Question

x=(2^n+3^n+4^n+5^n)/25 where n is a natural number.....
Find the value of n for which x is natural number too.

Answer 1

@mukushla

Answer 2

that was from me

Answer 3

Note that for x to be a natural number, \(2^n +3^n +4^n +5^n\) needs to be a multiple of 25.

Answer 4

YEP

Answer 5

its equal to solving\[k=\frac{2^n+3^n+4^n}{25}\]

Answer 6

U sure

Answer 7

yeah because 25 is a divisor of 5^n for n>1

Answer 8

n must be even

Answer 9

@mukushla is correct. Using modular arithmetic might help:
\[2^n mod 25+3^n mod25+4^n mod 25 + 5^n mod 25 =2^nmod25+3^nmod25 +4^nmod25\]

Answer 10

Really can solve each one individually to see when they are equal to \(0mod25\) and then see where there is an intersection.

Answer 11

\[2^n+3^n+4^n \equiv 2^{n+1}+1 \ \ \text{mod} \ 5\]

Answer 12

if n is even

Answer 13

what math is that @Traxter

Answer 14

we must have\[ 2^{2m+1}+1 \equiv 0\ \ \text{mod} \ 5\]but\[2^{2m+1}+1 \equiv 2(-1)^m+1\ \ \text{mod} \ 5\]so there is no answer

Answer 15

thats modular arithmetic

Answer 16

Yes there is no answer

Answer 17

what is ur reasoning?

Answer 18

2^1=2 3^1=3 4^1=4 5^1=5
2^2=4 3^2=9 4^2=16 5^2=25
2^3=8 3^3=27 4^3=64 5^3=125
2^4=6 3^4=81 4^4=256 5^4=625
2^5=12 3^5=163 4^5=1024
2^6=24 3^6=489 4^6=4096
.
.
.

Answer 19

See the last digits.....

Answer 20

So, the last digit will be
4 when n=1
6 when n=2
4 when n=3
8 when n=4
AND this pattern continues

Answer 21

But to be divisible by 25 the last digit must be 5.......Thus there is no answer for x.

Answer 22

@mukushla got my method

Answer 23

?

Answer 24

yeah :)

Answer 25

But I didnt get yours

Answer 26

np ... thats modular arithmetic ..u'll learn it :) do u know know about your courses at high school ?

Answer 27

I just checked my syllabes...... There is no modular arithmetic

Answer 28

oops