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Factors

Expert replies
by panda8989 » Fri Oct 23, 2009 1:19 pm
[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?
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Source: — Data Sufficiency |

by mehravikas » Fri Oct 23, 2009 2:09 pm
your approach is correct and a = -3, b = 4 are valid values. What you are missing is that '0' is between -3 and 4, the product of all integers i.e. x will be 0 and remember 0 is a multiple of every number.

Therefore you can say that 5 is a factor of x.

Hope that's clear
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by panda8989 » Fri Oct 23, 2009 2:24 pm
Ahh! Thanks mehravikas for clearing this up for me.

I checked to see if 5 is a factor of 0, and it is!
Another pitfall in my logic :D
Date: 10/21/2005 at 15:38:11
From: Doctor Vogler
Subject: Re: factors of zero

Hi Joseph,

Thanks for writing to Dr. Math. There are two common definitions for
"factors" of a number. One is natural numbers, and the other is
integers.

If you are only considering natural numbers, then you ask: What are
the natural numbers (factors) that can be multiplied by other natural
numbers (corresponding factors) to get n? For the natural number 12,
those are the ones you listed:

1, 2, 3, 4, 6, and 12.

But if you are considering integers, then you ask: What are the
integers (factors) that can be multiplied by other integers
(corresponding factors) to get n? For the integer 12, those are:

1, 2, 3, 4, 6, 12, -1, -2, -3, -4, -6, and -12.

You asked about the factors of 0. In a similar way, you could ask
about the factors of -5. In either case, these are not natural
numbers, so it doesn't make sense to use the first definition of a
factor. But if we use the second definition of factor, then the
factors of -5 are -5, -1, 1, and 5 (which are also the factors of 5),
and every integer is a factor of zero.

In fact, zero is the only integer that has infinitely many factors,
and zero is the only integer that has zero for a factor.

Finally, I should point out one more thing: I usually interpret
"natural number" to mean a positive integer, and many mathematicians
mean exactly this. But some mathematicians use "natural number" to
mean a nonnegative integer. So that means that zero is also a natural
number. If you use that definition for a natural number, then the
first definition of factor makes sense for zero (since zero is now a
natural number) and its factors are all natural numbers (including zero).

If you have any questions about this or need more help, please write
back, and I will try to clear things up.

- Doctor Vogler, The Math Forum
https://mathforum.org/dr.math/
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by mehravikas » Fri Oct 23, 2009 2:28 pm
Great :-)

Remember that 0 is a factor is no number and multiple of every number.
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