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Modulus help watch

1. Hi, looking back at my past notes i have found the following that i dont fully understand. Could someone please explain the following steps to me:

take k=145 (145 = 6x24+1) so that
k=1(mod24)
then theorem A gives 4^145=4(mod35)

so...
4= 4(mod35)
4^3=64 = 64-70 = -6(mod35)
4^6=36= 1(mod 35)
4^24(6)= 4^144 = 1(mod 35)

therefore 4^145 = 4(mod 35)

thanks for any help / explanation
2. (Original post by Hrov)
Hi, looking back at my past notes i have found the following that i dont fully understand. Could someone please explain the following steps to me:

take k=145 (145 = 6x24+1) so that
k=1(mod24)
then theorem A gives 4^145=4(mod35)

so...
4= 4(mod35)
4^3=64 = 64-70 = -6(mod35)
4^6=36= 1(mod 35)
4^24(6)= 4^144 = 1(mod 35)

therefore 4^145 = 4(mod 35)

thanks for any help / explanation
It seems fairly straight forward, so I am probably missing the point in where you're having the problem. Could you elaborate. It would also help if you said what "theorem A" is.
3. (Original post by Hrov)
Hi, looking back at my past notes i have found the following that i dont fully understand. Could someone please explain the following steps to me:

take k=145 (145 = 6x24+1) so that
k=1(mod24)
then theorem A gives 4^145=4(mod35)

so...
4= 4(mod35)
4^3=64 = 64-70 = -6(mod35)
4^6=36= 1(mod 35)
4^24(6)= 4^144 = 1(mod 35)

therefore 4^145 = 4(mod 35)

thanks for any help / explanation
from 4^6=36= 1(mod 35) you can deduce that 4^6x4^6 = 1(mod 35)

but thats the same as 4^(2x6) so you've basically shown 4^n6 = 1(mod 35)

so you've just put n = 24 to prove 4^144 = 1(mod 35)

i think*
4. (Original post by ghostwalker)
It seems fairly straight forward, so I am probably missing the point in where you're having the problem. Could you elaborate. It would also help if you said what "theorem A" is.
(Original post by Chaoslord)
from.
i dont know why we have to minus the 70 on the cubed line, i have done these before but not had to use odd powers apart from 1
5. (Original post by Hrov)
i dont know why we have to minus the 70 on the cubed line, i have done these before but not had to use odd powers apart from 1
Since we are working mod 35, we can add and subtract multiples of 35 with impunity, and subtracting 70 just makes the numbers easier to deal with.

We don't actually need that step.

You could go
4^3=64 (mod35)
4^6=64.64 = 4096 = 1(mod 35)

But larger numbers tend to be more difficult to follow.

Edit: Would you know that 4096 leaves a remainder of 1 when divided by 35, without some tedious work?
6. ok thanks for that, i guess i didnt realise that you can work with negative remainders
And yes i know that 4096 leave a remainder of one
cheers
7. (Original post by ghostwalker)
Since we are working mod 35, we can add and subtract multiples of 35 with impunity, and subtracting 70 just makes the numbers easier to deal with.

We don't actually need that step.

You could go
4^3=64 (mod35)
4^6=64.64 = 4096 = 1(mod 35)

But larger numbers tend to be more difficult to follow.

Edit: Would you know that 4096 leaves a remainder of 1 when divided by 35, without some tedious work?
Thanks for your help, i thought that i had understood but however, i have the following:

99=99(mod 713)
99^2= 532(mod713)
99^16=397(mod 713)

im trying to workout 99^19. But get 2 different answers depending on whether I multiply all at one, or in stages

e.g.
99^2 x 99^16 = 156(mod 713)
156 x 99 = 471(mod 713)

e.g.2)
99x532x397 = 309(mod 713)

Any idea what i am doing wrong?
8. (Original post by Hrov)
Thanks for your help, i thought that i had understood but however, i have the following:

99=99(mod 713)
99^2= 532(mod713)
99^16=397(mod 713)

im trying to workout 99^19. But get 2 different answers depending on whether I multiply all at one, or in stages

e.g.
99^2 x 99^16 = 156(mod 713)
156 x 99 = 471(mod 713)

e.g.2)
99x532x397 = 309(mod 713)

Any idea what i am doing wrong?
You've gone wrong on your multiplication, the bit in bold.

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