We can solve this problem using either logic or just picking numbers. I'll use a combination of both.

If we call c 15, and d 21, then the greatest common factor, m, is 3. Let's look at the answer choices:

A) c + d = 36, and the greatest common factor of c and c + d is 3, which is m. So using the numbers we've chose this appears to work. Logic is a bit more airtight, but tougher. But if you factor out the greatest common factor from c + d, you'd get m(leftovers of c + leftovers of d). We know that there are no factors in common in the leftovers of c and d since if there were, it would be included in m, so we therefore know that m is the GCF of c and c + d. So it's A.

B) Using our picked numbers, 2 + d = 23, which is a prime number and has no common factors with 15, so B is out. Logically, there's no reason to think that adding two to D will allow it to have a common factor with C.

C) cd = 15x21 = 315. The GCF of 15 and 315 is 15 itself. Logically, that makes sense: cd is simply some multiple of c, so c has to be the GCF of the two of them.

D) 2d = 42, and the GCF of 15 and 42 is 3, or m. But does it have to be? We've just put in an extra 2. What if c had had a 2 in it? For example, if we'd picked c = 6 and d = 21, their GCF is still 3, but the GCF of 6 and 2d, or 42, is now 6. So this one's out.

E) d^2 = 21^2 which is 441. (This should be on your list of things to memorize, but if you haven't, you could always just make it a smaller number that you do know the square of.) The GCF of 15 and 441 is 3, or m, so this looks good. But again, it doesn't have to work. What if c had had a square in it that d didn't have, but when you squared d it did have it? Let's say c = 45 and d = 21. So m is still 3 and d^2 is still 441. But now we know d^2 has 9 as a factor, and so does 45. Since 9 isn't m, E is out.

So it's A.
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Let’s let c = 4 and d = 6, so m = GCF(4, 6) = 2. Let’s analyze each choice.

A. c + d = 10, and GCF(4, 10) = 2, so A could be the answer.

B. 2 + d = 8, and GCF(4, 8) = 4, so B could not be the answer.

C. cd = 24, and GCF(4, 24) = 4, so C could not be the answer.

D. 2d = 12, and GCF(4, 12) = 4, so D could not be the answer.

E. d^2 = 36, and GCF(4, 36) = 4, so E could not be the answer.


Answer: A
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Can C in the above math equal M- where M is the greatest common factor?
Can we assume that C=2, D=8, M=2, or C and M are different integers ?
Please explain . Regards in advance

Assume c=6,d=3 (GCF = 3) as GCF of c and d is m
Now, evaluating the answer options and finding the GCF along with c=6

A. c + d = 9 (GCF remains 3)
B. 2 + d = 5 (GCF becomes 1)
C. cd = 18 (GCF becomes 6)
D. 2d = 6 (GCF becomes 6)
E. d^2 = 9 (GCF remains 3)

To eliminate between A and E, let c=12,d=6 (GCF = 6)

A. c + d = 18 (GCF remains 6)
E. d^2 = 36 (GCF becomes 12) (Option A)

It is assumed that the two numbers are different
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can someone prove this question for the pair (6,8). If not, why?

If c = 6 and d = 8, then we find that answer choices A, B, D and E all work.
That is, we can only eliminate C when we test c = 6 and d = 8
This means we have to now test another pair of values.


Likewise, if we use c = 2 and d = 2, then we find that EVERY answer choice works.

Cheers,
Brent
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