[From PR 1012 p.95]
If x is the product of all the integers between the integers a and b, non-inclusive, is 5 a factor of x?
(1) b - a = 7
(2) b = 9
Ans : A
- The explanation given in the book is as follow;
Plug In.
For statement (1), let a = 1 and b = 8.
Then x = 2 x 3 x 4 x 5 x 6 x 7, which has 5 as a factor.
Plug in again : if a = 15 and b = 22,
Then x = 16 x 17 x 18 x 19 x 20 x 21, which also has 5 as a factor.
Any five consecutive numbers has exactly one multiples of 5, so x will always have at least one multiples of 5 as a factor. Statement (1) is sufficient, so eliminate choices (B), (C), and (E).
statement (2) tells you that b = 9. Plug in 7 for a, making x = 8, which does not have 5 as a factor. Now plug 4 in for a, which makes x = 5 x 6 x 7 x 8. Here x does have 5 as a factor, so statement (2) is insufficient.
- My answer was C
The given explanation for the answer seems to ignore that a and/or b can be a negative number. If we set b = 4, a = -3 and then x = -2 x -1 x 0 x 1 x 2 x 3 = 0 which does not have 5 as a factor. But if (b) is true, a can not be a negative number (min possible number for a being 2) and it will always contain multiples of 5.
Am I mistaken? or is this a typo?
If x is the product of all the integers between the integers a and b, non-inclusive, is 5 a factor of x?
(1) b - a = 7
(2) b = 9
Ans : A
- The explanation given in the book is as follow;
Plug In.
For statement (1), let a = 1 and b = 8.
Then x = 2 x 3 x 4 x 5 x 6 x 7, which has 5 as a factor.
Plug in again : if a = 15 and b = 22,
Then x = 16 x 17 x 18 x 19 x 20 x 21, which also has 5 as a factor.
Any five consecutive numbers has exactly one multiples of 5, so x will always have at least one multiples of 5 as a factor. Statement (1) is sufficient, so eliminate choices (B), (C), and (E).
statement (2) tells you that b = 9. Plug in 7 for a, making x = 8, which does not have 5 as a factor. Now plug 4 in for a, which makes x = 5 x 6 x 7 x 8. Here x does have 5 as a factor, so statement (2) is insufficient.
- My answer was C
The given explanation for the answer seems to ignore that a and/or b can be a negative number. If we set b = 4, a = -3 and then x = -2 x -1 x 0 x 1 x 2 x 3 = 0 which does not have 5 as a factor. But if (b) is true, a can not be a negative number (min possible number for a being 2) and it will always contain multiples of 5.
Am I mistaken? or is this a typo?
