Question

1. find the gcd of the four numbers 1240, 6660, 15540, and 19980 without sage.

Explanation:

Step1: Recall the Euclidean Algorithm for GCD of multiple numbers

The GCD of multiple numbers can be found by first finding the GCD of two numbers, then finding the GCD of the result with the next number, and so on. The Euclidean Algorithm states that for two positive integers \(a\) and \(b\) (\(a > b\)), \(\gcd(a,b)=\gcd(b,a\bmod b)\), and we repeat this until the remainder is \(0\), at which point the non - zero remainder is the GCD.

Step2: Find \(\gcd(1240,6660)\)

Using the Euclidean Algorithm:
\(6660\div1240 = 5\) with a remainder. We calculate the remainder as \(6660-1240\times5=6660 - 6200 = 460\)
So, \(\gcd(1240,6660)=\gcd(1240,460)\)
Now, \(1240\div460 = 2\) with a remainder. The remainder is \(1240-460\times2=1240 - 920=320\)
So, \(\gcd(1240,460)=\gcd(460,320)\)
\(460\div320 = 1\) with a remainder. The remainder is \(460 - 320=140\)
So, \(\gcd(460,320)=\gcd(320,140)\)
\(320\div140 = 2\) with a remainder. The remainder is \(320-140\times2=320 - 280 = 40\)
So, \(\gcd(320,140)=\gcd(140,40)\)
\(140\div40 = 3\) with a remainder. The remainder is \(140-40\times3=140 - 120 = 20\)
So, \(\gcd(140,40)=\gcd(40,20)\)
\(40\div20=2\) with a remainder of \(0\). So, \(\gcd(1240,6660) = 20\)

Step3: Find \(\gcd(20,15540)\)

Using the Euclidean Algorithm: \(15540\div20 = 777\) with a remainder of \(0\). So, \(\gcd(20,15540)=20\)

Step4: Find \(\gcd(20,19980)\)

Using the Euclidean Algorithm: \(19980\div20=999\) with a remainder of \(0\). So, \(\gcd(20,19980) = 20\)

Answer:

The greatest common divisor (gcd) of 1240, 6660, 15540, and 19980 is \(\boldsymbol{20}\)