Find gcd(31415, 14142) by applying Euclid’s algorithm.b. Estimate how many times faster it will be to find gcd(31415, 14142) byEuclid’s algorithm compared with the algorithm based on checking con-secutive integers from min{m, n} down to gcd(m, n).

Sure, let's find the gcd(31415, 14142) using Euclid’s algorithm.

Euclid’s algorithm is based on the principle that the greatest common divisor of two numbers does not change if the larger number is replaced by its difference with the smaller number.

1. Start with the two numbers: 31415 and 14142.

2. Subtract the smaller number from the larger one: 31415 - 14142 = 17273.

3. Replace the larger number with the result from step 2: So now our two numbers are 17273 and 14142.

4. Repeat steps 2 and 3 until the numbers are equal:

   • 17273 - 14142 = 3131
   • Replace 17273 with 3131, so the numbers are now 3131 and 14142.
   • 14142 - 3131*4 = 1718
   • Replace 14142 with 1718, so the numbers are now 3131 and 1718.
   • 3131 - 1718 = 1413
   • Replace 3131 with 1413, so the numbers are now 1413 and 1718.
   • 1718 - 1413 = 305
   • Replace 1718 with 305, so the numbers are now 1413 and 305.
   • 1413 - 305*4 = 193
   • Replace 1413 with 193, so the numbers are now 193 and 305.
   • 305 - 193 = 112
   • Replace 305 with 112, so the numbers are now 193 and 112.
   • 193 - 112 = 81
   • Replace 193 with 81, so the numbers are now 81 and 112.
   • 112 - 81 = 31
   • Replace 112 with 31, so the numbers are now 81 and 31.
   • 81 - 31*2 = 19
   • Replace 81 with 19, so the numbers are now 19 and 31.
   • 31 - 19 = 12
   • Replace 31 with 12, so the numbers are now 19 and 12.
   • 19 - 12 = 7
   • Replace 19 with 7, so the numbers are now 7 and 12.
   • 12 - 7 = 5
   • Replace 12 with 5, so the numbers are now 7 and 5.
   • 7 - 5 = 2
   • Replace 7 with 2, so the numbers are now 2 and 5.
   • 5 - 2*2 = 1
   • Replace 5 with 1, so the numbers are now 2 and 1.
   • 2 - 1 = 1
   • Replace 2 with 1, so the numbers are now 1 and 1.

So, the gcd(31415, 14142) is 1.

For the second part of your question, the Euclidean algorithm is generally much faster than the method of checking consecutive integers. The Euclidean algorithm's time complexity is O(log min(a, b)) where a and b are the input numbers, while the time complexity of checking consecutive integers is O(min(a, b)).

So, if we take the smaller number of 14142, the Euclidean algorithm would be approximately log2(14142) = 13.8 times faster than the method of checking consecutive integers. This is a rough estimate and the actual speedup may vary depending on the specific numbers and the computational resources available.