There may be more brilliant solution than the following, but here are my thoughts.
We make use of Euclid's algorithm to help us out. Consider finding the hcf of A=2^(n+x)-1 and B=2^(n)-1.
If we repeated subtract B from A until the difference C is less than B (smaller number), the hcf between A and B is the same as the hcf between B and C.
For example, we would subtract 2^x times B from A, or C=A-2^xB=2^(n+x)-2^x(2^n-1)=2^(n+x)-2^(n+x)+2^n-1=2^n-1 By the Euclidean algorithm, hcf(A,B)=hcf(B,C)=hcf(2^n-1,2^x-1) If n is a multiple of x, then by repetition, we will end up with hcf(A,B)=hcf(2^x-1,2^x-1)=2^x-1
For the given example, n=100, x=20, so HCF(2^120-1, 2^100-1)=2^(120-100)-1=2^20-1=1048575 (since n=6x, a multiple of x).
How you do it is add the 44% of 132 to 132 (132 + (132×44%))
First you have to find out how much 44% is. Change it into decimal form (0.44) and then multiply the two together. 132 × 0.44 is <u>58.08 miles</u>. Then you add that to the rest of the trip, 132 miles. (132 + 58/08 is <u>190.08</u>)