What is the least number by which 10584 must be divided so that the quotient is a perfect cube?

(i) We have,

1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.

1536 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 3

So, in order to make it a perfect cube, it must be divided by 3.

Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.

(ii) We have,

10985 = 5 × 13 × 13 × 13

After grouping the prime factors in triplet, it’s seen that one factor 5 is left without grouping.

10985 = 5 × (13 × 13 × 13)

So, it must be divided by 5 in order to get a perfect cube.

Thus, the required smallest number is 5.

(iii) We have,

28672 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7

After grouping the prime factors in triplets, it’s seen that one factor 7 is left without grouping.

28672 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 7

So, it must be divided by 7 in order to get a perfect cube.

Thus, the required smallest number is 7.

(iv) 13718 = 2 × 19 × 19 × 19

After grouping the prime factors in triplets, it’s seen that one factor 2 is left without grouping.

13718 = 2 × (19 × 19 × 19)

So, it must be divided by 2 in order to get a perfect cube.

Thus, the required smallest number is 2.

Find the smallest number by which $26244$ may be divided so that the quotient is a perfect cube.

Nội dung chính

• Find the smallest number by which $26244$ may be divided so that the quotient is a perfect cube.
• The prime factors of 26244 are2×2×3×3×3×3×3×3×3×3=(3×3×3)×(3×3×3)×3×3×2×2Clearly,26244 must be divided by 3×3×2×2=36
• What is the smallest number by which 26244 must be divided to give a perfect cube also find the cube root of the quotient?
• What is the smallest number by which the following must be multiplied so that the product is a perfect cube 10584?
• What is the smallest number by which 6912 must be multiplied to make it a perfect cube?
• What is the smallest number by which must be multiplied so that the product is a perfect cube?

Answer

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Hint: To find the required smallest number, we will use the prime factorization method. We will write the given number $26244$ as the multiple of primes. After that it will be written in the form of a group of three if possible. Here we need to find the smallest number such that the quotient is a perfect cube. So, we cannot make a group of two primes.

Complete step-by-step answer:
To solve the given problem, we must know the prime factorization method. By using the method of prime factorization, we can express the given number as a product of prime numbers. Therefore, we will write the given number $26244$ as the product of primes. Let us do the prime factorization of $26244$. Note that $26244$ is an even number so we can start prime factorization with prime number $2$.

$2$	$26244$
$2$	$13122$
$3$	$6561$
$3$	$2187$
$3$	$729$
$3$	$243$
$3$	$81$
$3$	$27$
$3$	$9$
$3$	$3$
	$1$

Therefore, we can write $26244 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$. Here we are dealing with a perfect cube so we have to write the obtained factorization in the form of a group of three if possible. So, we can write $26244 = 2 \times 2 \times 3 \times 3 \times \left( {3 \times 3 \times 3} \right) \times \left( {3 \times 3 \times 3} \right) \cdots \cdots \left( 1 \right)$. To find the perfect cube root, we have to take one number from each group of three but here we can see that $2 \times 2$ and $3 \times 3$ cannot be written as a group of three.
Let us divide by $2 \times 2 \times 3 \times 3 = 36$ on both sides of the equation $\left( 1 \right)$. So, we can write
$\dfrac{{26244}}{{2 \times 2 \times 3 \times 3}} = \left( {3 \times 3 \times 3} \right) \times \left( {3 \times 3 \times 3} \right)$
$ \Rightarrow \dfrac{{26244}}{{36}} = \left( {3 \times 3 \times 3} \right) \times \left( {3 \times 3 \times 3} \right)$
Here we can take one number from each group of three. So we will get a quotient which is a perfect cube. Hence, the required smallest number is $36$.

Note: Here we can say that $26244$ is a perfect square because in the prime factorization of $26244$ we can see that each prime number can be written in the form of a group of two. If the sum of all digits of a number is divisible by $3$ then that number is also divisible by $3$.

26244 = 2×2×3×3×3×3×3×3×3×3. So, 2x2x 3x 3 = 36 is the smallest number by which 26244 must be divided so that the quotient is a perfect cube. So, dividing 26244 by 36 will give us the required quotient = 729. Therefore, the cube root of the quotient = ³√729= 9.

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Find the smallest number by which 26244 should be divided so that the quotient is a perfect cube.

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Solution

The prime factors of 26244 are2×2×3×3×3×3×3×3×3×3=(3×3×3)×(3×3×3)×3×3×2×2Clearly,26244 must be divided by 3×3×2×2=36

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What is the smallest number by which 26244 must be divided to give a perfect cube also find the cube root of the quotient?

26244 = 2×2×3×3×3×3×3×3×3×3. So, 2x2x 3x 3 = 36 is the smallest number by which 26244 must be divided so that the quotient is a perfect cube. So, dividing 26244 by 36 will give us the required quotient = 729. Therefore, the cube root of the quotient = ³√729= 9.

What is the smallest number by which the following must be multiplied so that the product is a perfect cube 10584?

Hence, the smallest number by which 10584 must be multiplied to obtain a perfect cube is 7.

What is the smallest number by which 6912 must be multiplied to make it a perfect cube?

Answer: No 6912 is not a perfect cube. It should be multiplied by 2. Thus , it should be multiplied by 2 to make it a perfect cube.

What is the smallest number by which must be multiplied so that the product is a perfect cube?

Therefore, 243 must be multiplied by 3 to make it a perfect cube. Here one factor 2 is required to make a 3's group. Therefore, 256 must be multiplied by 2 to make it a perfect cube.

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