Question

Find the remainder when 5^1998 is divided by 11
@ganeshie8

Answer 1

Let's look for a pattern in the powers

5^1 = `5`
5^2 = 25 = `3`
5^3 = 5^2*5 = 3*5 = 15 = `4`
5^4 = 3^2 = `9`
5^5 = 5^4*5 = 9*5 = 45 = `1`
5^6 = 5^5*5 = 1*5 = `5`

Answer 2

See the pattern in the remainders of 5^n ?

5, 3, 4, 9, 1, 5, ...

Answer 3

So the pattern is
5,3,4,9,1,5 .... and I just have to figure out what the number will be when n is 1998, right thanks

Answer 4

Okay one more thing
Find the value of m such that the equations
x³+mx-1=0
x³-3x+m = 0
have a common root

Answer 5

Let "a" be the common root.

Answer 6

Then we have

a^3 + ma - 1 = 0
a^3 - 3a + m = 0

Answer 7

subtract the equations and you can express "m" in terms of "a"

Answer 8

x = A
[1]-[2]
Am-1 +3A -m = 0
A(3+m) = m+1
A = m+1/m+3

3[1] +m[2]
3A³+3Am-3+A³m -3Am+m²=0
A³(3+m)+m²-3=0
A³= (3-m²)/3+m

A³= A³
(m+1/m+3)³ = (3-m²)/3+m
But this doesnt give me m = -2

Answer 9

@ganeshie8

Answer 10

Are you sure? Try plugging -2 in there again

Answer 11

@johnweldon1993 It also gives me some irrational numbers as roots too. What should I do about them?