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If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:00
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If x is an odd integer and \(y = x(x+1)(x1)\), which of the following must also be an integer? I. \(\frac{y}{2^3*3}\) II. \(\frac{y}{2^2*3^2}\) III. \(\frac{y}{2*3}\) (A) I only (B) III only (C) I and III (D) I, II, and III (E) none of the above
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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:23
If x is an odd integer and y=x(x+1)(x−1), which of the following must also be an integer?
y = x(x+1)(x−1) = (x−1)x(x+1), multiple of three consecutive integers, so must be multiple of 3, & x is odd, so both (x−1) & (x+1) must be even & one of them is multiple of 4, so y must be multiple of 2^3*3 I. y/(2^3∗3 ) > must be true( 2*3*4 = 2^3*3)
II. y/(2^2∗3^2) > not must be true( false: 2*3*4 = 2^3*3 & true: 8*9*10 = 2^2*3^2*k)
III. y/(2∗3)> must be true (same as I)
(A) I only (B) III only (C) I and III > correct (D) I, II, and III (E) none of the above




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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:06
i think the answer is C since x is odd integer. Say x =1 then y=x(x^21) = 0 if x = 3 then y = 3*8 For x =5 then y=5*24 and so on.
Hence answer is C since 1 and 3 are correct.



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If x is an odd integer and y = x(x+1)(x1), which of the following mus
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Updated on: 04 Jul 2019, 08:32
If x is an odd integer, (x1) and (x+1) are even.
This implies the product x(x1)(x+1) could be 0 or could have at least three two's and at least one three in its factors
Only I and III have denominators less than or equal to the above said powers of 2 and 3
Answer is (C)
Originally posted by firas92 on 04 Jul 2019, 08:07.
Last edited by firas92 on 04 Jul 2019, 08:32, edited 1 time in total.



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If x is an odd integer and y = x(x+1)(x1), which of the following mus
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Updated on: 05 Jul 2019, 06:25
If x is an odd integer and y=x(x+1)(x−1), which of the following must also be an integer?
Let's assume x=3 (smallest odd integer which will result in positive y)
I. \(\frac{y}{2^3*3}\) > integer
II. \(\frac{y}{2^2*3^2}\)> not an integer
III. \(\frac{y}{2*3}\)> Integer
So the answer is (c)
Originally posted by harry89 on 04 Jul 2019, 08:09.
Last edited by harry89 on 05 Jul 2019, 06:25, edited 3 times in total.



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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:13
Refer image attached. Ans. C
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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:17
The smallest x=1; so y=0 If x=3; y=2*3*4 so the minimum number of 2 is three (three 2's), and the minimum number of 3 is one
I. y/(2^3*3) Must be an integer
II. y/(2^2∗3^2) Not true (two 3's)
III. y/(2∗3) Must be an integer
Answer: (C) I and III



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If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:19
First things first. Let's rewrite this as (x1)x(x+1)  we know this is a product of three consecutive integers (given x is an integer), and now since x=odd, x1 and x+1 are two consecutive even integers, one of which will always be a multiple of 4 (i.e. 2^2).
Now, since we know that of three consecutive integers, 1 is always divisible by 3 (take 0,1,2; 1,2,3; 2,3,4; 3,4,5 for instance)
Thus, till the time the denominator has exactly one 3 and upto 2^3 (one 2 and one 4 (2^2)  as explained above), it will completely divide the numerator and hence, result in an integer.
The correct answer choice is, therefore, (C) I and III



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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:23
Okey, we have, y=x(x1)(x+1) reorganize it as, y=(x1)x(x+1), then what you see is that it is in consecutive order. Here is simple formula to remember now ,i.e. the product of two consecutive integers is always divisible by 2. the product of three consecutive integers is always divisible by the L.C.M. of 2&3 wich becomes 6. The product of two consecutive even integers is always divisible by 8. The product of three consecutive even integers is always divisible by 48. Finally the rule applicable for our question is> If the three consecutive integers are on the form that the middle term is always odd then the product of three such consecutive integers is always divisible by the L.C.M of 6 and 8 i.e. obviously 24.
So from here we know option first is must true. Option 2 is sometimes true sometimes false bcoz any number bigger than L.C.M is not always the perfect divisor. For third option, you can realize that 24= 2*2*2*3. From here you can see that 24 has atleast one 2 and one 3 on it. So it will always be the divisor of the Y.



