Ways to go to a Place
A, B, and C are three downs, each connected by a network of roads. There are 82 ways to get from A to B, including those routes that pass through C. There are 62 ways to get from B to C, including those routes that pass through A. The number of ways to get from A to C, including those ways passing through B, is less than 300. How many ways are there to get from A to C?
I took this problem of this site: http://mathproblems.info/
The answer is 46.
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1 solution
Direct route to A to B =x
Direct route to B to C = y
Direct route to A to C=z
Number of ways to A to B= x+Number of ways through C=x+zy-------(1)
Number of ways to B to C= y+number of ways through A= y+zx-------(2)
(1) + (2)=> (x+y)(z+1)=144----------(3)
(1)- (2) => (x-y)(z-1) = 20-------------(4)
factors of144=1,2,3,4,6,8,9,12,16,18,24,36,48,72,144
factors of 20= 1,2,4,5,10,20
If you write this as table, it will be easy to do the trial and error. z+1 is factor of 144 and z-1 is factor of 20, x+y should be greater than or equal to 12
So possible factors are 12, 16,18,24,36,72
possible values of z are 2,3,5,11,
when z=2, x+y=48 and x-y=20 , x=34, y=14 and substitute the values in z+xy<300
Continue the same procedure for z=3, 5 and 11. Only z=11 will be satisfying the inequality z+xy<300
where z=11, x+y=12, x-y=2 then x=7 and y= 5
total number of ways from A to C= z+xy=11+(7*5)=11+35=46