How many integer values of x satisfy x(x + 2)(x + 4)(x + 6)

Question

How many integer values of x satisfy  x(x + 2)(x + 4)(x + 6) < 200 ?

A. 6 
B. 7
C. 8
D. 9
E. 10

nick1816 · Accepted Answer

put x=a-3

(a-3)(a-1)(a+1)(a+3) < 200

(a^2-9) (a^2-1) <200

a^4 -10a^2 +9 <200

a^4 -10a^2 +9+16 <200+16

a^4 -10a^2 +25 <216

(a^2-5)^2 <216

- 15< a^2- 5 <15

-10 <a^2 <20 ........(1)

We know that square of any integer is non-negative, hence a^2>0.......(2)

From (1) and (2)

0<a^2<20

-\sqrt{20}<a< \sqrt{20}

-4.xyz < a < 4.xyz

a or x can take 9 integral values.

GMATWhizTeam · Answer

Solution

Number of integral solution of \(x(x+2)(x+4)(x+6)<200 =\) Number of integral solution of \(x(x+2)(x+4)(x+6) ≤ 0 \)+ Number of integral solution of \(0 < x(x+2)(x+4)(x+6) < 200\)

Integral solutions of \(x(x+2)(x+4)(x+6) ≤ 0\)

Integral solutions of \(0 < x(x+2)(x+4)(x+6) < 200\)

• This is one possible solution.

• This is not a possible solution.

• So, we no need to check further.

• This is a possible solution

• This is one possible solution.

• This is not a possible solution.

• So, we no need to check further.

Therefore, the number of integral solution of \(x(x+2)(x+4)(x+6)<200 = 6 +3 = 9\)
Thus, the correct answer is Option D.

yashikaaggarwal · Answer

IMO is D 9 values.
Starting from 1 to the left side of number line.
X(X+2)(X+4)(X+6)<200

Putting 2 as x, LHS will be 384>200(wrong)
Put x=1;105<200
Similarly putting x=0,-1,-2,-3,-4.....up to -7
All will give value less than 200
So Answer should be 9

Other way to do this is to put x=-6 in equation. You will get zero. So we know value of x from 1 to -6 is definitely less than 200
We just have to check for -7.

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LeenaPG · Answer

9

Steps:

first check with +6 as thats the greatest integer - not possible

Then check with -6 - possible, -7 is also possible.

0 to -7 is possible - 8 integers including +1

So totally 9 integers.

freedom128 · Answer

How many integer values of x satisfy: x(x + 2)(x + 4)(x + 6) < 200?

Let y=x+3

(y-3)(y-1)(y+1)(y+3) < 200
(y^2-9)(y^2-1) - 200 < 0
y^4 - 10y^2 - 191 < 0
(y^2-5)^2 < 216

-√216 < y^2-5 < √216, where √216 = ~15
-15 < y^2-5 < 15
-10 < y^2 < 20
---> 0 <= y^2 < 20, where y^2 must be non-negative
---> y = -4,-3,-2,-1, 0, 1, 2, 3, 4 (9 values)

Since y=x+3 and x is an integer, x also has 9 values: -7,-6,..., 0, 1

FINAL ANSWER IS (D)

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unraveled · Answer

How many integer values of x satisfy x(x+2)(x+4)(x+6)<200 ?

A. 6
B. 7
C. 8
D. 9
E. 10

x(x+2)(x+4)(x+6)<200

With four integers each 2 more than the previous multiplying, max. value of x is 1.
x = 1, 1*3*5*7 = 105 < 200
 x = 2, 2*4*6*8 = 384 < 200

Similarly, x will have other negative integer values including '0'. 
Values till x = -6 satisfy
x = -7, -7*-5*-3*-1 = 105 < 200
 x = -8, -8*-6*-4*-2 = 384 < 200

Total values of x = 1-(-7) + 1 = 9

Answer D.

exc4libur · Answer

x(x+2)(x+4)(x+6)<200 x=1: 1.3.5.7=35.3=105 x=2: 2.4.6.8=64.6=360+24>200 x=-7: 7.5.3.1=105 x=-8: 8.6.4.2=384>200 -8

MayankSingh · Answer

IMO D

x(x+2)(x+4)(x+6)<200

P= x(x+2)(x+4)(x+6) = Product of 4 consecutive number differing by 2

x= 2 , P= 2*4*6*8 > 200

x=1, P=1*3*5*7 <200
........
x= -7 , P = -7*-5*-3*-1 <200

x=-8 , P= -8*-6*-4*-2 >200

Total values of x = [-7, 1] = 9

D. 9

Archit3110 · Answer

substitute for values of x we get
that given relation is satisfied at x = -7,-6,-5,-4,-3,-2,-1,0,1,
total 9 values of x
OPTION D

How many integer values of x satisfy x(x+2)(x+4)(x+6)<200?

