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number properties

by GmatTakerNo.1 » Sun Apr 25, 2010 3:37 am
Hey, I got some problems with this one:

If k=m(m+4)(m+5) and k and m are positive integers. Which of the following always divide k evenly?
I.3
II.4
III.6
a) only I
b) only II
c) only III
d) both I and III
e) both II and III

The answer is D. Can somebody explain it to me?
Thanks
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by pradeepkaushal9518 » Sun Apr 25, 2010 3:56 am
k=m(m+4)(m+5)

it is seen that for every value of m the above expression is divisible by 6 so as with 3.
hence d

ex m=1
k= 1*5*6
m=2

k=2*6*7

m=3
3*7*8

and so on.....
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by Stuart@KaplanGMAT » Sun Apr 25, 2010 2:09 pm
GmatTakerNo.1 wrote:Hey, I got some problems with this one:

If k=m(m+4)(m+5) and k and m are positive integers. Which of the following always divide k evenly?
I.3
II.4
III.6
a) only I
b) only II
c) only III
d) both I and III
e) both II and III
Hi!

Since m is an integer, the three terms on the right side of the equation are also integers.

Roman numeral III occurs most frequently, so let's start with that one.

For an integer to be divisible by 6, it must have factors of both 2 and 3.

(m+4) and (m+5) are consecutive, so one of those must be even: that takes care of 2.

m is 3 away from (m+3), so for purposes of having 3 as a factor, m and (m+3) are identical.

Well, if our three terms had been (m+3),(m+4) and (m+5), we'd have 3 consecutive numbers, one of which must be a multiple of 3. Since m and (m+3) are the same when it comes to divisibility by 3, we now know that our product must be a multiple of 3.

So, it's a multiple of both 2 and 3: accordingly, it's a multiple of 6. Therefore, we want III as part of our answer: eliminate A and B.

We already proved that it's a multiple of 3, so we also need I as part of our answer: eliminate C and E. Only (D) left, no need to check II!

(By the way, picking numbers is almost certainly faster!)
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