Question

Determine all positive integers n such that 5^n−1 can be written as a product of an even
number of consecutive integers.

Answer 1

@tanjeetsarkar96

Answer 2

wait m thinking ..!!

Answer 3

is the total n-1 raised to power ??? @goformit100

Answer 4

ok

Answer 5

@Mandila Help me Sister

Answer 6

@tanjeetsarkar96 I think it would be\[5^{n}-1\]since that is always even and the product of an even number of integers is also always even, while \[5^{n-1}\]is only even if n=1.

Answer 7

Ok...

Answer 8

as n must be even let n belong to integers
then 2n will belong to even nos
as soln will be consecutive nos
equation becomes 2*n(2*(n+1))=5^n-1
2n(2n+2)=5^n-1
4n^2+4n=5^n-1
n^2+n-(5^n-1)/4=0
solving quadratic we get
n=-1+-(\[\sqrt{1^2-(-[5^n-1]/4)}\]
put n=1 you get 2=[-1+-{5^0.5}]/2 which is not soln
put n=2 you get 4=[-1+-{5^1}]/2 which gives one extraneous soln and other correct soln
put n=3 you get 6=[-1+-{5^1.5}]/2 which is not soln
put values gap gets wider and wider
therefore solution is 2 only.....

Answer 9

@go formit is my answer ryt

Answer 10

im like clueless here! my mind is filled with too much econ i guess!

Answer 11

@mandila pls go thru my answer is it ryt

Answer 12

(Y)

Answer 13

ty @Mandila

Answer 14

question gives a different meanin! it has to be product of 2 consecutive even numbers! i was soo confused! otherwise solving this would be like magic! :)

Answer 15

to me, it cannot be done.
product of two even consecutive number is an even number
let 2x is the first one, (it is an even, so, it must be formed as 2x )
and the second one is 2x+2
the product is 2x*(2x+2) = 4 (x)(x+1) is an even number
whereas 5 ^(n-1) is odd numbers with all values of n
how can an even number = an odd number?

Answer 16

-1 is not in the index part right?

Answer 17

@Mandila you ask me or the asker?

Answer 18

if n is odd--> n-1 is even
if n is even--> n -1 is odd.
however, no matter what it is, 5^ ( odd) is an odd, and 5^(even) is an odd, too

Answer 19

asker

Answer 20

@goformit100 you can put it in logic by \[5^1 = 5 mod 10\]
\[5^2 =5 mod 10\]
\[5^3 =5 mod 10\]
................................
\[5^{n-1}=5 mod 10\]
so, the last digit of 5^(n-1) is 5 , that shows it is an odd number
that logic for the first part,
the second part is even part as I show above.
then combine them to give out conclusion.

Answer 21

Thankyou

Answer 22

yw