find the least positive integer N so that 1<gcd(N, 271) < gcd(N, 2014). Explain how you find N.

Thanks for your help

271 is prime, so N is a multiple of 271

2014 = 2*19*53,

If N = 271*2*2014, GCD(N,2014) = 2*271 and GCD(N,271) = 271

So, it appears to me that 271*2*2014 is the smallest N that fits the requirements

To find the least positive integer N that satisfies the given condition, we need to find the smallest value of N for which the greatest common divisor (gcd) of N and 271 is strictly greater than the gcd of N and 2014.

Here's how to find N step by step:

1. Start by finding the gcd of 271 and 2014. In this case, gcd(271, 2014) = 67. This value will be important for comparison later.

2. Begin with N = 1. We'll increment N until we find the desired value.

3. For each value of N, calculate the gcd of N and 271 using Euclid's algorithm. If the gcd is greater than 67 and less than the gcd of N and 2014, we have found the desired N.

4. If the gcd(N, 271) > 67, check if gcd(N, 271) < gcd(N, 2014). If the condition is satisfied, we have found the least positive integer N. If not, increment N by 1 and repeat the previous step.

5. Repeat steps 3 and 4 until we find the least positive integer N that meets the condition.

Let's calculate the value of N:

N = 1
gcd(1, 271) = 1
Since 1 is not greater than 67, we increment N.

N = 2
gcd(2, 271) = 1
Again, 1 is not greater than 67, so we increment N.

N = 3
gcd(3, 271) = 1
Once again, 1 is not greater than 67, so we continue incrementing N.

N = 4
gcd(4, 271) = 1
Still, 1 is not greater than 67, so we go to the next value.

N = 5
gcd(5, 271) = 1
1 is not greater than 67, so we increment N.

N = 6
gcd(6, 271) = 1
Again, 1 is not greater than 67, so we increase N.

N = 7
gcd(7, 271) = 1
1 is not greater than 67, so we move on.

N = 8
gcd(8, 271) = 1
Still, 1 is not greater than 67, so we increment N.

N = 9
gcd(9, 271) = 1
1 is not greater than 67, so we increase N.

N = 10
gcd(10, 271) = 1
Once again, 1 is not greater than 67, so we increment N.

Continuing this process, we find that gcd(N, 271) remains 1 until we reach N = 68.

N = 68
gcd(68, 271) = 17
Now, we have a gcd of 17 which is greater than 67 (gcd of 271 and 2014) and less than the gcd of N and 2014. Therefore, the least positive integer that satisfies the given condition is N = 68.

Note: It's worth mentioning that since 1 is not greater than 67, it is not necessary to check for the second condition (gcd(N, 271) < gcd(N, 2014)). It simplifies the process and helps to find the desired value faster.