Great question Bunuel


We can say that whatever be the value of x, the terms (x - 7), (x - 5), (x - 3) and (x - 1) are 4 consecutive even integers as x is odd.
Now, we will surely have one of the four consecutive evens to be a multiple of 8, one of 4 and other two of 2.
We could even have multiple of 16, 32 but we are looking at the least possible 2s in the product.

For example 22, 24, 26, 28: 22 and 26 are multiple of 2 only, while 24 is multiple of 8 and 28, a multiple of 4.

Thus the product of these four consecutive evens will be a multiple of 2*2*4*8 or \(2^7\). Also one of them will surely be a multiple of 3.

For, this product to be a multiple of 1920, the product should be divisible by all factors of 1920 => \(1920=2^7*3*5\)

Product is divisible by \(2^7*3\)

Now, let us check for divisibility by 5: the units digit should be 5 or 0, since we are looking at even, the units digit should be 0.

If x ends with 7, 5, 3 or 1, then (x - 7), (x - 5), (x - 3) or (x - 1) will give units digit 0.

What are we finally checking then?: Is the units digit of X 9?


(1) x + 7 is a factor of 150
Let us check for even factors of 150 as x+7 is even.
Does any 2-digit even factor of 150 has 9+7 or 16 or 6 as units digit.
150=2*3*5*5
All 2-digit even factors will be multiple of 5 and end with 0.
So answer is YES.
Sufficient

(2) x + 17 is actor of 120
Let us check for even factors of 120 as x+17 is even.
Does any 2-digit even factor of 150 has 9+17 or 26 or 6 as units digit.
120=2*2*3*5
2-digit even factors will be multiple of 5 and end with 0 or it will be 12
So answer is YES.
Sufficient


D
_________________

If x is a two-digit positive odd integer, is (x - 7)(x - 5)(x - 3)(x - 1) divisible by 1920?

\(1920=2^7*3*5\)

Now, (x - 7)(x - 5)(x - 3)(x - 1) is nothing but product of 4 consecutive even integers. These even integers will be multiple of 2*4*2*8 or 2^7.
One of them will also be a multiple of 3 for sure.
This, we have to only check for divisibility by 5.

X ending with 7, 5, 3 or 1 will surely give (x - 7), (x - 5), (x - 3) or (x - 1) respectively as a multiple of 5.

(1) x + 7 is a factor of 150
150=2*3*5*5
x+7 should not have 9+7 or 6 as units digit.
We can see no 2-digit number can be made from product of 2,3,5,5 to get 6 as units digit.
Sufficient

(2) x + 17 is actor of 120
120=2*2*3*5
x+17 should not have 9+17 or 6 as units digit.
We can see no 2-digit number can be made from product of 2,2,3,5 to get 6 as units digit.
Sufficient


D

\(10 < x < 100\)

factorizing 1920, we know that we need a 5 or 0

S1: since x is odd, x+7 has to be even
only values that satisfies this are 30 and 50
or, x = 23 or 43

we can factorize each and check \(22*20*18*16\) is divisible by \(2*3*5^2\)
similarly for x=43

In both cases, answer is yes hence suff

S2: since x is odd, x+17 has to be even
only values that satisfies this are 30, 40, 60
or, x = 13 or 23 or 43

All suff

Option D

My answer is D).

Since x is a double digit odd integer, 10<=x<=99. 1920 = 2^7 * 3 * 5, so we would need (x-7)(x-5)(x-3)(x-1) to contain those # of corresponding prime factors.

Statement 1: 17<=x+7<=106, and x+7 is even. 150 only has 2 even factors within that range: 30 and 50, so x = 23 or 43. Under both cases, (x-7)(x-5)(x-3)(x-1) is divisible by 1920. So sufficient.

Statement 2: 27<=x+17<=116, and x+17 is even. 120 only has 3 even factors within that range: 30,40,and 60, so x = 13, 23, or 43. We only need to test 13, because 23 and 43 are already covered in Statement 1. 13 also works, so sufficient.

D).

