Bunuel
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
Great question
BunuelWe 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