The GCD (or GCF) of three natural numbers can be computed by replacing any number with the difference it makes with any smaller number. If two (or more) numbers are the same, all but one can be discarded. The last positive number remaining is the GCF, since GCF(a) =a for any natural number a. Use the pairwise difference method to compute these GCDs and GCFs. (a) $$ \operatorname{GCF}(12,30,90) $$ (b) $$ \operatorname{GCD}(42,63,105) $$ (c) $$ \operatorname{GCD}(32,48,80,128) $$ (d) $$ \operatorname{GCF}(37,23,29,31) $$ (a) $$ \operatorname{GCF}(12,30,90)=6 $$ (b) $$ \operatorname{GCD}(42,63,105)=21 $$ (c) $$ \operatorname{GCD}(32,48,80,128)= $$

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**(c) Compute $$\operatorname{GCD}(32,48,80,128)$$ using the pairwise difference method**

We start with the set:  
$$
\{32,\;48,\;80,\;128\}
$$

1. **Sort in increasing order:**  
   $$
   32,\;48,\;80,\;128
   $$

2. **Compare 48 and 32:**  
   Replace $$48$$ by the difference:  
   $$
   48 - 32 = 16
   $$  
   New set (with $$32$$ unchanged):  
   $$
   \{32,\;16,\;80,\;128\}
   $$  
   Sorting gives:  
   $$
   16,\;32,\;80,\;128
   $$

3. **Compare 32 and 16:**  
   Replace $$32$$ by the difference:  
   $$
   32 - 16 = 16
   $$  
   This gives:  
   $$
   \{16,\;16,\;80,\;128\}
   $$  
   Since two numbers are the same, discard one:  
   $$
   \{16,\;80,\;128\}
   $$

4. **Compare 80 and 16:**  
   Replace $$80$$ by the difference:  
   $$
   80 - 16 = 64
   $$  
   New set:  
   $$
   \{16,\;64,\;128\}
   $$

5. **Compare 64 and 16:**  
   Replace $$64$$ by:  
   $$
   64 - 16 = 48
   $$  
   New set:  
   $$
   \{16,\;48,\;128\}
   $$

6. **Compare 48 and 16:**  
   Replace $$48$$ by:  
   $$
   48 - 16 = 32
   $$  
   New set:  
   $$
   \{16,\;32,\;128\}
   $$

7. **Compare 32 and 16:**  
   Replace $$32$$ by:  
   $$
   32 - 16 = 16
   $$  
   New set:  
   $$
   \{16,\;16,\;128\}
   $$  
   Discard one duplicate:  
   $$
   \{16,\;128\}
   $$

8. **Compare 128 and 16:**  
   Replace $$128$$ by the difference computed repeatedly or observe that the process is equivalent to computing the GCD of $$16$$ and $$128$$. Since  
   $$
   128 = 16 \times 8,
   $$  
   the greatest common divisor is  
   $$
   16.
   $$

Thus,  
$$
\operatorname{GCD}(32,48,80,128)=16.
$$

---

**(d) Compute $$\operatorname{GCF}(37,23,29,31)$$ using the pairwise difference method**

We have the set:  
$$
\{37,\;23,\;29,\;31\}
$$

1. **Sort in increasing order:**  
   $$
   23,\;29,\;31,\;37
   $$

2. **Take differences with the smallest number $$23$$:**  
   - $$29 - 23 = 6$$  
   - $$31 - 23 = 8$$  
   - $$37 - 23 = 14$$

Now, we have the numbers:  
   $$
   \{23,\,6,\,8,\,14\}
   $$

3. **Notice that none of the original prime numbers share any common factors, and the intermediate numbers (6, 8, 14) are not multiples of a common prime factor that divides $$23$$.**

4. **The only common divisor among prime numbers (if they are all different) is $$1$$.**

Thus,  
$$
\operatorname{GCF}(37,23,29,31)=1.
$$

---

**Final Answers:**  
$$(c)$$ $$\operatorname{GCD}(32,48,80,128)=16$$  
$$(d)$$ $$\operatorname{GCF}(37,23,29,31)=1$$
**(c) $$\operatorname{GCD}(32,48,80,128)=16$$ (d) $$\operatorname{GCF}(37,23,29,31)=1$$

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