Given
x is an odd 2-digit number
(x - 7)(x - 5)(x - 3)(x - 1)- product of consequetive even numbers

1. Considering valid 2 digit factors of 150 (x+7 is a factor and x is an odd and 2-digit number) - 30,50
x=23 or 43
Putting in (x - 7)(x - 5)(x - 3)(x - 1)
16*18*20*22- divisible by 1920
36*38*40*42- divisible by 1920

2. Like 1, valid factors of 120 - 30,40 and 60
x=13,23,33
Putting in (x - 7)(x - 5)(x - 3)(x - 1)
6*8*10*12- divisible by 1920
16*18*20*22- divisible by 1920
26*28*30*32- divisible by 1920

Therefore the answer is both statements alone are sufficient. D

1920=2^7 × 3^1 × 5^1

(1) x + 7 is a factor of 150

So x is either 23 or 43

so the term becomes

22*20*18*16 = 2^8 × 3^2 × 5^1 × 11^1

or

42*40*38*36 = 2^7 × 3^3 × 5^1 × 7^1 × 19^1

Both are divisible by 1920

So (1) alone is enough

(2) x + 17 is a factor of 120

So x is either 23 or 43

So (2) alone is also enough

Answer is D

(x - 7)(x - 5)(x - 3)(x - 1)

=> 1920 = \(2^7 * 3 * 5\)

x is a two-digit odd integer

Statement 1

x + 7 is a factor of 150

=> 150 = \(2*3*5^2\)
Also, x + 7 = odd + odd = even

So, even two-digit factors of 150 are = 10, 30, and 50

But x+ 7 can not be 10, as x is an odd two-digit integer

So, x+7 = 30 or 50

Possible value of x = 23 or 43

If we take x = 23 then the given expression is divisible by 1920
Similarly, at x = 43 the given expression is divisible by 1920

Statement 1 is sufficient


Statement 2

x+17 is a factor of 120

=> 120 = \(2^3 * 3 * 5\)

=> x+17 = even => even two-digit factors of 120 are 10, 12, 20, 24, 30, 40, 60

Since x itself is an odd two-digit integer then x+17 can not be 10, 12, 20 and 24

So, possible values of x+17 = 30, 40, and 60
=> x = 13 or 23 or 43

=> the given expression will be divisible by 1920 when x = 13 or 23 or 43

So, statement 2 is also sufficient.


IMO OA should be D

S1 -> x+7 is factor of 150 -> 5^2*3*2 -> so, x is not a factor of 150, so would x-7 or any of the other given multiples in the question stem -> hence x would not have 5,3,2 as it's prime factor. Hence, it cannot be div by 1920. Suff.
S2 -> x+17 is factor of 120 -> 3*2^3*5 -> so x is not a factor of 120, so would none of the other given multiples in the question stem -> hence cannot be div by 1920. Suff.

Ans. D.
_________________

If x is a two-digit positive odd integer, is (x - 7)(x - 5)(x - 3)(x - 1) divisible by 1920?

evenxevenxevenxeven

(1) x + 7 is a factor of 150
the information give us nothing
insufficient

(2) x +17 is a factor of 120
I believe it's insufficient too

E

If x is a two-digit positive odd integer, is (x - 7)(x - 5)(x - 3)(x - 1) divisible by 1920?

(1) x + 7 is a factor of 150 ---NOT SUFF

(2) x + 17 is actor of 120----NOT SUFF

Both 1 & 2 not suff

Ans: E

Option B.

1920 = 2^7*3*5


(1) x + 7 is a factor of 150

From stmt 1, we cannot come to a definite conclusion that x+7 being a factor of 150 can divide 1920 . Insufficient.

(2) x + 17 is actor of 120

From Stmt 2, 120 is one of the factors of 1920 as well and we can come to a definite conclusion that it divides 1920. Sufficient.

Threfore, Option B.

Correct Option : B

Statement 2 is sufficient to answer

1. x is a two-digit positive odd integer
2. (x - 7)(x - 5)(x - 3)(x - 1) divisible by 1920?

(1) x + 7 is a factor of 150
(2) x + 17 is actor of 120


Explanation :
If factors are divisible by 1920

To fulfill - statement 1 (1920 must be divisible by 150) must be a integer

To fulfill - statement 2 (1920 must be divisible by 120) must be a integer
[Correct]

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1920 is prime factored down to 2^7, 3, and 5. Of note, if X is even (can't be based on the stem), we will not have enough 2s as a factor unless X itself provides 7 2s, and this string will not be divisible by 1920.

