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# When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]

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Manager
Joined: 15 Jan 2011
Posts: 88
When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]  [#permalink]

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06 May 2012, 07:28
6
14
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Difficulty:

55% (hard)

Question Stats:

52% (01:18) correct 48% (01:14) wrong based on 366 sessions

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When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4] = ?

A. [22]
B. [44]
C. [45]
D. [88]
E. [90]
Math Expert
Joined: 02 Sep 2009
Posts: 64249
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]  [#permalink]

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06 May 2012, 09:13
6
5
Galiya wrote:
When X is even, [x]=a/2+1, when X is odd [x]=2a+1 then [7]*[4]=?

A. [22]
B. [44]
C. [45]
D. [88]
E. [90]

i went with C and was happy till didn't see the official answer

[7]*[4]=(2*7+1)(4/2+1)=45.

Now, notice that the answer choices are also in boxes, so we have that [x]=45, which means that for even x it should be 45=x/2+1 --> x=88=even and for odd x it should be 45=x*2+1 --> x=22=even, so not a valid solution.

Similar question to practice: for-all-positive-integers-m-3m-when-m-is-odd-and-123618.html

Hope it helps.
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##### General Discussion
Manager
Joined: 15 Jan 2011
Posts: 88
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]  [#permalink]

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06 May 2012, 09:18
thanks, Bunuel! i was hoping for such an easy explanation
Manager
Joined: 15 Jan 2011
Posts: 88
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]  [#permalink]

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26 Jun 2012, 02:42
Bunuel
was reviewing my posts and got a question (im pretty sure, after the explanation ill think- OMG it was so obvious, but anyway):
looking at the question above, why do we substitute [x], e.g. [7] or [4], instead of a and b?
x and a (or b) are not the same numbers.
Math Expert
Joined: 02 Sep 2009
Posts: 64249
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]  [#permalink]

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26 Jun 2012, 02:47
1
Galiya wrote:
Bunuel
was reviewing my posts and got a question (im pretty sure, after the explanation ill think- OMG it was so obvious, but anyway):
looking at the question above, why do we substitute [x], e.g. [7] or [4], instead of a and b?
x and a (or b) are not the same numbers.

It should be: "When X is even, [x]=x/2+1, when X is odd [x]=2x+1 then [7]*[4]=?"
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Manager
Joined: 15 Jan 2011
Posts: 88
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]  [#permalink]

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26 Jun 2012, 03:00
+1 thank you for this and a lot of previous answers&explanations
Intern
Joined: 12 Dec 2013
Posts: 15
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]  [#permalink]

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03 Nov 2014, 07:47
Bunuel wrote:
Galiya wrote:
When X is even, [x]=a/2+1, when X is odd [x]=2a+1 then [7]*[4]=?

A. [22]
B. [44]
C. [45]
D. [88]
E. [90]

i went with C and was happy till didn't see the official answer

[7]*[4]=(2*7+1)(4/2+1)=45.

Now, notice that the answer choices are also in boxes, so we have that [x]=45, which means that for even x it should be 45=x/2+1 --> x=88=even and for odd x it should be 45=x*2+1 --> x=22=even, so not a valid solution.

Similar question to practice: for-all-positive-integers-m-3m-when-m-is-odd-and-123618.html

Hope it helps.

Hi Bunuel,

Why do we not consider 22 as an answer? Is it because 22 as an answer was even when equated to an odd equation, i.e 2a+1? Does the solution also
have to be odd when equated to an odd equation (2a+1)?
I hope my question makes sense ... not sure if I was able to explain myself clearly? =P Please let me know!

Thanks as always!!
Math Expert
Joined: 02 Sep 2009
Posts: 64249
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]  [#permalink]

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03 Nov 2014, 07:58
1
Rca wrote:
Bunuel wrote:
Galiya wrote:
When X is even, [x]=a/2+1, when X is odd [x]=2a+1 then [7]*[4]=?

A. [22]
B. [44]
C. [45]
D. [88]
E. [90]

i went with C and was happy till didn't see the official answer

[7]*[4]=(2*7+1)(4/2+1)=45.

Now, notice that the answer choices are also in boxes, so we have that [x]=45, which means that for even x it should be 45=x/2+1 --> x=88=even and for odd x it should be 45=x*2+1 --> x=22=even, so not a valid solution.

Similar question to practice: for-all-positive-integers-m-3m-when-m-is-odd-and-123618.html

Hope it helps.

Hi Bunuel,

Why do we not consider 22 as an answer? Is it because 22 as an answer was even when equated to an odd equation, i.e 2a+1? Does the solution also
have to be odd when equated to an odd equation (2a+1)?
I hope my question makes sense ... not sure if I was able to explain myself clearly? =P Please let me know!

Thanks as always!!

We got that [7]*[4] = 45.

Now, [22], since 22 is even, is 22/2 + 1 = 12, not 45.

Does this make sense?
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Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]  [#permalink]

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15 Mar 2020, 00:24
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Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4]   [#permalink] 15 Mar 2020, 00:24