A number theory problem by Paola Ramírez

{ A = 2 C = A B D = B C E = C D F = D E F = 6075000 \begin{cases} A=2 \\ C=AB \\ D=BC \\ E=CD \\ F= DE \\ F=6075000 \end{cases}

Given that six positive integers A , B , C , D , E A,B,C,D,E and F F meet the conditions above, find the value of B + C + D + E B+C+D+E .


The answer is 13995.

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1 solution

. .
Feb 27, 2021

{ A = 2 C = A B D = B C E = C D F = D E F = 6075000 \begin{cases} A = 2 & \quad \\ C = AB & \quad \\ D = BC & \quad \\ E = CD & \quad \\ F = DE & \quad \\ F = 6075000 & \quad \end{cases}

A = 2 , C = 2 B , D = 2 B 2 , E = 4 B 3 , F = 8 B 5 , F = 6075000 A = 2, C = 2B, D = 2B^2, E = 4B^3, F = 8B^5, F = 6075000

B 5 = 6075000 8 = 759375 B = 759375 5 = 15 B^5 = \frac { 6075000 } { 8 } = 759375 \rightarrow B = \sqrt[5]{759375} = 15

C = 30 , D = 450 , E = 13500 C = 30, D = 450, E = 13500

B + C + D + E = 15 + 30 + 450 + 13500 = 13995 B + C + D + E = 15 + 30 + 450 + 13500 = \boxed { 13995 }

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