If k is an even integer, what is the smallest possible value of k

Question

If k is a positive even integer, what is the smallest possible value of k such that 3675*k is the square of an integer?

A.3
B.9
C.12
D.15
E.20

acegmat123 · Answer

3675*k= 7^2*5^2*3*k

Since k is even and the integer should be a square.

k=2^2*3=12

C

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ScottTargetTestPrep · Answer

If 3675*k is a perfect square, then its prime factorization must contain only even exponents. Let’s begin by prime factoring 3,675.
 
3675 = 25 x 147 = 5 x 5 x 49 x 3 = 5 x 5 x 7 x 7 x 3 = 5^2 x 7^2 x 3^1 
 
We can see that 3,675 is not a perfect square because its prime factorization contains an odd exponent (that is, 3^1). If k only has to be an integer, then the smallest value k can be is 3, since 3675*k would be 5^2 x 7^2 x 3^2, a perfect square. However, since the requirement is that k must be an even integer, we need k to be divisible by an even perfect square (notice that 3675 doesn’t have any even prime factors). Since the smallest even perfect square is 2^2 = 4, the smallest possible value of k is 3 x 4 = 12, so that 3,675*k is a perfect square. 
In fact, if k = 12, then 3,675*k = 5^2 x 7^2 x 3^2 x 2^2, which is a perfect square.

Answer: C

stonecold · Answer

Absolutely Spot On.!!
P.S=> Edited the Question.

Abhishek009 · Answer

3675 = 3^1 * 5^2 * 7^2

Now , is the most important part of this question

K must be even ( negate options A, B and D )  W e are left with C and E

Try one to confirm the answer...

Let's try (E)

20 = 2^2 * 5^1

3675*20 = 2^2 * 3^1 * 5^3 * 7^2  ( Can not square of an Integer )

So, Answer must be (C)....

Double check to be sure -

3675*12 = 2^2 * 3^2 * 5^2 * 7^2  (Square of an Integer )
  
Hope this helps !!

stonecold · Answer

You didn't fall for it this time :twisted:

Great

PKay · Answer

I feel like killing myself for choosing A. If k is a positive even integer

EVEN digit ...grrr....

gracie · Answer

3675=35*105
if k were odd, then 3 would give us 3*35*105=105^2
so we need an even multiple of 3
12 is only option
4*3*35*105)=(2*105)^2
12
C

stonecold · Answer

Here are my 2 cents on this one ->

Given info => K is an even integer.
3675*k=square of an integer.

In a perfect square the power of each prime factor is an even number.
Breaking down 3675*k=> 5^2*7^2*3*k => Perfect square.
Now the general expression for k will be 3*(Any-prime)^even
If the question had not stated that k is even => A would have been sufficient.
But here as k is even => Minimum value of k=> 3*2^2 (why 2^2? => Beacuse 2 is the smallest positive even number )
Hence k=>2^2*3=>12

Hence C.

sealberg · Answer

Ugh, I fell into the trap of forgetting that k is even, and chose A: k=3. I need to remember to always double check the constraints!

stonecold · Answer

We all did my friend. We all did. P.S -> Here is the Link to the entire Mock series -> https://gmatclub.com/forum/stonecold-s- ... 60-40.html

payal1496 · Answer

It is mentioned that K is even integer so eliminate options A, B, D,

so consider option c-12*3675=44100 which is a perfect square of 210.

Ans is- C-12

bumpbot · Answer

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If k is an even integer, what is the smallest possible value of k

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If k is an even integer, what is the smallest possible value of k

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