math - Geometrical progression with any number row -

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i can have number row consists 2 10 numbers. , row, have geometrical progression.

for example: given number row: 125 5 625 have answer 5. row: 128 8 512 have answer 4.

can give me hand? don't ask program, hint, want understand myself , write code myself, damn, have been thinking whole day , couldn't figure out.

thank you.

don't write whole program!

guys, don't it, can't simple make division. have geometrical progression + show numbers. in 128 8 512 row numbers be: 8 32 128 512

seth's answer right one. i'm leaving answer here elaborate on why answer 128 8 512 4 because people seem having trouble that.

a geometric progression's elements can written in form c*b^n b number you're looking (b greater 1), c constant , n arbritrary number.

so best bet start smallest number, factorize , @ possible solutions writing in c*b^n form, using b on remaining numbers. return largest result works.

so examples:

125 5 625

start 5. 5 prime, can written in 1 way: 5 = 1*5^1. b 5. can stop now, assuming know row in fact geometric. if need determine whether it's geometric test b on remaining numbers.

128 8 512

8 can written in more 1 way: 8 = 1*8^1, 8 = 2*2^2, 8 = 2*4^1, 8 = 4*2^1. have 3 possible values b, few different options c. try biggest first. 8 doesn't work. try 4. works! 128 = 2*4^3 , 512 = 2*4^4. b 4 , c 2.

3 15 375

this 1 bit mean because first number prime isn't b, it's c. you'll need make sure if first b-candidate doesn't work on remaining numbers have @ next smallest number , decompose it. here you'd decompose 15: 15 = 15*?^0 (degenerate case), 15 = 3*5^1, 15 = 5*3^1, 15 = 1*15^1. answer 5, , 3 = 3*5^0, works out.


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