If m is a positive integer and f and g are factors of m

You can consider any two factors of m. What makes you think that the product of the two factors must necessarily total m?

f and g are factors of m.

We are not told anything about f and g, one of them could easily be m itself and other something greater than 1.

I. \(\frac{m}{f+g}\)=\(\frac{m}{m+g}<1\)…..example \(\frac{6}{6+3}\)

II. \(\frac{m}{fg}=\frac{m}{mg}=\frac{1}{g}<1\) …..example \(\frac{6}{6*3}\)

III. \(\frac{f}{g}=\frac{f}{m}<1\) …..example \(\frac{3}{6}\)

None

A

I) m=6, f=2,g=3; 6/(2+3)=Non-Int
II) m=6,f=2,g=6 ; m/fg=Non-integer
III) f=2,g=3, f/g=Non-Int

A)

m=6
f,g from the set {1,2,3,6}
then found that statement I and II are wrong, ruling out options BCDE.
No need to calculate statement III.

M = f x g

Take M = 10, factors of 10 = 1, 2, 5, 10. take f = 5, g = 10

M/f+g = 10/15 – not an integer
m/fg = 10/50 – not an integer
f/g = 5/10 – not an integer

Ans: None must be an integer

Is this type pf question where we must just plug in? Until u get the answer?

Hey people, interesting question. For this one, perhaps using a DISPROVE approach may help.

One thing you could start off with is creating a table to get different factors for a specific m value (e.g. 20)

1 times 20
2 times 10
4 times 5

We have a lot of different factors to work with, but we're trying to disprove the statements, so trying to pick factors that would likely lean towards doing so could be helpful. Perhaps you quickly note that f = 5 and g = 10 may do so.

Let m = 20, f = 5, g = 10

When it comes to Statement 1, inputting these figures doesn't result in an integer.

\(\frac{20}{5+10}\)

When it comes to Statement 2, if we use these same figures again we don't end up with an integer.

\(\frac{20}{ (5) (10)}\)

When it comes to Statement 3, similarly again we don't end up with an integer.

\(\frac{5}{10}\)

Conceptually, we know for something to result in an integer, the numerator has to be more than (or at least equal) to the denominator. Technically, we therefore didn't have to plug anything in for Statement 3.

For each expression:

I. m / (f + g) is not always an integer. Example: m = 6, f = 2, g = 3 → 6 / (2 + 3) = 6 / 5.

II. m / (fg) is not always an integer. Example: m = 6, f = 2, g = 6 → 6 / (2 * 6) = 6 / 12.

III. f / g is not always an integer. Example: f = 2, g = 3 → 2 / 3.

None of these expressions must always be an integer.

Answer: A (None)

I have identified a unique approach to solve it, instead of trial and error where coul be taken up:

Let f=m/a; g=m/b (Because f and g are factors of m)

Use these two expressions in the options given. In every option, the denominator will never be 1.

Therefore, none of the options.