Numbers divisible by a cube

If an integer is divisible by a cube larger than or equal to 2, then it is a multiple of that cube. So, we are looking for the number of distinct multiples of cubes between 1 and than 1000.

Let us determine the number of multiples of the cubes of prime numbers within our range. We only determine the multiples of the cubes of prime numbers because they encompass the multiples of the cubes of composite numbers as well.

The smallest cube we need to consider is $2^3$ , so we determine how many multiples of $2^3=8$ there are between 1 and 1000.

$\lfloor 999÷8 \rfloor=124$

So, there are 124 multiples of 8 less than 1000.

Using this method, we find the number of multiples of the cubes of the other primes.

$\lfloor999÷3^3\rfloor=37$

There are 37 multiples of $3^3=27$ less than 1000.

$\lfloor999÷5^3\rfloor=7$

There are 7 multiples of $5^3=125$ less than 1000.

$\lfloor999÷7^3\rfloor=2$

There are 2 multiples of $7^3=343$ less than 1000.

We can stop at $7^3$ because the next prime's cube, $11^3$ , is greater than 1000 so there will be no multiples of it that are within our range.

Adding these multiples together, we get $124+37+7+2=170$ integers.

Now we need to subtract from our total any duplicates that we may have counted. If an integer is divisible by the cube of a non-prime number, then it is divisible by the cube of each of its prime factors. So, we will have counted this integer multiple times, once as a multiple for each distinct prime factor's cube. Consider all composite numbers between 2 and 11 with 2 or more distinct prime factors. A quick look will yield 2 candidates, 6 and 10. But $10^3$ is out of the range we are considering, so we only need to consider 6. If we determine the number of multiples of $6^3$ there are between 1 and 1000, we can subtract that number from our total, because each of these multiples will have been counted twice, once as a multiple of $2^3$ and once as a multiple of $3^3$ .

$6^3=216$

$\lfloor999÷216\rfloor=4$

There are 4 multiples of $6^3=216$ within our range, each which we counted twice. We subtract this from our previous total to get our final answer of 166.

Here, at first we will come to know that we need to check cube of 2 to 9, as 9^3=729 after which 10^3=1000 which more than the given limit. So, now we need to check all multiples of 8,27,64,125,216,343,512 and 729. but as 64 and 512 are multiple of 8,also 729 is a multiple of 27 then this are excluded. Now, multiple of 8 less than 1000 are 124. multiple of 27 less than 1000 are 37. but 8*27=216 whose multiples are repeating so we will subtract it from the total.. Now, multiple of 125 and 343 are 7 and 2 respectively. So, finally 124+37+7+2-4=166

If a integer between 1 and 1000 can be fully divided by the cube of another integer, we can say that the number being divided is a multiple of that integer and of its cube. Every integer can be factorized into Prime numbers. The biggest prime we can cube that will be less than 1000 is 7, since 11^3 is 1331. So, our number we are looking for is a multiple either of 2^3, 3^3, 5^3 or 7^3. Why should'nt we consider, for an example, 9^3? It is because we will be already count that number on the 3^3 multiples. Dividing 999 by 8 we get 124,875, so it means there are 124 multiples of 2^3 between one and 999. We do the same with 3^3 and find 37 multiples; with 5^3 we get 7; and finally, with 7^3 we get 2. But, we can still think of a multiple of both or more prime factors. The only couple we can make is 8 \times 27, wich is equal to 216. Dividing 999 by 216 we can assume there are four multiples of 216 between 1 and 999. It means that four of the multiples of 8 we counted we counted them aswell on the 27 multiples. We got to take those out. The final sum gets 124+37+7+2-4=166.

We need to find the first few cubes of prime numbers that are less than 1000. These are $2^3 = 8$ , $3^3=27$ , $5^3=125$ , and $7^3=343$ . There are 124 multiples of 8, 37 multiples of 27, 7 multiples of 125, and 2 multiples of 343 from 1 to 1000 (exclusive). However, we have over counted because the product of 8 and 27 have 4 common multiples in that range. Therefore, the answer is $124+37+7+2-4=166$ .

Cubes of numbers larger than or equal to 2 : 8,27,64,125,216,343,512,729. (Between 1 to 1000)

Cubes which are participating in division process: 8,27,125,343.

64,512 are rejected because multiples of these two numbers are considered in division by 8.

Similarly, 729 is rejected because it is counted in multiples of 27.

216 is rejected as its multiples are counted in division by 8 and 27 both.

Therefore, Nos. which are divisible by 8 = (1000/8) - 1 = 125 - 1 = 124. (As 1000 is excluded.) Nos. which are divisible by 27 = (1000/27) = 37.03. (~37) Nos. which are divisible by 125 = (1000/125) - 1 =8-1 = 7. (As 1000 is excluded.) Nos. which are divisible by 343 = (1000/343) =2.91. (~2)

Total Nos. = (124+37+7+2) - 4=166. 4 is subtracted because multiples of 216 are counted twice while counting multiples of 8 and 27. Therefore, one set of multiples are reduced as (1000/216) = 4.62 (~4).

We first observe that $10^3 = 1000$ , so any number less than 1000 that is divisible by a cube is divisible by the cube of a number less than 10. Looking at the numbers from 2 to 9, we know that any number that is divisible by $4^3$ or $8^3$ is also divisible by $2^3$ , (and no other square) so we don't need to count those divisible by $4^3$ or $8^3$ . This is similarly true for numbers divisible by $9^3$ . That leaves us with checking for numbers divisible by $2^3, 3^3, 5^3, 6^3, 7^3$ . Since numbers divisible by $6^3$ are also divisible by $2^3$ and $3^3$ , we will use the principle of inclusion and exclusion to count the numbers we want.

There are $\lfloor \frac{999}{2^3} \rfloor =124$ numbers divisible by $2^3$ , $\lfloor \frac{999}{3^3} \rfloor = 37$ numbers divisible by $3^3$ , $\lfloor \frac{999}{5^3} \rfloor = 7$ numbers divisible by $5^3$ , $\lfloor \frac{999}{6^3} \rfloor = 4$ numbers divisible by $6^3$ , and $\lfloor \frac{999}{7^3} \rfloor = 2$ numbers divisible by $7^3$ . Hence, applying PIE, this gives a total of $124 + 37 + 7 - 4 + 2 = 166$ numbers less than 1000 divisible by a cube larger than 1.