A number theory problem by Morgan Blake

First, find the factors of $24 = (2 \times 2 \times 2 \times 3)$ .

So the first multiple of 24 which is a perfect square is $(2 \times 2 \times 2 \times 3) \times 2 \times 3 = 144$

Now, any multiple of 144 is a multiple of 24 too. So, the numbers which are perfect squares are

$144 \times 1^{2}$

$144 \times 2^{2}$

$144 \times 3^{2}$

... and so on, such that $144 \times k^{2} \leq 10^{6}$

Solving this, we get $k \leq 83.333$

Hence, the answer is $83$ .

In order to have this possible the prime factorization of the perfect square has to be 2x2x2x3. So for the original number we need at least 2x2x3 so that we can reach at least 2^3 and 3 when its squared . That just gives us all multiples of 12 until a 1000. So 1000/12 floored equals to 83.

Solution in Python:

answer=0
n=1
while n<1000000:
    if (n**0.5)%1==0 and n%24==0:
        answer+=1
        n+=1
    else:
        n+=1
print answer
>>> 83

$24=2^3 \cdot 3$ , so its smallest perfect square multiple is $2^3 \cdot 3 \cdot 2 \cdot 3 = 2^4 \cdot 3^2 = 144$ . If we divide $10^6=1,000,000$ by $144$ , we would get $6944.444 \ldots \approx 6944$ . Then, we calculate the greatest prefect square smaller than 6944, and we get $83^2=6889$ . Therefore, the number of perfect square multiples of 24 less than 1,000,000 is $\boxed{83}$ .

• 8 | k^2 , entonces 16| k^2, análogamente 9| k^2
• k^2 = 144*(m^2) < 10^6 , no se da la igualdad por ser multiplo de 3, de ahi simplificas y m < 83,333 , entonces m toma desde el 1 hasta el 83, por lo tanto son 83 valores

{ root (10^6 / 144 ) } = 83.something so 83 is the ans
details
24 = 2 x 2 x 2 x 3 = 2^2 x 2 x 3
the lowest perfect square which divided by 24 is = 2^2 x 2^2 x 3^2 = 144
the 2nd lowest perfect square which divided by 24 is = 2^2 x 2^2 x 2^2 x 3^2 = 144 x 2^2
.........3rd.................................................................................................=144 x 3^2
.................................................................................................................
.................................................................................................................=144 x 83^2 = 992016 ..................................................................................................................=144 x 84^2 = 1016064 > 10^6

As we are talking about perfect squares $≤10^{6}$ , the last number whose square can be included is $1000$ .

Now , by observing some starting square numbers , we can see that $12^{2}$ is the first number which can be divided by $24$ .........The next would be $24^{2}$ .........so on.

Hence , we can conclude that the multiples of $12$ less than $1000$ will be the number of perfect squares divisible by $24$ .

The number which is divisible by $12$ just before $1000$ is $996$ .

Therefore ,

Number if such numbers is ,

$\frac{996}{12}=\boxed{83}$

see 24=2 2 2*3 and as we have to find squares divisible by it therefore if we find nos divisble by 12 that no nos i.e. sq of those nos will itself get divided by 12 and btw 24 too as if sq no it should have repeated factors and as the sq no shld be less than 10^6 therefore their roots shld be less than 1000 and ur done by jst finding the no nos divisible by 12 below 1000

The prime factorization of 24 is 2^3.3. Thus, each square must have at least 3 factors of 2 and 1 factor of 3 and its square root must have 2 factors of 2 and 1 factor of 3. This means that each square is in the form (12c)^2, where c is a positive integer less than sqrt{10^6}. that gives us 1000/12=83

12^2 is 144 which is a multiple of 24, 12 2^2 is a multiple of 24, ................................................ till 12 x^2 is a multiple of 24, such that x takes the value less than square root of 10^6... on solving we get x=83.333 therefore ans is 83

24 can be split into 6 and 4, so for a number to be a multiple of 24 it has to be multiples of both 6 and 4 at the same time. This is to make sure we dont miss out any perfect numbers, which will be explained later.

