Question

Please help me ---- [1]

Answer 1

Not getting the sollution

Answer 2

Do you agree 5 can be represented as 2k+1 ?

Answer 3

yes as it is odd

Answer 4

what about 5^5 ?
is it still odd ?

Answer 5

5^{2m+1}

Answer 6

so, yeah odd

Answer 7

5*5*5*5*5*... = odd number

yes ?

Answer 8

yes

Answer 9

it doesn't matter how many times you multiply an odd prime, it always stays odd.

Answer 10

good thats the only thing you need to know

Answer 11

can we say \[5^{5^{5^{5^{\cdots}}}} = 5^{2m+1}\]

for some integer \(m\)

Answer 12

yes

Answer 13

good, so we reduced the problem to finding the remainder of a simple friendly number : \(5^{2m+1}\)

Answer 14

\[5^{2m+1} = 5^{2m}\cdot 5^1 = 25^m\cdot 5\]

still yes ?

Answer 15

25 = 5^2
5= 5^1

25^m .5 = 5^{2m+1}
so yes

Answer 16

divide by 24 and use binomial theorem to find the remainder

Answer 17

\[\dfrac{25^m\cdot 5}{24}\]

Answer 18

\[\dfrac{(24+1)^m\cdot 5}{24}\]

Answer 19

\[\dfrac{(24X+1)\cdot 5}{24}\]

Answer 20

GOT it X is any integer right

Answer 21

yes

Answer 22

thank you

Answer 23

\[(24+1)^m = 24^m + 24^{m-1} + \cdots + 24^2+24+1 = 24X+1 \]