Question
The number 920 is divisible by
Answers
GPT-4o mini
To determine the numbers by which 920 is divisible, we can start by finding its prime factorization.
First, we divide 920 by 2 (the smallest prime number):
920 ÷ 2 = 460
460 ÷ 2 = 230
230 ÷ 2 = 115
Now that 115 is not divisible by 2, we try the next prime number, which is 3.
115 ÷ 3 is not a whole number, so we try the next prime, which is 5:
115 ÷ 5 = 23
Now, 23 is a prime number.
Putting it all together, the prime factorization of 920 is:
\[ 920 = 2^3 \times 5 \times 23 \]
Next, to find all the divisors of 920, we can use the prime factorization:
The possible exponents for each factor can range from 0 up to their respective maximum in the factorization:
- For \(2\): 0 to 3 (4 options: \(2^0, 2^1, 2^2, 2^3\))
- For \(5\): 0 to 1 (2 options: \(5^0, 5^1\))
- For \(23\): 0 to 1 (2 options: \(23^0, 23^1\))
Calculating the total number of divisors:
\[ (3 + 1)(1 + 1)(1 + 1) = 4 \times 2 \times 2 = 16 \]
Now, let's list the divisors:
1. \(2^0 \times 5^0 \times 23^0 = 1\)
2. \(2^0 \times 5^0 \times 23^1 = 23\)
3. \(2^0 \times 5^1 \times 23^0 = 5\)
4. \(2^0 \times 5^1 \times 23^1 = 115\)
5. \(2^1 \times 5^0 \times 23^0 = 2\)
6. \(2^1 \times 5^0 \times 23^1 = 46\)
7. \(2^1 \times 5^1 \times 23^0 = 10\)
8. \(2^1 \times 5^1 \times 23^1 = 230\)
9. \(2^2 \times 5^0 \times 23^0 = 4\)
10. \(2^2 \times 5^0 \times 23^1 = 92\)
11. \(2^2 \times 5^1 \times 23^0 = 20\)
12. \(2^2 \times 5^1 \times 23^1 = 460\)
13. \(2^3 \times 5^0 \times 23^0 = 8\)
14. \(2^3 \times 5^0 \times 23^1 = 184\)
15. \(2^3 \times 5^1 \times 23^0 = 40\)
16. \(2^3 \times 5^1 \times 23^1 = 920\)
Thus, the complete list of the divisors of 920 is:
1, 2, 4, 5, 8, 10, 20, 23, 40, 46, 92, 115, 184, 230, 460, and 920.
In conclusion, 920 is divisible by all these numbers.
First, we divide 920 by 2 (the smallest prime number):
920 ÷ 2 = 460
460 ÷ 2 = 230
230 ÷ 2 = 115
Now that 115 is not divisible by 2, we try the next prime number, which is 3.
115 ÷ 3 is not a whole number, so we try the next prime, which is 5:
115 ÷ 5 = 23
Now, 23 is a prime number.
Putting it all together, the prime factorization of 920 is:
\[ 920 = 2^3 \times 5 \times 23 \]
Next, to find all the divisors of 920, we can use the prime factorization:
The possible exponents for each factor can range from 0 up to their respective maximum in the factorization:
- For \(2\): 0 to 3 (4 options: \(2^0, 2^1, 2^2, 2^3\))
- For \(5\): 0 to 1 (2 options: \(5^0, 5^1\))
- For \(23\): 0 to 1 (2 options: \(23^0, 23^1\))
Calculating the total number of divisors:
\[ (3 + 1)(1 + 1)(1 + 1) = 4 \times 2 \times 2 = 16 \]
Now, let's list the divisors:
1. \(2^0 \times 5^0 \times 23^0 = 1\)
2. \(2^0 \times 5^0 \times 23^1 = 23\)
3. \(2^0 \times 5^1 \times 23^0 = 5\)
4. \(2^0 \times 5^1 \times 23^1 = 115\)
5. \(2^1 \times 5^0 \times 23^0 = 2\)
6. \(2^1 \times 5^0 \times 23^1 = 46\)
7. \(2^1 \times 5^1 \times 23^0 = 10\)
8. \(2^1 \times 5^1 \times 23^1 = 230\)
9. \(2^2 \times 5^0 \times 23^0 = 4\)
10. \(2^2 \times 5^0 \times 23^1 = 92\)
11. \(2^2 \times 5^1 \times 23^0 = 20\)
12. \(2^2 \times 5^1 \times 23^1 = 460\)
13. \(2^3 \times 5^0 \times 23^0 = 8\)
14. \(2^3 \times 5^0 \times 23^1 = 184\)
15. \(2^3 \times 5^1 \times 23^0 = 40\)
16. \(2^3 \times 5^1 \times 23^1 = 920\)
Thus, the complete list of the divisors of 920 is:
1, 2, 4, 5, 8, 10, 20, 23, 40, 46, 92, 115, 184, 230, 460, and 920.
In conclusion, 920 is divisible by all these numbers.