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 painwise 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) \( \operatorname{GCD}(42,63,105)=\square \)

To solve \( \operatorname{GCD}(42,63,105) \) using the pairwise difference method, we can start by taking the differences: 1. \( 63 - 42 = 21 \) 2. \( 105 - 63 = 42 \) 3. \( 105 - 42 = 63 \) Now we have the numbers \( 21, 42, \) and \( 63 \). We can apply the same reasoning to these numbers: 1. \( 42 - 21 = 21 \) 2. \( 63 - 42 = 21 \) Now we only have the number \( 21 \) remaining. Thus, \( \operatorname{GCD}(42,63,105)=21 \). For part (c) \( \operatorname{GCD}(32,48,80,128) \): 1. Start with: \( 48 - 32 = 16 \) 2. \( 80 - 48 = 32 \) 3. \( 128 - 80 = 48 \) 4. \( 80 - 32 = 48 \) Now, we work with \( 16, 32, 48, 128 \). Re-applying the method: 1. Use \( 32 - 16 = 16 \) 2. \( 48 - 32 = 16 \) 3. \( 128 - 80 = 48 \) Continue to work down until we arrive at the value which consistently becomes smaller: Ultimately, \( \operatorname{GCD}(32,48,80,128)=16 \). So, we find: (b) \( \operatorname{GCD}(42,63,105)=21 \) (c) \( \operatorname{GCD}(32,48,80,128)=16 \) Time to rejoice! You just crunched through some math magic, finding the GCD using pairwise differences! Keep those numbers organized, and you’ll always have your GCD on speed dial! Want to tackle more number mysteries? Let’s dive into the wonderful world of mathematics!