What is the smallest number to be multiplied to 768 to be a perfect square?

Question 7: The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

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• What is the perfect square of 768?
• What is the prime factorization of 768?
• Why is 16 a perfect square?
• Why 729 is a perfect square?

Answer: We need to calculate the square root of 2401 to get the solution.
By prime factorisation of 2401, we get
`2401 = 7 xx 7 xx 7 xx 7`

Or, `2401=7^2xx7^2`

Or, `sqrt(2401)=sqrt(7^2xx7^2)`

`=7xx7=49`

There are 49 students, each contributing 49 rupees
Thus, Answer = 49

Question 8: 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square.

\therefore252\times7=1764

And (i) \sqrt{1764}=2\times3\times7=42

(ii) 180 = 2 x 2 x 3 x 3 x 5

Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.

\therefore180\times5=900

And \sqrt{900}=2\times3\times5=30

(iii) 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7

Here, prime factor 7 has no pair. Therefore 1008 must be multiplied by 7 to make it a perfect square.

\therefore1008\times7=7056

And \sqrt{7056}=2\times2\times3\times7=84

(iv) 2028 = 2 x 2 x 3 x 13 x 13

Here, prime factor 3 has no pair. Therefore 2028 must be multiplied by 3 to make it a perfect square.

\therefore2028\times3=6084

And \sqrt{6084}=2\times2\times3\times3\times13\times13=78

(v) 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3

Here, prime factor 2 has no pair. Therefore 1458 must be multiplied by 2 to make it a perfect square.

Ex 6.3, 5 For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained. (vi) 768 Prime factorizing 768 By prime factorization, 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 Since 3 does not occur in pair we multiply by 3 to make it a pair So, our number becomes 768 × 3 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 Square root of 2304 ∴ √2304 = 2 × 2 × 2 × 2 × 3 = 4 × 4 × 3 = 48 ∴ The smallest whole number to be multiplied = 3 and square root of new number = 48

What is the perfect square of 768?

It is the positive solution of the equation x2 = 768. We can express the square root of 768 in its lowest radical form as 16 √3. ... Square Root of 768..

Ex 6.3, 5 For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained. (vi) 768 Prime factorizing 768 By prime factorization, 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 Since 3 does not occur in pair we multiply by 3 to make it a pair So, our number becomes 768 × 3 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 Square root of 2304 ∴ √2304 = 2 × 2 × 2 × 2 × 3 = 4 × 4 × 3 = 48 ∴ The smallest whole number to be multiplied = 3 and square root of new number = 48

Question 7: The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Answer: We need to calculate the square root of 2401 to get the solution.
By prime factorisation of 2401, we get
`2401 = 7 xx 7 xx 7 xx 7`

Or, `2401=7^2xx7^2`

Or, `sqrt(2401)=sqrt(7^2xx7^2)`

`=7xx7=49`

There are 49 students, each contributing 49 rupees
Thus, Answer = 49

Question 8: 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square.

\therefore252\times7=1764

And (i) \sqrt{1764}=2\times3\times7=42

(ii) 180 = 2 x 2 x 3 x 3 x 5

Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.

\therefore180\times5=900

And \sqrt{900}=2\times3\times5=30

(iii) 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7

Here, prime factor 7 has no pair. Therefore 1008 must be multiplied by 7 to make it a perfect square.

\therefore1008\times7=7056

And \sqrt{7056}=2\times2\times3\times7=84

(iv) 2028 = 2 x 2 x 3 x 13 x 13

Here, prime factor 3 has no pair. Therefore 2028 must be multiplied by 3 to make it a perfect square.

\therefore2028\times3=6084

And \sqrt{6084}=2\times2\times3\times3\times13\times13=78

(v) 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3

Here, prime factor 2 has no pair. Therefore 1458 must be multiplied by 2 to make it a perfect square.

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