Solution: 51. 

Before jumping in to this problem, consider a simpler example. Assume that there are three direct routes between A and B, two direct routes between B and C, and one direct route between A and C, as shown below.

Then, there are:

• 3 + 2 × 1 = 5 routes between A and B
• 2 + 3 × 1 = 5 routes between B and C
• 1 + 2 × 3 = 7 routes between A and C

That is, the number of routes between two cities is equal to the number of direct routes between those cities, plus the product of direct routes between the other cities.

So, for this problem, let x equal the number of direct routes between Arlington and Bedford, y the number of direct routes between Bedford and Cambridge, and z the number of direct routes between Arlington and Cambridge. This gives the following equations:

                A→B: 59 = x + yz 

                B→C: 39 = y + xz 

Adding and subtracting these equations gives

                98 = (x + y)(z + 1)

                20 = (y – x)(z – 1)

The proper factors of 98 are 1, 2, 7, 14. The proper factors of 20 are 1, 2, 5, 10. The above equations imply that z + 1 divides evenly into 98 and z – 1 divides evenly into 20; because these two numbers differ by two, it must be that z + 1 = 7 and z – 1 = 5, so z = 6.

Substituting z = 6 into the above equations, we get

                59 = x + 6y 

                39 = y + 6x, 

which yields x = 5 and y = 9. Therefore, the total number of routes between Arlington and Cambridge, given by z + xy, is 6 + 5 × 9 = 51.