Answer :
Let's solve each part of the question step-by-step:
### (a) Determine if 15 and 16 are relatively prime
Two numbers are relatively prime if their greatest common divisor (GCD) is 1. In this case, the GCD of 15 and 16 is 1.
Thus, 15 and 16 are relatively prime.
### (b) Express 225 as a sum of odd numbers
To express 225 as a sum of odd numbers, observe the pattern where consecutive odd numbers add up to perfect squares. For 225, we can write it as:
[tex]\[ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = 225 \][/tex]
### (c) Find [tex]\((35)^2\)[/tex] without actual multiplication
We can use the identity [tex]\((a + b)^2 = a^2 + 2ab + b^2\)[/tex]:
[tex]\[ 35^2 = (30 + 5)^2 = 30^2 + 2 \cdot 30 \cdot 5 + 5^2 \][/tex]
[tex]\[ 35^2 = 900 + 300 + 25 = 1225 \][/tex]
So, [tex]\(35^2 = 1225\)[/tex].
### (d) Find the Pythagorean triplet if one member is 20
Using the Pythagorean theorem [tex]\(a^2 + b^2 = c^2\)[/tex], we can find a Pythagorean triplet where one of the members (a) is 20. If we choose [tex]\(b\)[/tex] as 21, then:
[tex]\[ a = 20 \][/tex]
[tex]\[ b = 21 \][/tex]
[tex]\[ c = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29 \][/tex]
So, the Pythagorean triplet is (20, 21, 29).
### (e) If [tex]\( 14^2 = 196\)[/tex], what is [tex]\( 18 \times \sqrt{196} \)[/tex]?
Since [tex]\( 14^2 = 196\)[/tex], we know that [tex]\( \sqrt{196} = 14 \)[/tex].
Therefore:
[tex]\[ 18 \times \sqrt{196} = 18 \times 14 = 252 \][/tex]
### (f) Find the smallest number to be multiplied by 2352 to get a perfect square. Also, find the perfect square obtained.
To determine the smallest multiplier that makes 2352 a perfect square, we perform prime factorization:
[tex]\[ 2352 = 2^4 \times 3 \times 7^2 \][/tex]
For 2352 to become a perfect square, each exponent in its prime factorization must be even. The exponent for 3 is odd (1), so we need another factor of 3 to make it even. Therefore, the smallest multiplier is 3.
Thus:
[tex]\[ 3 \times 2352 = 7056 \][/tex]
Now, 7056 is a perfect square.
### (g) Find [tex]\(\sqrt{36864}\)[/tex]. How many digits are in this number?
The square root of 36864 is 192.
So, the number of digits in 192 is 3.
In summary:
- (a) 15 and 16 are relatively prime.
- (b) [tex]\(225\)[/tex] can be expressed as the sum of: [tex]\([1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29]\)[/tex]
- (c) [tex]\(35^2 = 1225\)[/tex]
- (d) The Pythagorean triplet is [tex]\((20, 21, 29)\)[/tex]
- (e) [tex]\(18 \times \sqrt{196} = 252\)[/tex]
- (f) The smallest number to be multiplied by 2352 to get a perfect square is 3. The perfect square obtained is 7056.
- (g) [tex]\(\sqrt{36864} = 192\)[/tex], and the number of digits in 192 is 3.
### (a) Determine if 15 and 16 are relatively prime
Two numbers are relatively prime if their greatest common divisor (GCD) is 1. In this case, the GCD of 15 and 16 is 1.
Thus, 15 and 16 are relatively prime.
### (b) Express 225 as a sum of odd numbers
To express 225 as a sum of odd numbers, observe the pattern where consecutive odd numbers add up to perfect squares. For 225, we can write it as:
[tex]\[ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = 225 \][/tex]
### (c) Find [tex]\((35)^2\)[/tex] without actual multiplication
We can use the identity [tex]\((a + b)^2 = a^2 + 2ab + b^2\)[/tex]:
[tex]\[ 35^2 = (30 + 5)^2 = 30^2 + 2 \cdot 30 \cdot 5 + 5^2 \][/tex]
[tex]\[ 35^2 = 900 + 300 + 25 = 1225 \][/tex]
So, [tex]\(35^2 = 1225\)[/tex].
### (d) Find the Pythagorean triplet if one member is 20
Using the Pythagorean theorem [tex]\(a^2 + b^2 = c^2\)[/tex], we can find a Pythagorean triplet where one of the members (a) is 20. If we choose [tex]\(b\)[/tex] as 21, then:
[tex]\[ a = 20 \][/tex]
[tex]\[ b = 21 \][/tex]
[tex]\[ c = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29 \][/tex]
So, the Pythagorean triplet is (20, 21, 29).
### (e) If [tex]\( 14^2 = 196\)[/tex], what is [tex]\( 18 \times \sqrt{196} \)[/tex]?
Since [tex]\( 14^2 = 196\)[/tex], we know that [tex]\( \sqrt{196} = 14 \)[/tex].
Therefore:
[tex]\[ 18 \times \sqrt{196} = 18 \times 14 = 252 \][/tex]
### (f) Find the smallest number to be multiplied by 2352 to get a perfect square. Also, find the perfect square obtained.
To determine the smallest multiplier that makes 2352 a perfect square, we perform prime factorization:
[tex]\[ 2352 = 2^4 \times 3 \times 7^2 \][/tex]
For 2352 to become a perfect square, each exponent in its prime factorization must be even. The exponent for 3 is odd (1), so we need another factor of 3 to make it even. Therefore, the smallest multiplier is 3.
Thus:
[tex]\[ 3 \times 2352 = 7056 \][/tex]
Now, 7056 is a perfect square.
### (g) Find [tex]\(\sqrt{36864}\)[/tex]. How many digits are in this number?
The square root of 36864 is 192.
So, the number of digits in 192 is 3.
In summary:
- (a) 15 and 16 are relatively prime.
- (b) [tex]\(225\)[/tex] can be expressed as the sum of: [tex]\([1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29]\)[/tex]
- (c) [tex]\(35^2 = 1225\)[/tex]
- (d) The Pythagorean triplet is [tex]\((20, 21, 29)\)[/tex]
- (e) [tex]\(18 \times \sqrt{196} = 252\)[/tex]
- (f) The smallest number to be multiplied by 2352 to get a perfect square is 3. The perfect square obtained is 7056.
- (g) [tex]\(\sqrt{36864} = 192\)[/tex], and the number of digits in 192 is 3.