A. 6
B. 7
C. 8
D. 9
E. 10

Kinshook · Answer

Asked: How many integer values of x satisfy  x(x + 2)(x + 4)(x + 6) < 200 ?

if x =1; 1*3*5*7 = 105 < 200
If x =2; 2*4*6*8 = 2^4*4! = 16*24 = 384 > 200
If x=0; 0 < 200
If x=-1; -1*1*3*5 = - 15 < 200
..
..
If x=-7; -7*-5*-3*-1 = 105<200
x = {1,0,-1,-2,-3,-4,-5,-6,-7}; 9 solutions

IMO D

ScottTargetTestPrep · Answer

If  x is 0, -2, -4, or -6 , we see that the value of the expression on the left hand side of the expression will be 0 and thus less than 200.

If  x is -1 , then one factor (the first one) is negative while the other three factors are positive. Therefore, the value of the expression is negative and less than 200. Similarly, if  x is -5 , then one factor (the last one) is positive while the other three factors are negative. Therefore, the value of the expression is negative and less than 200.

If  x is -3 , then the first two factors are negative and the last two are positive. Therefore, the value of the expression is positive. It might or might not be less than 200, so we have to check:

-3(-1)(1)(3) = 9 < 200 → Yes!

If x is greater than 0, then all the factors are positive. Therefore,the value of the expression is positive. It might or might not be less than 200, again we have to check:

x = 1 : 1(3)(5)(7) = 105 < 200 → Yes!

x = 2: 2(4)(6)(8) = 8(48) < 200 → No!

We see that we don’t need to check any integer values greater than 2 since the value of the expression is already greater than 200 when x = 2. Last but not least, if x is less than -6, then all the factors are negative. Therefore,the value of the expression is positive. It might or might not be less than 200, so we have to check:

x = -7 : -7(-5)(-3)(-1) = 105 < 200 → Yes!
 
x = -8: -8(-6)(-4)(-2) = 48(8) < 200 → No!

Again, we see that we don’t need to check any integer values greater than -8 since the value of the expression is already greater than 200 when x = -8. Therefore, there are 9 integer values (-7 to 1, inclusive) that satisfy the given inequality.

Answer: D

pappal · Answer

x can't be less than -7 and greater than 1.
as the value will then be for x=-8 as -8*-6*-4*-2>200 
and for x=2 as 2*4*6*8>200
for any value of x from -7 through 1 the value of the expression is less than 200
hence x={-7,1} i.e. total 09 values 
D

KarishmaB · Answer

Here are a few things one can see: If any factor is 0, the product goes to 0 and is less than 200. 
The factors are 4 alternate numbers. Taking x as 1, gives us 1*3*5*7 which is 105 < 200
But taking x as 2 gives us 2*4*6*8 which is 8*48 much greater than 200.

We know that the product can be 0 or negative too. At this point, it is a good idea to draw the number line.

(-7)....(-6).....(-5)....(-4)....(-3)....(-2)....(-1)...(0)....(1)....(2)....(3)....(4)....(5)....(6)....(7)

x can take all values from -7 to 1. It cannot be take -8 or less and it cannot take 2 or more (since product increases to more than 200 in those cases)

Hence total 9 cases.

Answer (D)

Kinshook · Answer

How many integer values of x satisfy x(x + 2)(x + 4)(x + 6) < 200 ? x=0; x(x + 2)(x + 4)(x + 6) = 0 < 200 x = 1; x(x + 2)(x + 4)(x + 6) = 1*3*5*7 = 105 < 200 x = 2; x(x + 2)(x + 4)(x + 6) = 2*4*6*8 = 384 > 200 x = 0 & x = 1 satisfy x(x + 2)(x + 4)(x + 6) < 200 x=-1; x(x + 2)(x + 4)(x + 6) = -1*1*3*5 = -15 < 200 x=-2; x(x + 2)(x + 4)(x + 6) = -2*0*2*4 = 0 < 200 x=-3; x(x + 2)(x + 4)(x + 6) = -3*-1*1*3 = 9 < 200 x=-4; x(x + 2)(x + 4)(x + 6) = -4*-2*0*2= 0 < 200 x=-5; x(x + 2)(x + 4)(x + 6) = -5*-3*-1*1 = 15 < 200 x=-6; x(x + 2)(x + 4)(x + 6) = -6*-4*-2*0= 0 < 200 x=-7; x(x + 2)(x + 4)(x + 6) = -7*-5*-3*-1 = 105 < 200 x=-8; x(x + 2)(x + 4)(x + 6) = -8*-6*-4*-2= 384 > 200 x = {-7,-6,-5,-4,-3,-2,-1,0,1} : 9 integer values satisfy x(x + 2)(x + 4)(x + 6) < 200 IMO D

How many integer values of x satisfy x(x + 2)(x + 4)(x + 6) < 200?

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