If x is an odd integer and y = x(x+1)(x-1), which of the following mus

Okey,
we have, y=x(x-1)(x+1)
reorganize it as, y=(x-1)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.

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

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



Since y = (x-1)*x*(x+1)
I e. Product of 3 consecutive integers starting with an even number

Therefore one of three numbers must be divisible by 2
Other must be divisible by 4
And one of them must be divisible by 3

I.e. y must be a multiple of 2*4*3= 24

Hence I and III will be integers but we can't say about II due to 3^2

Answer option C

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y=(x-1)*x*(x+1)
1) y is the multiple of 3 consecutive numbers. Hence 1 out of 3 numbers must be multiple of 3.
2) x is an odd number. It implies that both x-1 and x+1 are even numbers. As x-1 and x+1 are consecutive even number, one of them is odd multiple of 2 and other number is even multiple of 2.

y=2^3*3*k, where k is an integer

I. y/(2^3*3) = 2^3*3*k/(2^3*3)=k
Always an integer

II) y/(2^2∗3^2)= 2^3*3*k/(2^2*3^2)=k/3
If k is multiple of 3, y/(2^2∗3^2) will be an integer
If k is not a multiple of 3, y/(2^2∗3^2) will not be an integer.

III. y/2∗3= 2^3*3*k/(2*3)= 2^2*k
4k is always an integer, as k is an integer.

IMO C

x is an odd integer and y=x(x+1)(x−1)

I. y/2^3∗3 = y/24

y will always be a product of 3 consecutive integers.... 1 will be odd and 2 will be even (because x is odd, x+1 and x_1 will always be even)

examples of numerator (y):
x=1 : 0*1*2
x=3 : 2*3*4
X=5 : 4*5*6
x=7 : 6*7*8
x=9 : 8*9*10
x=17 : 16*17*18
x=-3 : -2*-3*-4

All numerators ALWAYS have 2,3,and4 as factors when not 0... so numerator(y) will always be divisible by 24... so y/24 = integer (0 is also an integer)

I MUST BE INTEGER


II. y/(2^2∗3^2) = y/36
when x is 3, y=2*3*4 = 24
24/36 is not an integer

II MAY NOT BE INTEGER

III. y/2∗3 = y/6
In I. we saw that y is always divisible by 24. So, y is also always divisible by 6.

III MUST BE INTEGER

Ans. C - I and III

x is an odd integer and y=x(x+1)(x−1)
x = ....-3,-1,1,3,5,7.....

y = -3(-3+1)(-3-1) = -3*2*\(2^2\)
y = -1(-1+1)(-1-1) = 0
y = 1(1+1)(1-1) = 0
y = 3(3+1)(3-1) = 3*\(2^2\)*2
y = 5(5+1)(5-1) = 5*2*3*\(2^2\)
y = 7(7+1)(7-1) = 7*\(2^3\)*2*3

except for y=0
every value of y have \(2^3\)*3

I. \(\frac{y}{2^3*3}\)
must be integer for y=0 and every value 0f y

II. \(\frac{y}{2^2*3^2}\)
this can be true for some values such as y=0 y=\(3^2\)*\(2^4\)*5 (x=9)
but not always true if y=3*\(2^2\)*2 then it will not be an integer
since this is must be true question so out

III. \(\frac{y}{2*3}\)
for every value of y, this must be an integer

(C) I and III
C is the correct answer

If x is odd then y = odd * even * even

x could be 3 y = 2 * 3 * 4
x could be 5 y = 4 * 5 * 6
x could be 1 y = 0 * 1 * 2

Is Y divisible by 24 ?
y = 2 * 3 * 4 = Yes
y = 4 * 5 * 6 = Yes
y = 0 * 1 * 2 = Yes

Is Y divisible by 36 ?
y = 2 * 3 * 4 = No
y = 4 * 5 * 6 = No
y = 0 * 1 * 2 = Yes

Is Y divisible by 6 ?
y = 2 * 3 * 4 = Yes
y = 4 * 5 * 6 = Yes
y = 0 * 1 * 2 = Yes

I & III - Looks good , C

Solution:

This is a must be question, Therefore, all the values we get for y must satisfy the equation.

y = x(x+1) (x-2)
This tells us that Y is the product of 3 consecutive integers, since x is an odd integer, we can test some values for x, x can be 1,3,5,7 etc.
if x= 1, then y= 1 X 2 X 0 = 0. therefore all the 3 given equations are satisfied.

if x=3, then y = 3 X 4 X 2 = 24 , 1 & 3 can satisfy ( division results into an integer)

if x=5 then y = 5 X 6 X 4 = 120, 1 & 3 satisfy .

if x=7 then y = 7 X 8 X 6 = 336, 1 & 3 satisfy.

with this pattern going on , 1 & 3 always comes out as an integer . Hence the answer is C.

While there are many ways we could think about this question - I think taking it conceptually and thinking about the number properties at play simplifies the process.

In this case, if we simplify the question stem itself, we can see that the question is really asking us "what factors y must contain," if y is the product of even*odd*even consecutive integers.

If given this structure, we know that any set of even*odd*even integers will contain: a factor of 3 (as any set of 3 consecutive numbers will contain a factor of 3), a factor of 2 (one of the even numbers), and a factor of 4 (the other of the even numbers, as every other even number contains at least 2 factors of 2). Thus, we know we at the very least have divisibility by 24, or 2^3*3. I and III must be true, but we cannot be certain of II, thus our answer is (C)

(*note - this complies even if one of the values is zero, making the product zero, as zero is divisible by each of the listed values as well.)

The key to this question was effectively breaking down the question, as well as cues from the options, to better understand what the question was attempting to ask us, and what "tools in our toolkit" might be helpful in allowing us to determine what "must be" an integer out of our options.

I hope this helps!

y = (x - 1) * (x) * (x + 1) ----- which is the product of 3 Consecutive Integers, given that X = ODD Integer


y = Even * Odd * Even


In the Product of 3 Consecutive Integers that takes this FORM, the following must be true:

1st) At least 1 of the Factors must be a Multiple of 3 ---- therefore, the Product Y = Multiple of 3

2nd) Since an Even and an Odd Factor are being Multiplied ---- the Product Y = Multiple of 6

3rd) The Product of 2 Consecutive Even Factors will Always Contain a Multiple of 2 and a Multiple of 4 ----therefore the Product Y = Multiple of 8

4th)Since Y must contain at least 1 Factor that is a Multiple of 3 and the 2 Consecutive Even Factors will produce a Multiple of 8 ----- Y will also at the least be a Multiple of (8 * 3) = 24





I. Y will definitely be Divisible by 24

II. We do not know for sure whether Y is a Multiple of 36

III. Y will definitely by Divisible by 6


I and III are Correct

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