pr. 10, test 6

[irs031]

When the integer b is divided by 12, the remainder is 4, but when the integer c is divided by 12, the remainder is 6. If b and c are both positive, what is the largest integer that must be a factor of the product, bc?

Explanation:
We can use Remainder notation to write equations for both b and c. If the remainder is 4 when the integer b is divided by 12, we can write b as b = 12 p + 4, where p is an integer greater than or equal to 0. We can do the same for c: c = 12 q + 6, where q is also an integer greater than or equal to 0. (Note that we must use different multiplier variables, p and q, because we do not know that these will be the same for b and c.)
Now, the product bc is (12 p + 4)(12 q + 6). We know that b and c are both positive, so p and q must be greater than or equal to 0, but there is no way to know their specific values. Therefore, we can only determine factors of the product bc through factoring:
(12 p + 4)(12 q + 6) = (4)(3 p + 1)(6)(2 q + 1) = (24)(3 p + 1)(2 q + 1)
The greatest integer that must be a factor of the product bc is therefore 24.
The correct answer is 24.


I am a little bit confused with this problem.
The first one was the wording. The largest factor of a number is the number itself. For example the largest factor of 100 is 100, because 100/100 = 1.
It appears as though the problem is asking to find the actual value bc (not possible, given that we have two variables and one equation - (24)(3p+1)(2q+1)).
However let's assume that 24 is the correct answer and backsolve.
b/12 = p + 4
b could be 8, thus 8/12 = 0 + 4/12
c/12 - q + 6
c could be 6, thus 6/12 = 0 + 6/12
8x6 = 48
In this case 24 is a factor of 48. In fact it is the largest factor of 48, after 48 itself of course.


But lets pick different numbers:
b could be 16, thus 16/12 = 1 + 4/12
c could be 18, this 18/12 = 1 + 6/12
16x18 = 288
24 is a factor of 288, however it is not the largest factor of 288.
The largest factor of 288 (after 288 itself) would be 144, not 24.

I spent quite a bit of time on this problem. It is a devilish problem, so there still might be something that I am missing. Thanks for the help in advance.

[tommywallach]

Hey IRS,

You're right that you're mistaking the logic, but all your math is correct. They're asking "what is the largest integer that MUST be a factor of bc," not "What is the largest factor of bc". Do you see the difference? Here's a new question

Integer x is a multiple of 8. What is the largest integer that must be a factor of x?

The answer here is 8. Yes, x could be 16, but the largest integer that MUST be a factor of x, no matter what x is, is 8. You see? It wouldn't be 16, because if x is only 8, then 16 doesn't go into 8. Similarly, if bc is 24, then 144 and 288 don't go into it, which means they can't be described as factors that must go into bc.

-t

[irs031]

Tommy,

Thanks. That helped a lot.

I believe that the biggest challenge on these devilish gre math problems is not solving them correctly, but correctly interpreting what the problem is asking you to solve.

[tommywallach]

Absolutely!

-t