We have to find the smallest number by which 675 must be multiplied to obtain a perfect cube Solution 675 On prime factorisation of 675 we get, 675 = 3×3×3×5×5 By assorting the factors in triplets of equal factors, 675 = (3×3×3) x 5×5 Here, 5 cannot be arranged into triplets of equal factors. ∴ We will multiply 675 by 5 to get perfect cube. Answer The smallest number by which 675 must be multiplied to obtain a perfect cube is 5.
Answer  Hint: First, we need to know about the concept of the perfect square and perfect cube.Since we just need to know such things about the square root numbers and perfect square numbers, A perfect square is the numbers that obtain by multiplying any whole numbers (zero to infinity) twice, or the square of the given numbers yields a whole number like $\sqrt {25} = 5$or $25 = {5^2}$Similarly, the perfect cube can be expressed as $\sqrt[3]{3} = 27,27 = {3^3}$ Complete step-by-step solution: Since from the given, we have the number $2560$ and we need to find its perfect cube which will be multiplied with the unknown number to make it.Hence, we will use the prime factorization method to simplify the given number.Prime numbers are the numbers that are divisible by themselves and $1$ only or also known as the numbers whose factors are the given number itself.But the composite numbers which are divisible by themselves, $1$ and also with some other numbers (at least one number other than $1$ and itself)Every composite number can be represented in the form of prime factorization.Since $2560$ it's not the prime number and hence we can use the prime factor method as $ 2\left| \!{\underline {\, {2560} \,}} \right. \\ 2\left| \!{\underline {\, {1280} \,}} \right. \\ 2\left| \!{\underline {\, {640} \,}} \right. \\ 2\left| \!{\underline {\, {320} \,}} \right. \\ 2\left| \!{\underline {\, {160} \,}} \right. \\ 2\left| \!{\underline {\, {80} \,}} \right. \\ 2\left| \!{\underline {\, {40} \,}} \right. \\ 2\left| \!{\underline {\, {20} \,}} \right. \\ 2\left| \!{\underline {\, {10} \,}} \right. \\ 5\left| \!{\underline {\, 5 \,}} \right. \\ 1\left| \!{\underline {\, 1 \,}} \right. \\ $Hence, we have rewritten it as $2560 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$Clearly, we see that the number $2$ is repeated $9$ times which means the number $2$ is a perfect cube because it can be expressed in the form of ${2^3},{2^3},{2^3}$But the number $5$ is occurring one time only.Hence, we need to multiply the same number twice to get the perfect cube.Which is $5 \times {5^2}$ will make the perfect cube.Hence the smallest number by $2560$ number must be multiplied so that the product is a perfect cube is $25$. Note: If we multiplied the given number $2560$ by $25$ then we get $64000$ which is a perfect cube because $64000 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5$ and all primes are perfect cubes. Since don’t write the number ${2^2} = 4$ because $4$ is the composite number, and thus ${2^2},{3^2},{5^2}$ is the repeated prime number while in the process of prime factorization.We can find whether the given number is prime or composite by the trial-and-error methods. Divide the number with the prime numbers less than the given number. if the number is exactly divisible by the prime number, it is the composite number, if not then it is the prime number.The only even prime number is $2$ and all other prime numbers are odd.Text Solution Solution : By using prime factorization we could find:<br>(i)`675`<br>`675=3^3times5^2`<br>As we can see that `5` is not cubed.<br>Hence, `5` is the smallest number by which `675` should be multiplied to make it a perfect cube.<br><br>(ii)`1323`<br>`1323=3^3times7^2`<br>As we can see that `7` is not cubed.<br>Hence, `7` is the smallest number by which `1323` should be multiplied to make it a perfect cube.<br>(iii)`2580`<br>`2560=2^9times5`<br>As we can see that `5` is not cubed.<br>Hence, `5times5=25` is the smallest number by which `2560` should be multiplied to make it a perfect cube. Text Solution Solution : By using prime factorization we could find:<br>(i)`7803`<br>`675=3^3times17^2`<br>As we can see that `17` is not cubed.<br>Hence, `17` is the smallest number by which `7803` should be multiplied to make it a perfect cube.<br><br>(ii)`107811`<br>`107811=3^3times3times11^3`<br>As we can see that `3` is not cubed.<br>Hence, `3` is the smallest number by which `107811` should be multiplied to make it a perfect cube.<br>(iii)`35721`<br>`35721=3^6times7^2`<br>As we can see that `7` is not cubed.<br>Hence, `7` is the smallest number by which `35721` should be multiplied to make it a perfect cube.
Question 10 Cube and Cube Roots Exercise 4.1 Next
Answer:
(i) 675 First find the factors of 675 675 = 3 × 3 × 3 × 5 × 5 = 33 × 52 ∴To make a perfect cube we need to multiply the product by 5. (ii) 1323 First find the factors of 1323 1323 = 3 × 3 × 3 × 7 × 7 = 33 × 72 ∴To make a perfect cube we need to multiply the product by 7. (iii) 2560 First find the factors of 2560 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 = 23 × 23 × 23 × 5 ∴To make a perfect cube we need to multiply the product by 5 × 5 = 25. (iv) 7803 First find the factors of 7803 7803 = 3 × 3 × 3 × 17 × 17 = 33 × 172 ∴To make a perfect cube we need to multiply the product by 17. (v) 107811 First find the factors of 107811 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11 = 33 × 3 × 113 ∴To make a perfect cube we need to multiply the product by 3 × 3 = 9. (vi) 35721 First find the factors of 35721 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7 = 33 × 33 × 72 ∴To make a perfect cube we need to multiply the product by 7.
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