How many positive integers less than 1,000,000 are both multiples of 7

We want to find perfect cubes that have a factor of 72. 72 = 8*9, 8 is already a cube but 9 isn't, we can multiply by another 3 to "complete the cube". So if a cube has 72 as a factor, it must have 72*3 as a factor as well (since the factor 3 must come in triples).

Another way of thinking about this is 72 suggests the cube root is 6, so the complete cube 72 came from would be \(6^3 = 216\).

Therefore we are asking how many perfect cubes, with a factor of 216, are less than 1 million. 216 is already a cube so to make another cube, we have to multiply by a cube. Putting this in algebra that is we want all \(i\) such that \(216*i^3 < 1000000\).
\(216*i^3 < 10^6\), cuberoot both sides.
\(6*i < 10^2\)
\(i < 100/6 = 16.67\). So 1 through 16.

Ans: B

Hi abhinavsodha800 ,

1 million is \(10^6\), if we cube root \(10^6\) the exponent gets divided by 3. So the result of cube rooting one million is one hundred, this tells us the cubes of 1 through 99 are all under one million but when we reach \(100^3\) that is exactly one million.

Hi Raffaele,

You can't put 2 and 3 out of sqrt as 3 is power 2 (not 3).

Can you please re-elaborate accordingly?

Tks

Hi ism1994,

Firstly I assumed that every possible "k" had to contain at least a "3" factor, otherwise there was no way for the squrt to simplify the equation...So you can notice how there is also the solution for K=1, which corresponds to the trivial solution k=3.

Does it sound better now?

chetansharma, Bunuel, I would love to know your solution to this question.

this can be solved with a little calculation as well.
72 factors = 2^3 * 3^2
the no. which is divisible by 72 and a perfect cube
has to be a perfect cube to the least
72 will be a perfect cube if multiplied by 3
so 216 (6^3) is the lowest no. which is a multiple of both 72 and a perfect cube.

Now,
divide 1,000,000 by 216
you will get 4630 approx
we have to find perfect cubes </= 4630
15^3 = 3375 (knowing squares will help)
16^3 = 4096
17 cube will take a hike of 4800+ (given the difference between preceding consecutive no. cubes are less than the succeeding consecutive no. cubes difference)
16^3 is the most we can take,
Hence answer is B

Hi kntombat

We are looking for cube roots under 1,000,000, that is positive integers less than 100 as 100^3=1,000,000

Now second condition given is that they have to be multiple of 72 or 2*2*2*3*3. Now make groups of 3 of same type as we are looking at cube roots => (2*2*2)*(3*3)
If the cube is to be multiple of 72, the cube root has to be a multiple of 2, which has come out of 2*2*2, and 3, which has come out of 3*3.
So our cube root should be a multiple of 2 and 3 or 6.
There are 16 multiples of 6 below 100, as 6*16=96.

Hence 16

B

I understand the first step that the perfect cube must be less than 100.
But, why does the cube root HAVE to be a multiple of 6? Can you elaborate on this further?

Think that your goal is to take out of the roots those 2³*3² , so just multiply by 3 and take 3*2 out, and then keep on multiplaying by perfect cubes as 2³,3³..16³

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