To make things easier, take the LCM of the 6 and 4, which would be 12. Making it multiples of 12 would mean that they are always divisible by 6 or 4 at the same time.

4,8,12

6,12

Instead of trying to guess the perfect squares like 144 or 576, it is easier to start before the number is squared

12^2, 24^2, 36^2, 48^2 ..... and so on

All these will be multiples of 24, as they are divisible by 6 and 4 at the same time. So it continues until it reaches 1000^2 since it cant reach past a million. Bear in mind it needs to start squaring from multiples of 12 and not start squaring from multiples of 24 as you miss numbers like 36^2, 60^2 and others when u start with 24.

To find the number of multiples of 12 that are before a thousand simply take 1000 and divide by 12.

1000/12 = 83.333333

So you have,

(12 x 1)^2, (12 x 2)^2 ......... until (12 x 83)^2 all of which are multiples of 24

1-83 --> 83 numbers

The prime factorization of 24 is 2^3 \cdot 3. Thus, each square must have at least 3 factors of 2 and 1 factor of 3 and its square root must have 2 factors of 2 and 1 factor of 3. This means that each square is in the form (12c)^2, where c is a positive integer less than \sqrt{10^6}. There are \left\lfloor \frac{1000}{12}\right\rfloor = 083

Nearest squared number multiple of 24 is 144. So the numbers have to be squared number multiples of 144 to be a perfect square. (10^6/144)^0.5 = 83.3. So answer = 83

• If a number squared is less than $10^6$ , the number itself will be less than $10^3$ .
• If its square's a multiple of $24=2^3\times3$ , the number itself shall be a multiple of $12=2^2\times2$ .
• Number of numbers less than $10^3$ that are divisible by $12$ is $\left\lfloor\frac{1000}{12}\right\rfloor=\boxed{83}$

All of the perfect squares that are multiples of 24 are going to be squares of the multiples of 12, so all you really have to do is found out how many times $12$ fits into $\sqrt{10^6} = 10^3 = 1000$ . That can easily be done by dividing 1000 by 12, and getting the floor of that value (rounding down).

$\frac{1000}{12} = 83.333... = \boxed{83}$

24 can be written as = $2 \times 2 \times 2 \times 3$ = $2^{3} \times 3$

Any multiple of 24 which is a perfect square should have its prime factorization as $2^{4} \times 3^{2} \times n^{2}$ where n is a natural number.

So, the numbers which we have to find have to be in the form of $144 \times n^{2}$ < 1,000,000 = $n^{2}$ < $\frac{1000000}{144} = n^{2}$ < 6944.44

$83^{2}$ < 6944.44 < $84^{2}$ .

So, there will be 83 positive perfect squares less than $10^{6}$ which would be multiple of 24.

Whenever there is a question on perfect squares, it is a good idea to factor the number given:
$24 = 2^{3} \times 3^{1}$
In order to be a perfect square, all the exponents in the prime factorization of the number must be even. The lowest number that is a multiple of 24 that satisfies this condition is $2^{4} \times 3^{2} = 144$ .
Now, every number that satisfies the condition of being a perfect square and a multiple of 24 must be in the form of: $144 \times n$
where n is a perfect square. Note: A perfect square times a perfect square is another perfect square.
Given the bound: $0 <144 n < 10^6$ , the number of solutions for n is the number of perfect squares under $\frac{10^{6}}{144}$ :
$n= \lfloor\sqrt{\frac{10^{6}}{144}} \rfloor = \boxed{83}$ .

we can break down 24 into 2 2 2 3... since each prime factor of positive perfect squares has to come in pair, the first positive perfect squares which is a multiple of 24 is 144 ( 2 2 2 3 ) * ( 2 3 )... to find the next number, we just need to multiply 144 with another perfect squares... the nearest number to 10^6 is 144 (83)^2... thus the quantity of the approptiate number is 83