Find the square root of the 20657025 by prime factorisation method

4545

Step-by-step explanation:

1. Factor the following integer:

20657025

The last digit of 20657025 is 5, which means it is divisible by 5

20657025 = 5×4131405:

20657025 = 5×4131405

The last digit of 4131405 is 5, which means it is divisible by 5

4131405 = 5×826281:

20657025 = 5×5×826281

The sum of the digits of 826281 is 8 + 2 + 6 + 2 + 8 + 1 = 27, which is divisible by 9. This means 826281 is too

826281 = 9×91809:

20657025 = 5×5×9×91809

9 = 3^2:

20657025 = 5×5×3^2×91809

The sum of the digits of 91809 is 9 + 1 + 8 + 0 + 9 = 27, which is divisible by 9. This means 91809 is too

91809 = 9×10201:

20657025 = 5×5×3^2×9×10201

9 = 3^2:

20657025 = 5×5×3^2×3^2×10201

Because 10201 is odd, only test odd numbers for divisibility

10201 is not divisible by 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97 or 99

10201 = 101×101 which means 10201 is divisible by 101:

20657025 = 5×5×3^2×3^2×101×101

3 is prime:

20657025 = 5×5×3^2×3^2×101×101

5 is prime:

20657025 = 5×5×3^2×3^2×101×101

Because 101 is odd, only test odd numbers for divisibility

101 is not divisible by 3, 5, 7 or 9

Since 101 is not divisible by any integer up to 10, it is prime:

20657025 = 5×5×3^2×3^2×101×101

There are 4 copies of 3, 2 copies of 5 and 2 copies of 101 in the product:

20657025 = 3^4×5^2×101^2

2. Simplify the following:

sqrt(3^4×5^2×101^2)

| | 1 | 0 | 1

× | | 1 | 0 | 1

| | 1 | 0 | 1

| 0 | 0 | 0 | 0

1 | 0 | 1 | 0 | 0

1 | 0 | 2 | 0 | 1:

sqrt(3^4×5^2×10201)

5^2 = 25:

sqrt(3^4×25×10201)

3^4 = (3^2)^2:

sqrt((3^2)^2 25×10201)

3^2 = 9:

sqrt(9^2×25×10201)

9^2 = 81:

sqrt(81×25×10201)

81×25 = 2025:

sqrt(2025×10201)

2025×10201 = 20657025:

sqrt(20657025)

sqrt(20657025) = sqrt(81×255025) = sqrt(3^4×505^2):

sqrt(3^4 505^2)

sqrt(3^4 505^2) = sqrt(3^4) sqrt(505^2) = 3^(4/2)×505^(2/2) = 9×505:

9×505

9×505 = 4545:

4545

(6, 3 ) is not on the line

Step-by-step explanation:

Substitute x = 6 into the right side of the equation and if the value is equal to the y- coordinate of the point, then it is on the line

y = - 6 + 3 = - 3 ≠ 3

Thus (6, 3 ) is not on the line