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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:25
If x is an odd integer and y=x(x+1)(x−1) , which of the following must also be an integer? I. y/2^3∗3 II. y/2^2∗3^2 III. y/2∗3 (A) I only (B) III only (C) I and III (D) I, II, and III (E) none of the above I. y/2^3∗3 For 3 consecutive integers out of the 3 numbers, one is be divisible by 3. Since middle number is odd, one of the other 2 numbers will be divisible by 4 and other will be divisible by 2. Exception is when 0 is a number but 0 divided by anything yields and integer 0. y will be divisible by 3, 4 & 2 => y/8*3 = y/2^3*3 will be an integer. MUST BE TRUE. II. y/2^2∗3^2 For 3 consecutive integer, one will be divisible by 3 but no necessarily by 9. e.g 6,7,8 => 6 is divisible by 3 but NOT by 9. NOT NECESSARILY TRUE. III. y/2∗3 For 3 consecutive integers out of the 3 numbers, one is be divisible by 3. Since middle number is odd, one of the other 2 numbers will be divisible by 4 and other will be divisible by 2. Exception is when 0 is a number but 0 divided by anything yields and integer 0. y will be divisible by 3, 4 & 2 => y/8*3 = y/8*3 will be an integer. => 4y/8*3 = y/2*3 must also be an integer. MUST BE TRUE. IMO C
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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:26
x is odd x+1 is even x1 is even y is odd x even x even
y when divided by 3 cannot give an integer.
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If x is an odd integer and y = x(x+1)(x1), which of the following mus
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Updated on: 05 Jul 2019, 09:12
Since x is odd therefore y = odd * even * even y = even hence, divisible by 2 since (x1) (x) (x+1) are consecutive integers the product must be divisible by 3 I & III
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Originally posted by GmatPrime on 04 Jul 2019, 08:29.
Last edited by GmatPrime on 05 Jul 2019, 09:12, edited 1 time in total.



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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:29
Here is my approach for the above problem.
Lets say x=1,then y=x*(x+1)*(x1)product of consecutive three numbers=1*2*0=0
Lets say x=3 then y=3*4*2=3*(2^3)
So from options we can safely assume 1 and 3 will be true.
Hence I would go for option C .



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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:35
X is odd.
So, X can be 1, 3, 5 ...
Y = (x1) *x *(x+1) So, if x=1 => y=0
And all three options will be 0 which is an integer.
Now, If x=3 Y=2*3*4 Y=24
In this case, 1 & 3 are an integer.
So, Ans should be (C)



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If x is an odd integer and y = x(x+1)(x1), which of the following mus
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Updated on: 04 Jul 2019, 22:28
x is an odd integer and y=x(x+1)(x−1) Since X is odd, for X = 1, y would be 0, But if X is 3, y = 3*4*2 = 2^3*3 would be a multiple of 2,2,2,3. x = 5, y = 5*6*4 = 2^3*3*5 Similarly for every odd integer x, y is a multiple of 2^3*3 Therefore I and III would be integers. II would not be an integer since y is divided by 3^2. So C would be the answer. Posted from my mobile device
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Originally posted by prashanths on 04 Jul 2019, 08:37.
Last edited by prashanths on 04 Jul 2019, 22:28, edited 1 time in total.



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If x is an odd integer and y = x(x+1)(x1), which of the following mus
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Updated on: 04 Jul 2019, 08:40
If x is an odd integer and y=x(x+1)(x−1), which of the following must also be an integer?
I. \(\frac{y}{2^3∗3}\)
II.\(\frac{y}{2^2∗3^2}\)
III.\(\frac{y}{2∗3}\)
(A) I only (B) III only (C) I and III (D) I, II, and III (E) none of the above
x(x+1)(x−1) means three consecutive numbers given x is odd Hence x+1 and x1 will be even now three consecutive number are always divisible are by 6 ( 2 and 3) and even 2 are even they are divisible 24 ( 2 * 2* 2 *3) for instance take\(2*3*4\) or \(16*17*18\) Hence 1 and 3 are always true middle number might or might not be divisible by 9 such as 2*3*4 not divisible by 9 ;\(16*17*18\) divisible by 9 Hence C
Originally posted by globaldesi on 04 Jul 2019, 08:39.
Last edited by globaldesi on 04 Jul 2019, 08:40, edited 1 time in total.



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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:39
x is an odd integer and y=x(x+1)(x−1)y=x(x+1)(x−1) provides 1)means y=Odd*even*even 2) any three consecutive integers will be divisible by 3, and 3) product of any two consecutive even numbers will be divisible by 8 ( one will be a multiple of 2 and other a multiple of 4, so product will be a multiple of 8) So Y will be divisible always by 3*2^3. hence I and III will always be integers. IMO C (if X is 1 then the product will be zero, which is divisible by all integers)
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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:39
Note: The product of 3 consecutive integers is divisible by 6. Here y=(x1)x(x+1) represents the product of 3 consecutive integers. Hence it must be divisible by 6(=2*3). Hence the expression III. y/(2*3) must yield an integer. Ans. (B)
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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:40
let x=3 so y = 2^3*3 out of given options 1 & 3 are correct IMO C If x is an odd integer and y=x(x+1)(x−1)y=x(x+1)(x−1), which of the following must also be an integer? I. y23∗3y23∗3 II. y22∗32y22∗32 III. y2∗3y2∗3 (A) I only (B) III only (C) I and III (D) I, II, and III (E) none of the above
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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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04 Jul 2019, 08:43
Hi, Given x is an odd integer, that means x can be (2a+1) or (2a1). So, if x = (2a+1),then y= (2a+1)(2a+2)(2a) => y = \(2^2\) * a * (a+1) * (2a+1)
and if x = (2a1), then y = (2a1)(2a)(2a2) => y = \(2^2\) * a * (a1) * (2a1)
So, for the expression to be an integer, the denominator must contain two 2 and one 3 or less than that.
let`s take, a = 2, then y = 2^3 * 3 * 5 or y = 2^3 * 1 * 3
So only choice C is possible.
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Re: If x is an odd integer and y = x(x+1)(x1), which of the following mus
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