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If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes

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If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes [#permalink] New post   11 Mar 2020, 02:33
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If \(0 < x < 1000\), and \([\frac{x}{2}] + [\frac{x}{3}] + [\frac{x}{5}] = \frac{31x}{30}\) where [x] denotes the greatest integer less than or equal to x, then the number of possible values of x will be


A. 30
B. 33
C. 43
D. 44
E. 45

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If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes [#permalink] New post   11 Mar 2020, 02:48
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If \(0<x<1000\), and \([\frac{x}{2}]+[\frac{x}{3}]+[\frac{x}{5}]=\frac{31x}{30}\) where [x] denotes the greatest integer less than or equal to x, then the number of possible values of x will be


The value of \([\frac{x}{2}]+[\frac{x}{3}]+[\frac{x}{5}]\) will be always INTEGER.
--> The value of \(\frac{31x}{30}\) must be integer too.

In order the value of \(\frac{31x}{30}\) to be integer, \(x\) must be \(30\), \(60\), \(90\)..... \(990\)

\(x_1= 30\)

\(x_n = 990\)

--> \(n = \frac{(990-30)}{30} +1 = 33\)

Answer (B)
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Re: If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes [#permalink] New post   11 Mar 2020, 02:57
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Any number can be written as sum of integral part and fraction part.

A=[A]+{A}
[] represents integral part
{} represents fraction part

[A]=A-{A}

[x/2] + [x/3] + [x/5]= x/2-{x/2}+x/3-{x/3}+x/5-{x/5}

31x/30 - {x/2}-{x/3}-{x/5}= 31x/30

{x/2}+{x/3}+{x/5}=0

sum of 3 non-negative numbers is zero, if each of them is zero

or x is a multiple of 2,3 and 5

x= LCM(2,3,5)k= 30k, where k is an integer

0<30k<1000

0<k<33.xyz

k can take 33 values



Bunuel wrote:
If \(0 < x < 1000\), and \([\frac{x}{2}] + [\frac{x}{3}] + [\frac{x}{5}] = \frac{31x}{30}\) where [x] denotes the greatest integer less than or equal to x, then the number of possible values of x will be


A. 30
B. 33
C. 43
D. 44
E. 45

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Re: If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes [#permalink] New post   03 Jun 2020, 11:05
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See the attachment. Answer is B
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If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes [#permalink] New post   Updated on: 11 Mar 2020, 02:52
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Bunuel wrote:
If \(0 < x < 1000\), and \([\frac{x}{2}] + [\frac{x}{3}] + [\frac{x}{5}] = \frac{31x}{30}\) where [x] denotes the greatest integer less than or equal to x, then the number of possible values of x will be


A. 30
B. 33
C. 43
D. 44
E. 45




\([\frac{x}{2}] + [\frac{x}{3}] + [\frac{x}{5}] = \frac{31x}{30}\) = Integer

I.e. x Must be a multiple of 30 for 31x/30 to be an Integer

at x = 30, \([\frac{30}{2}] + [\frac{30}{3}] + [\frac{30}{5}] = 15+10+6 = 31 =\frac{31x}{30}\)

Since all options suggest that total possible values of x must be 30 or more

where total values of x as per range 0 < x < 1000 = 1000/30 = 33

Let's try with biggest value of x

at x = 990, \([\frac{990}{2}] + [\frac{990}{3}] + [\frac{990}{5}] = 495+330+198 = 1023 = \frac{31x}{30}\)

I.e. Total possible values of x must be 33 hence

ANswer: Option B

All values of x from 30 to 990 satisfy the given relation

Originally posted by GMATinsight on 11 Mar 2020, 02:45.
Last edited by GMATinsight on 11 Mar 2020, 02:52, edited 1 time in total.
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Re: If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes [#permalink] New post   01 Jul 2023, 03:39
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Re: If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes [#permalink]
01 Jul 2023, 03:39
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