Respuesta :
Answer:
19747 = 7^2×13×31
Step-by-step explanation:
Factor the following integer:
19747
Since sqrt(19747)≈140.5, it is sufficient to test 19747 for divisibility by primes less than or equal to 140.:
Test primes up to 140 for factors of 19747.
19747 is not divisible by 2 since 19747 is odd and 2 is even:
19747 = 19747 (19747 is not divisible by 2)
The sum of the digits of 19747 is 1 + 9 + 7 + 4 + 7 = 28, which is not divisible by 3. This means 19747 is not divisible by 3:
19747 = 19747 (19747 is not divisible by 2 or 3)
The last digit of 19747 is not 5 or 0, which means 19747 is not divisible by 5:
19747 = 19747 (19747 is not divisible by 2, 3 or 5)
Divide 7 into 19747:
| | 2 | 8 | 2 | 1 | (quotient)
7 | 1 | 9 | 7 | 4 | 7 |
- | 1 | 4 | | | |
| | 5 | 7 | | |
| - | 5 | 6 | | |
| | | 1 | 4 | |
| | - | 1 | 4 | |
| | | | 0 | 7 |
| | | | - | 7 |
| | | | | 0 | (remainder)
19747 = 7×2821 (2821 is not divisible by 2, 3 or 5 since 19747 is not)
Divide 7 into 2821:
| | 4 | 0 | 3 | (quotient)
7 | 2 | 8 | 2 | 1 |
- | 2 | 8 | | |
| | 0 | 2 | |
| | - | 0 | |
| | | 2 | 1 |
| | - | 2 | 1 |
| | | | 0 | (remainder)
19747 = 7×7×403 (403 is not divisible by 2, 3 or 5 since 2821 is not)
Divide 7 into 403:
| | 5 | 7 | (quotient)
7 | 4 | 0 | 3 |
- | 3 | 5 | |
| | 5 | 3 |
| - | 4 | 9 |
| | | 4 | (remainder)
403 is not divisible by 7:
19747 = 7×7×403 (403 is not divisible by 2, 3, 5 or 7)
The alternating sum of the digits of 403 is 4 - 0 + 3 = 7, which is not divisible by 11. This means 403 is not divisible by 11:
19747 = 7×7×403 (403 is not divisible by 2, 3, 5, 7 or 11)
Divide 13 into 403:
| | | 3 | 1 | (quotient)
1 | 3 | 4 | 0 | 3 |
| - | 3 | 9 | |
| | | 1 | 3 |
| | - | 1 | 3 |
| | | | 0 | (remainder)
19747 = 7×7×13×31 (31 is not divisible by 2, 3, 5, 7 or 11 since 403 is not)
31 is not divisible by any prime less than sqrt(31)≈5.5 (namely 2, 3 or 5). Therefore 31 is prime:
19747 = 7×7×13×31
There are 2 copies of 7, 1 copy of 13 and 1 copy of 31 in the product:
Answer: 19747 = 7^2×13×31