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
