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 Seventh Term Settlement (Posted on 2016-07-22)
Each of A(1), A(2),..., A(7) is a positive integer such that:

(i) A(6) = 144, and

(ii) A(n+3) = A(n+2)(A(n+1)+A(n)), for n = 1,2,3,4

Find A(7)

 No Solution Yet Submitted by K Sengupta No Rating

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 Analytical Solution | Comment 7 of 10 |
let a(1)=x, a(2)=y, a(3)=z
then
a(6)=z^2*(x+y)(x+y+1)(y+z)
144=2^4*3^2
z^2(x+y)(x+y+1)(y+z)=144 (1)
divisors:
1,2,3,4,6,8,9,12,16,18,24,36,48,72,144
(x+y) and (x+y+1) are consecutive integers
only pairs of divisors which are consecutive integers is
(1,2) (2,3) (3,4) and (8,9)

case 1:
x+y=1
impossible since x and y are each positive

case 2:
x+y=2
thus x=1 and y=1 leaving (1) to be
z^2*1*2*(z+1)=144
z^2(z+1)=72
which has no integer solutions

case 3:
x+y=3
so either x=1,y=2 or x=2,y=1
subcase 1:
x=1,y=2
giving (1) to be
z^2*3*4*(z+2)=144
again, no integer solutions leaving
subcase 2:
x=2,y=1
giving (1) to be
z^2*3*4*(z+1)=144
only integer solution is z=2
thus x=2,y=1,z=2 which gives a(7)=3456

case 4:
x+y=8
so this gives us
8*9*z^2*(z+y)=144
z^2*(z+y)=2
thus z=1 leaving
y+1=2
y=1
which gives x=7
thus we have x=7,y=1,z=1 which also gives a(7)=3456

thus a(7)=3456
resulting from either starting values (2,1,2) or (7,1,1)

*edited to include the x+y=8 case I originally missed*

Edited on July 24, 2016, 6:09 am
 Posted by Daniel on 2016-07-23 06:43:50

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