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) \( \operatorname{GCF}(12,30,90)=\square \)

To compute the GCF of \(12, 30, 90\) using the pairwise difference method, we start by replacing 90 with the difference between 90 and 30, which gives us \(90 - 30 = 60\). Now, we have \(12, 30, 60\). Next, we replace 60 with \(60 - 30 = 30\), leaving us with \(12, 30\). We can now replace 30 with \(30 - 12 = 18\), resulting in \(12, 18\). Finally, we replace 18 with \(18 - 12 = 6\), giving us \(12, 6\). Discarding the duplicate numbers, we replace 12 with \(12 - 6 = 6\), resulting in the last remaining positive number as \(6\). So, \( \operatorname{GCF}(12,30,90) = 6 \). To calculate \( \operatorname{GCD}(42,63,105) \), we first replace 105 with \(105 - 63 = 42\), resulting in \(42, 63, 42\). We can discard one of the 42s and now we have \(42, 63\). Next, we replace 63 with \(63 - 42 = 21\), which gives us \(42, 21\). Now, we replace 42 with \(42 - 21 = 21\). At this point, we have two 21s remaining, adding up to the last positive number, which is \(21\). Hence, \( \operatorname{GCD}(42,63,105) = 21 \). For \( \operatorname{GCD}(32,48,80,128) \), we will first consider pairwise differences among the numbers. Start with \(128 - 80 = 48\); thus we have \(32, 48, 48, 32\). Discard the duplicates, getting \(32, 48\). Next, we can replace \(48 - 32 = 16\) to have \(32, 16\) left. By replacing \(32 - 16 = 16\), we’re down to \(16, 16\), resulting in \( \operatorname{GCD}(32,48,80,128) = 16 \). To solve \( \operatorname{GCF}(37,23,29,31) \), all numbers are unique, so we need to create differences. We start with \(37 - 31 = 6\); therefore, now we have \(6, 23, 29, 31\). Next, \(31 - 29 = 2\) results in \(6, 23, 2\). Then \(23 - 6 = 17\) gives us \(17, 6, 2\). Subtracting \(17 - 6 = 11\) results in \(11, 6, 2\). Continuing on, after several steps, notice all numbers are reduced until we find a safe pair of 6 and 2 at \(6 - 2 = 4\). Finally, we continue until we narrow it down and confirm the only positive integers left, which allows us to conclude \( \operatorname{GCF}(37,23,29,31) = 1 \).