# number theory

• Sep 22nd 2012, 10:12 AM
Petrus
number theory
i prob post on wrong sector but idk where i should post it, sorry
the quest is :
Which (positive) residual obtained when 11^30+5^31-6 divided by 24?

i have got this far
11^30=(11^2)^15=121^15 ==1^15 (mod 24)
i cant solve rest :S
• Sep 22nd 2012, 10:51 AM
MaxJasper
Re: number theory
Quote:

Originally Posted by Petrus
i prob post on wrong sector but idk where i should post it, sorry
the quest is :
Which (positive) residual obtained when 11^30+5^31-6 divided by 24?

i have got this far
11^30=(11^2)^15=121^15 ==1^15 (mod 24)
i cant solve rest :S

$\left(11^{30}+5^{31}-6\right) \bmod 24 =11^{30}\text{mod} 24 +5^{31}\text{mod} 24 -6 \bmod 24=11^{2+28}\text{mod} 24+5^{1+30}\text{mod24}-6=121*11^{28}\text{mod24}+5*5^{30}\text{mod} 24-6 \bmod 24 = 1+5-6=0$
• Sep 22nd 2012, 11:01 AM
Petrus
Re: number theory
Sorry but i did not get it with 5^31
• Sep 22nd 2012, 12:26 PM
Soroban
Re: number theory
Hello, Petrus!

You are on the right track . . .

Quote:

$\text{What is the remainder when }11^{30}+5^{31}-6\text{ is divided by 24?}$

$11^{30} \;=\; (11^2)^{15} \;=\;(121)^{15} \;\equiv\; 1^{15}\text{ (mod 24)} \;\equiv\;1\text{ (mod 24)}$

$5^{31} \;=\;5\cdot5^{30} \;=\;5(5^2)^{15} \;=\;5(25)^{15} \;\equiv\; 5(1)^{15}\text{ (mod 24)} \;=\; 5\text{ (mod 24)}$

Therefore: . $11^{30} + 5^{31} - 6 \;\equiv\;1 + 5 - 6 \text{ (mod 24)} \;\equiv\;0\text{ (mod 24)}$

The remainder is zero.

• Sep 22nd 2012, 12:39 PM
Petrus
Re: number theory
ty sorban ur was clearly and i did understand :P
maxjasper sorry u did not have alot space kinda confused me :P
i was thinking like soo but thought it was wrong