b = 1.414)

Are you talking about statement 1 or 2?

Since you are using ab = 2 there, then I guess you are talking about statement 1. Still not clear what are you trying to say... I got that [a] + [b] cannot equal to 1, which means that we have a NO answer to the question (so statement 1 is sufficient). In your examples [a] + [b] also does not equal to 1. So, they do not contradict what I wrote...

I was actually about statement 1, your inequality has left the possibility that a can be equal to 1 and b can be equal to 2. Though the result will still remain the same.
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Re: If [x] denotes the greatest integer less than or equal to x for any nu  [#permalink]

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01 Nov 2014, 05:10
honchos wrote:
Bunuel wrote:
honchos wrote:

I think Bunuel you missed something for statement 2 what if a=1 and b=2 in this case ab =2 and [a] + [b] =3

what if X IS ROOT of 2, root of 2 X root of 2 = 2

In that case [a] + [b] = 2 (a=b = 1.414)

Are you talking about statement 1 or 2?

Since you are using ab = 2 there, then I guess you are talking about statement 1. Still not clear what are you trying to say... I got that [a] + [b] cannot equal to 1, which means that we have a NO answer to the question (so statement 1 is sufficient). In your examples [a] + [b] also does not equal to 1. So, they do not contradict what I wrote...

I was actually about statement 1, your inequality has left the possibility that a can be equal to 1 and b can be equal to 2. Though the result will still remain the same.

Dear honchos, please read carefully what I wrote. For [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that $$0\leq{a}<1$$ and $$1\leq{b}<2$$ (or vise-versa). So, for [a] + [b] = 1 to hold true, b CANNOT BE 2!
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Re: If [x] denotes the greatest integer less than or equal to x for any nu  [#permalink]

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13 Mar 2018, 02:14
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