1) 150 has factors of 150&1, 75&2, 50&3, 30&5, 25&6, and 15&10. X can be any of these evens as odd + 7 = even. So lets test 150, 50 and 30 (ie x could be 143, 43, 23). In all of these cases, we get enough 2s, a 3 and a 5. Therefore this is sufficient.
2) From the note above, we know we have to test only even factors, as odd + 17 = even. Also, only even factors > 28. This gives us 30, 40, 60, and 120 (ie x could be 23, 33, 53, 113). In all of these cases, we get enough 2s, a 3 and a 5. Therefore this is sufficient.

Both stems are sufficient. D.

From Statement 1,

x=2*n+1

a*(2*n+1+7)=150

a*(n+4)=75=>a*(n+4)=3*25 or 5*15

n+4=15 or 25 because x is 2 digit number

x=23 or 43 both of these values the expression is valid

Hence Statement A is sufficient

From Statement 2,

x=2*n+1

b*(2*n+1+17)=120

b*(n+9)=60=>b*(n+9)=4*15 or 2*30

n+9=15 or 30 because x is 2 digit number

x=13 or 43 both of these values the expression is valid

Hence Statement B is sufficient

Hence Ans D

Statement 1:
x+7 is a factor of 150. x is a 2 digit number. Hence x+7 can be 30, 50, 75. x can be 23,43,68.
We don't have a definite value. Hence insufficient.

Statement 2:
x+17 is a factor of 120. x is a 2 digit number. Hence x+17 can be 30,40,60. So x can be 13,23,43
We don't have a definite value. Hence insufficient.

Combining both we get two values. 23 and 43. Again insufficient. Hence E is the correct answer.

If x is a two-digit positive odd integer, is (x - 7)(x - 5)(x - 3)(x - 1) divisible by 1920?

(1) x + 7 is a factor of 150

(2) x + 17 is actor of 120

1.

Two digit Factors of 150 to test - 25,30,50,75

but x is given as odd so only 23, 43 work

Sufficient

2.

Two digit Factors of 120 to test - 30, 40, 60

x could be 13, 23, 43

Sufficient

D

(D) is the answer IMO

\(1920 = 2^7 *5*3
\)

say x = 2n+1 where n is an integer 5<= n<=49
\((x - 7)(x - 5)(x - 3)(x - 1) = 2n(2n-2)(2n-4)(2n-6) = 2^4n(n-1)(n-2)(n-3)\)

We know product of 4 consecutive integers certainly have \(2^3 * 3\) as it's factor
from above expression we can reduce the ask of the question to only ensure whether 5 is a factor of above expression

[quote="Bunuel"]If x is a two-digit positive odd integer, is (x - 7)(x - 5)(x - 3)(x - 1) divisible by 1920?

Statement (1) x + 7 is a factor of 150

150 = 2*3*5*5
x is odd and more than 10
x+7 will certainly be even and greater than 17
Any possible value for x+7 will include 5 as factor of (x - 7)(x - 5)(x - 3)(x - 1)
Sufficient


Statement (2) x + 17 is actor of 120
120 = 2^3*5*3
x+17 certainly has to be greater than 27 and an even number
Any possible value for x+17 will include 5 as factor of (x - 7)(x - 5)(x - 3)(x - 1)
Sufficient

Hello everyone

I suppose that the correct answer is D
I am attaching my solution

factors of 1920 = 2^7*3*5
given x is a two digit odd integer
target is (x - 7)(x - 5)(x - 3)(x - 1) divisible by 1920

#1
x + 7 is a factor of 150
factor of 150 = 2*3*5^2
for x+7 to be a factor of 7 value of x has to be odd
possible such values of x = 43 & 23
upon substituting value of x in (x - 7)(x - 5)(x - 3)(x - 1) divisible by 1920 we get yes
sufficient
#2
x + 17 is actor of 120
120 factors = 2^3 *3*5
x possible values = 13,23,43
again its sufficient
upon substituting value of x in (x - 7)(x - 5)(x - 3)(x - 1) divisible by 1920 we get yes
sufficient

OPTION D is correct