Question

1. The following numbers are not perfect squares. Give reason.
i. 1857
ii. 56000
v. 4298
iv. 1600000
1 with digite of the squares of the following​

Answer 1

Answer:

These numbers are not perfect squares because they are not the product of the same number multiplied twice. Also, these numbers end with 7, 8, three 0s and five 0s. To be a perfect sqaure the numbers must end with 1, 4, 5, 6, 9 or an even number of 0s.

Answer 2

Answer:

Here given numbers are 1857,56000,4298,1600000

By option test we can solve this problem.

Option -1:

$1857 = 3 \times 619$

So,this is not a perfect square.

Option -2:

$56000 \\ = 2 \times 2 \times 2 \times 7 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \\ = {(2 \times 2 \times 2)}^{2} \times 7 \times {(5 \times5)}^{2}$

So,this is not a perfect square.

Option -3:

$4298 = 2 \times 7 \times 307$

So,this is not a perfect square.

Option -4:

$1600000 \\ = 4 \times 4 \times 10 \times 10 \times 10 \times 10 \times 10 \\ = {4}^{2} \times {10}^{5}$

So,this is not a perfect square.

Some important Square value :

${1}^{2} = 1 \\ {2}^{2} = 4 \\ {3}^{2} = 9 \\ {4}^{2} = 16 \\ {5}^{2} = 25 \\ {6}^{2} = 36 \\ {7}^{2} = 49 \\ {8}^{2} = 64 \\ {9}^{2} = 81 \\ {10}^{2} = 100 \\ {11}^{2} = 121 \\ {12}^{2} = 144 \\ {13}^{2} = 169 \\ {14}^{2} = 196 \\ {15}^{2} = 225 \\ {16}^{2} = 256 \\ {17}^{2} = 289 \\ {18}^{2} = 324 \\ {19}^{2} = 361 \\ {20}^{2} = 400 \\ {21}^{2} = 441 \\ {22}^{2} = 484 \\ {23}^{2} = 529 \\ {24}^{2} = 576 \\ {25}^{2} = 625 \\ {27}^{2} = 729 \\ {28}^{2} = 784 \\ {29}^{2} = 841 \\ {30}^{2} = 900$

Square root related two more questions:

https://brainly.in/question/17174336

https://brainly.in/question/5282843

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