k, m, and n are integers. Is k/(m^2) an integer? 1) k/(n^2)

Question

k, m, and n are integers. Is k/(m^2) an integer?

1) k/(n^2) is an integer.

2) n/m is an integer

HongHu · Answer

(I) k=ln^2 not sufficient
(II) n=km not sufficient

Combine:
k=l(km)^2=lk^2m^2
k/m^2=lk^2 is an integer
sufficient

(C)

christoph · Answer

C ?

1) k>=n => insuff

2) n>=m => insuff

1)+2) k>m and k is factor of m => suff

mdf2 · Answer

agree with c, 
together sufficient

tareks · Answer

I think i follow the logic here, BUT the way the equations are set up with all the tiny letters isnt as clear. 

1. No relationship with k/m2- INS

2. Again no relationship - INS

Combined:

n/m= Int. so n= (Int)(m)

Substitute k/(Int x m)2= INTEGER SUFFICIENT

Does that seem right? Ian77 any feedback or simpler solution. I followed your previous example, just looked at the relationships then when combining I substituted. Anything else to keep in mind

ian7777 · Answer

That's basically right. The way I read it is like this:

1. k/n^2 is an integer has nothing to do with m.
2. n/m is an integer has nothing to do with k.

Together, I see that m divides perfectly into n, and n perfectly into k. Since m is inside m, then m must be inside n when n^2 is going into k, so m^2 must also go into k.

Think about the fact that 24/12 is an integer and 12/6 is an integer. That must mean that there's a 6 in 12, and since everything that's in 12 cancels out with 24, then the 6 must have cancelled out, so 24/6 must also be an integer.

Basically, if a number is divisible by another number, it will also be divisible by all of the second number's factors.

k, m, and n are integers. Is k/(m^2) an integer? 1) k/(n^2)

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