What is the smallest number by which 1600 be divided to make it a perfect cube?

To make the quotient a perfect cube we divide it by 172 = 289, which gives 27 as quotient and we know that 27 is a perfect cube ∴ 289 is the required smallest number. (vi) 107811 First find the prime factors of 107811 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11 = 33 × 113 × 3 Since 107811 is not a perfect cube. To make the quotient a perfect cube we divide it by 3, which gives 35937 as quotient and we know that 35937 is a perfect cube. ∴ 3 is the required smallest number. (vii) 35721 First find the prime factors of 35721 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7 = 33 × 33 × 72 Since 35721 is not a perfect cube. To make the quotient a perfect cube we divide it by 72 = 49, which gives 729 as quotient and we know that 729 is a perfect cube ∴ 49 is the required smallest number. (viii) 243000 First find the prime factors of 243000 243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 23 × 33 × 53 × 32 Since 243000 is not a perfect cube. To make the quotient a perfect cube we divide it by 32 = 9, which gives 27000 as quotient and we know that 27000 is a perfect cube ∴ 9 is the required smallest number.

To make the quotient a perfect cube we divide it by 52 = 25, which gives 27 as quotient

where, 27 is a perfect cube.

∴ 25 is the required smallest number.

(ii) 8640

First find the prime factors of 8640

8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5

= 23 × 23 × 33 × 5

Since 8640 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5, which gives 1728 as quotient and

we know that 1728 is a perfect cube.

∴5 is the required smallest number.

(iii) 1600

First find the prime factors of 1600

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

= 23 × 23 × 52

Since 1600 is not a perfect cube.

To make the quotient a perfect cube we divide it by 52 = 25, which gives 64 as quotient

and we know that 64 is a perfect cube

∴ 25 is the required smallest number.

(iv) 8788

First find the prime factors of 8788

8788 = 2 × 2 × 13 × 13 × 13

= 22 × 133

Since 8788 is not a perfect cube.

To make the quotient a perfect cube we divide it by 4, which gives 2197 as quotient and

we know that 2197 is a perfect cube

∴ 4 is the required smallest number.

(v) 7803

First find the prime factors of 7803

7803 = 3 × 3 × 3 × 17 × 17

= 33 × 172

Since 7803 is not a perfect cube.

To make the quotient a perfect cube we divide it by 172 = 289, which gives 27 as quotient and we know that 27 is a perfect cube ∴ 289 is the required smallest number. (vi) 107811 First find the prime factors of 107811 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11 = 33 × 113 × 3 Since 107811 is not a perfect cube. To make the quotient a perfect cube we divide it by 3, which gives 35937 as quotient and we know that 35937 is a perfect cube. ∴ 3 is the required smallest number. (vii) 35721 First find the prime factors of 35721 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7 = 33 × 33 × 72 Since 35721 is not a perfect cube. To make the quotient a perfect cube we divide it by 72 = 49, which gives 729 as quotient and we know that 729 is a perfect cube ∴ 49 is the required smallest number. (viii) 243000 First find the prime factors of 243000 243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 23 × 33 × 53 × 32 Since 243000 is not a perfect cube. To make the quotient a perfect cube we divide it by 32 = 9, which gives 27000 as quotient and we know that 27000 is a perfect cube ∴ 9 is the required smallest number.

Video transcript

"Hello kids. Welcome to little homework. My name is Emma. And I'm Lido tutor here. We have a question. The question is by which smallest number must be the following numbers be divided so that the coefficient is perfect Cube. So they have given eight numbers to solve and this numbers are not perfect Cube. We have to make them perfect Cube and dividing with the smallest number. So let's go with the first number, which is 675 So the green light 675 prime factor, so 675 will be equal to 3 x 3 x 3 X 5 x 5 if you see here we have to make a group of three numbers when you're calculating for the perfect Cube. This should be a group of three numbers. For example, this is forming a one group in the second group. Only two numbers are there are two factors are there so this one doesn't form a group. So we have to divide 675 with 25 and that will be equal to 27 and we know that Seven is a perfect Cube. So therefore 25 is a smallest number or smallest the required number we can say so therefore. 25 is required. Smallest number So that's it. We got the first one. So therefore now we'll go to the second one. The second number is 8640. So will I down the number second one? 8640 now 8640 can be written as So we're going to multiply your too. Six times so this is 3 x 2 x 2 certain six times here and no to multiply even three three times and then multiply by 5 now. We'll start making a group of Threes. So this form of one group this forms another group this forms another group and here five a single number. So we have to eliminate this file. So from the VIN number 8640 we have to divide with Phi and this gives a value 1728. So this is a perfect Cube and we are to divide a fine from the given number so we can say Phi is a smallest a required number so therefore Five is a required smallest number now. Let's go to the third one. Third number is 1600 now will I turn 1600 and a light its prime factors? So 1600 is can be written as now we're going to multiply again to six times. And then we'll multiply 5 2 times this other prime factors of 1,600 now, we'll make a group of three. So this is a one group. This is a second group and here in the last one. Only two factors are there so which is Phi phi's are 25. So from 1600 we have to divide 25 and this gives 64 and we know that 64 is a perfect Cube It's a cube of for so 25 the smallest required. But here so therefore we can conclude that. 25 Is a smallest required number now we'll go to the fourth. Number one. The fourth one is eight. Seven eight eight. So let's write down the number. Eight eight and this number can be written as so this here. We're going to multiply to 2 times and then multiply 13 three times. So this is 13 x 13 x 13. Now we'll make a group of three. So this is a one group. Not this one. The last one 13 is forming a one group so we can take this as a one group and here if you see the first one only two factors are there so we have to divide these two factors from the given number so 2 x 2 is 4 so from 8 7 8 8 we have to divide 4 and that gives to 1970 which is a Perfect Cube. So therefore for is a required smallest number 4. For is a required smallest number. Let's go to V 1 V 1 is 780 3 so we'll right on the number first. So the number is 780 3 This number can be written as 3 x 3 x 3 x 17 x 17 this are the prime factors of 7 803 now we'll make a group of Threes. So this is a forming of one group and the last there are two factors of 17 re 17 x 17, which is 289. So we have to divide 78023 by to 18. / 289 and this will be equal to 27 and we know that 27 is a perfect Cube. It is a cube of 3. So therefore 289 is the smallest a required number or to organize the required smallest number. So 289 is a required smallest number now, we'll go to the next one that is sixth problem in the sixth. They have given number 107 811 so we'll write down the number first."

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