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Bunuel
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Bunuel
How many positive integers less than 1,000,000 are both multiples of 72 and perfect cubes?

A. 15
B. 16
C. 17
D. 18
E. 19


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
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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.
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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



raffaeleprio
Hi!

Since 1,000,000 is the perfect cube of 100, we already know that we have 100 perfect cubes within 1,000,000.
Furthermore, 72 can be factorized as \(2^3*3^2\).
Hence our perfect cube divisible by 72 has the following form \(n=\sqrt[3]{2^3*3^2*k}\), with \(n\leq{100}\)

Now the trivial one is for k=3...and simplifying and putting 2 and 3 out of the root we get \(n=2*3*\sqrt[3]{k}\), with \(n\leq{100}\)
Hence k can assume all cube powers which ensure that \(n\leq{100}\) ---> k= \(1^3\), \(2^3\), \(3^3\),\(4^3\),\(5^3\)...up to \(16^3\)

This leads to answer B.

IMO B!
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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?
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chetansharma, Bunuel, I would love to know your solution to this question.
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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
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Bunuel
How many positive integers less than 1,000,000 are both multiples of 72 and perfect cubes?

A. 15
B. 16
C. 17
D. 18
E. 19



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
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chetansharma
Bunuel
How many positive integers less than 1,000,000 are both multiples of 72 and perfect cubes?

A. 15
B. 16
C. 17
D. 18
E. 19



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?
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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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