GMAT CLUB OLYMPICS: If x is a two-digit positive odd integer, is (x -

Question

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

M36-10

cs15811581 · Answer

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

chetan2u · Answer

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

Bunuel · Answer

Official Solution:

If  x  is a two-digit positive odd integer, is  (x - 7)(x - 5)(x - 3)(x - 1)  divisible by 1920? 
 
First of all, factor 1920:  1920=2^7*3*5 
 
Next, notice that since  x  is odd, then  (x - 7)(x - 5)(x - 3)(x - 1)  will be the product of four consecutive  even  integers:  (2k)(2k+2)(2k+4)(2k+6)=2^4(k)(k+1)(k+2)(k+3) . Now,  (k)(k+1)(k+2)(k+3)  is the product of four consecutive integers, thus one of them will be a multiple of 2, one of them will be a multiple of 4 ( 2^2 ) and at least one of them will be a multiple of 3:  (x - 7)(x - 5)(x - 3)(x - 1)=2^4(k)(k+1)(k+2)(k+3)=2^4*2*2^2*3*something=2^7*3*something . So, basically all we need to find out is whether  (x - 7)(x - 5)(x - 3)(x - 1)  is divisible by 5 (because we know that it's for sure is divisible by  2^7*3 ).
 
(1)  x + 7  is a factor of 150
 
Since  x  is a two-digit positive odd integer, then  x + 7  is a two-digit positive even integer. Even two-digit factors of 150 are 10, 30 and 50 only. So,  x  is either 23 or 43 ( x  cannot be 7 because we are told that  x  is a two-digit number). In both of these cases  (x - 3)  will be a multiple of 5, thus (x - 7)(x - 5)(x - 3)(x - 1)  will be divisible by 5. Sufficient.
 
(2)  x + 17  is actor of 120
 
Since  x  is a two-digit positive odd integer, then  x + 17  is a two-digit positive even integer. Even two-digit factors of 120 are 10, 12, 20, 24, 30, 40, and 60 only. So,  x  is either 13, 23 or 43 ( x  cannot be -7, -5, 3, or 7 because we are told that  x  is a two-digit positive number). In all of these cases  (x - 3)  will be a multiple of 5, thus (x - 7)(x - 5)(x - 3)(x - 1)  will be divisible by 5. Sufficient.

Answer: D

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GMAT CLUB OLYMPICS: If x is a two-digit positive odd integer, is (x -

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